On the Poisson Difference Distribution Inference and Applications

On The Poisson Difference Distribution Inference and Applications

by Abdulhamid A. Alzaid, Maha A. Omair Department of Statistics and Operations Research King Saud University

Abstract:

The distribution of the difference between two independent Poisson random variables involves the modified Bessel function of the first kind. Using properties of this function, maximum likelihood estimates of the parameters of the Poisson difference were derived. Asymptotic distribution property of the maximum likelihood estimates is discussed. Maximum likelihood estimates were compared with the moment estimates in a Monte Carlo study. Hypothesis testing using likelihood ratio tests was considered. Some new formulas concerning the modified Bessel function of the first kind were provided. Alternative formulas for the probability mass function of the Poisson difference distribution are introduced. Finally, two new applications for the Poisson difference distribution are presented. The first is from the Saudi stock exchange (TASI) and the second from Dallah hospital.

Key Words: Poisson difference distribution, Skellam distribution, Bessel function, regularized hypergeometric function, maximum likelihood estimate, likelihood ratio test.

1. Introduction

The distribution of the difference between two independent Poisson random variables was derived by Irwin (1937) for the case of equal parameters. Skellam (1946) and Prekopa (1952) discussed the case of unequal parameters. The distribution of the difference between two correlated Poisson random variables was recently introduced by Karlis and Ntzoufras (2006) who proved that it reduces to the Skellam distribution (Poisson difference of two independent Poisson). Strakee and van der Gon (1962) presented tables of the cumulative distribution function of the Poisson difference distribution to four decimal places for some combinations of values of the two parameters. Their tables also show the differences between the normal approximations (see Fisz(1953)). Romani (1956) showed that all the odd cummulants of the Poisson difference distribution ( PD()θ1,θ2 ) equal to θ1 −θ2 , and that all the even cummulant equal to θ1 +θ2 . He also discussed the properties of the maximum likelihood estimator of

E()X1 − X 2 = θ1 −θ2 . Katti (1960) studied E X1 − X 2 . Karlis and Ntzoufras (2000) discussed in details properties of the Poisson difference distribution and obtained the

1 maximum likelihood estimates via the EM algorithm. Karlis and Ntzoufras (2006) derived Bayesian estimates and used the Bayesian approach for testing the equality of the two parameters of the Poisson difference distribution. The Poisson difference distribution has many applications in different fields. Karlis and Ntzoufras (2008) applied the Poisson difference distribution for modeling the difference of the number of goals in football games. Karlis and Ntzoufras (2006) used the Poisson difference distribution and the zero inflated Poisson difference distribution to model the difference in the decayed, missing and filled teeth (DMFT) index before and after treatment. Hwang et al (2007) showed that the Skellam distribution can be used to measure the intensity difference of pixels in cameras. Strackee and van der Gon (1962) state, “In a steady state the number of light quanta, emitted or absorbed in a definite time, is distributed according to a Poisson distribution. In view thereof, the physical limit of perceptible contrast in vision can be studied in terms of the difference between two independent variates each following a Poisson distribution”. The distribution of differences may also be relevant when a physical effect is estimated as the difference between two counts, one when a “cause” is acting, and the other a “control” to estimate the “background effect”. For more applications see Alvarez (2007). The aim of this paper is to obtain some inference results for the parameters of the Poisson difference distribution and give application on share and occupancy modeling. Maximum likelihood estimates of θ1 and θ2 are obtained by maximizing the likelihood function (or equivalently the log likelihood), using the properties of the modified Bessel function of the first kind. A Monte Carlo study is conducted to compare two estimation methods, the method of moment and the maximum likelihood. Moreover, since regularity conditions hold, asymptotic distribution of the maximum likelihood estimates is obtained. Moreover, hypothesis testing using Likelihood ratio test for equality of the two parameters is introduced and Monte Carlo study is presented with the empirical power being calculated. For simplification alternative formulas of the Poisson difference distribution are presented for which Poisson distribution and negative of Poisson distribution can be shown by direct substitution to be special cases of the Poisson difference distribution. These formulas are used for estimation and testing. The applications considered in this study are such that only the difference of two variables could be estimated while each one by its own is not easily estimated. Our considered data could take both positive and negative integer values. Hence, Poisson difference distribution could be a good candidate for such data. The first is from the Saudi stock exchange (TASI) and the second from Dallah hospital at Riyadh. The remainder of this paper proceeds as follows: properties of the Poisson difference distribution are revised with some properties of the modified Bessel function of the first kind and new formulas for the Bessel function are derived in section 2. In section 3, new representation of the Poisson difference distribution is presented. Maximum likelihood estimates are considered in details with their asymptotic properties in section 4. In section 5, likelihood ratio tests for equality of means and for testing if one of the parameters has zero value are presented. A simulation study is conducted in section 6. Finally, two new applications of the Poisson difference distribution are illustrated in section 7.

2 2. Definition and basic Properties

Definition 2.1:

For any pair of variables ()X ,Y that can be written as X = W1 +W3 and Y = W2 +W3 with W1 ~ Poisson(θ1 ) independent of W2 ~ Poisson(θ 2 ) and W3 following any distribution, the probability mass function of Z = X − Y is given by z 2 ⎛ θ ⎞ −θ1 −θ 2 ⎜ 1 ⎟ P(Z = z) = e ⎜ ⎟ I z ()2 θ1θ2 , z = K,−1,0,1,K (2.1) ⎝θ2 ⎠ k ⎛ x2 ⎞ ⎜ ⎟ y ∞ ⎜ ⎟ ⎛ x ⎞ ⎝ 4 ⎠ where I y ()x = ⎜ ⎟ ∑ is the modified Bessel function of the first kind and Z ⎝ 2 ⎠ k =0 k!()y + k ! is said to have the Poisson difference distribution (Skellam distribution) denoted by

PD(θ1 ,θ 2 ). (Karlis and Ntzoufras (2008)) An interesting property is a type of symmetry given by

P()Z = zθ1,θ2 = P()Z = −zθ2 ,θ1 . The moment generating function is given by t −t M Z ()t = exp[− (θ1 +θ 2 )+θ1e +θ 2e ] (2.2)

The expected value is E()Z = θ1 −θ2 , while the variance is V (Z ) = θ1 +θ2 . The odd cummulants are equal to θ1 −θ2 while the even cummulants are equal to θ1 +θ2 . The

θ1 −θ2 skweness coefficient is given by β1 = 3 2 , that is the distribution is positively ()θ1 +θ2 skewed when θ1 > θ2 , negatively skewed when θ1 < θ2 and symmetric when θ1 = θ2 . The 1 kurtosis coefficient β2 = 3 + . As either θ1 or θ 2 tends to infinity kurtosis θ1 +θ2 coefficient tends to 3 and for a constant difference θ1 −θ2 , skeweness coefficient tends to zero implying that the distribution approaches the normal distribution. The Poisson difference distribution is strongly unimodal.

If Y1 ~PD(θ1 ,θ 2 ) independent of Y2 ~PD(θ 3 ,θ 4 ) then

1. Y1 + Y2 ~ PD(θ1 +θ 3 ,θ 2 +θ 4 )

2. Y1 − Y2 ~ PD(θ1 +θ 4 ,θ 2 +θ 3 ). More properties of the Poisson difference distribution can be found in Karlis and Ntzoufras(2006).

The following are some known identities for the modified Bessel function of the first kind (Abramowitz and Stegun (1970)): For any θ > 0 and y ∈ Z

I. I y ()θ = I − y ()θ . (2.3) ∞ θ II. ∑ I y ()θ = e . (2.4) y=−∞

3 ∞ III. ∑ yI y ()θ = 0 . (2.5) y=−∞ y ⎛θ ⎞ ~⎛ θ 2 ⎞ ⎜ ⎟ IV. I y ()θ = ⎜ ⎟ 0 F1⎜ y +1, ⎟ , (2.6) ⎝ 2 ⎠ ⎝ 4 ⎠ ~ ∞ z k where 0 F1 ();b; z = ∑ is the regularized hypergeometric function and Γ(x) k =0 k!Γ()b + k is the gamma function ∂I ()θ y V. y = I ()θ + I ()θ . (2.7) ∂θ θ y y+1 2()y +1 VI. I ()θ = I ()θ + I ()θ . (2.8) y θ y+1 y+2

In the following proposition other relations for the Bessel function which can be easily driven from (2.1) and (2.2) are presented.

Proposition 1:

For anyθ > 0 , θ1 > 0 and θ 2 > 0 then y 2 ∞ ⎛ θ ⎞ ⎜ 1 ⎟ θ1 +θ2 I. ∑ ⎜ ⎟ I y ()2 θ1θ2 = e (2.9) y=−∞ ⎝θ2 ⎠ y 2 ∞ ⎛ θ ⎞ ⎜ 1 ⎟ θ1 +θ2 II. ∑ y⎜ ⎟ I y ()2 θ1θ 2 = ()θ1 −θ 2 e (2.10) y=−∞ ⎝θ 2 ⎠ y 2 ∞ ⎛ θ ⎞ 2 2 ⎜ 1 ⎟ θ1 +θ2 III. ∑ y ⎜ ⎟ I y ()2 θ1θ 2 = ()θ1 +θ 2 + ()θ1 −θ 2 e (2.11) y=−∞ ⎝θ 2 ⎠ ∞ 2 θ IV. ∑ y I y ()θ = θe for any θ > 0 (2.12) y=−∞ ∞ 4 θ V. ∑ y I y ()θ = θe (3θ +1 ) for any θ > 0 (2.13) y=−∞

Proof:

(I) is obtained from the fact that (2.1) is a probability mass function. (II) and (III) follow form the mean and the variance representations. (IV) is a special case θ of (III) by setting θ = θ = . 1 2 2 2 (V) The fourth cummulant K4 = μ4 − 3μ2 ⎛θ θ ⎞ If Y ~ PD⎜ , ⎟, then the fourth cummulant is θ and μ2 = θ . ⎝ 2 2 ⎠ E(Y 4 )= θ + 3θ 2

4 ∞ 4 θ ∑ y I y ()θ = θe (3θ +1 ) for any θ > 0 . y=−∞

3. New representation of the Poisson Difference distribution

~ The regularized hypergeometric function 0 F1 is linked with the modified Bessel function of the first kind through the identity given by equation (2.6). It has the property ~ y ~ 0 F1()();−y +1;θ = θ 0 F1 ; y +1;θ (3.1) Using (2.6) and (3.1) the Poisson difference distribution can be expressed using any of the following equivalent formulas: y 2 ⎛ θ ⎞ −θ1 −θ 2 ⎜ 1 ⎟ P()Y = y = e ⎜ ⎟ I y ()2 θ1θ2 , y = K,−1,0,1,K( formula I) ⎝θ2 ⎠

−θ1 −θ 2 y ~ P()Y = y = e θ1 0 F1 (y +1,θ1θ2 ) , y = K,−1,0,1,K( formula II)

−θ1 −θ 2 − y ~ P()Y = y = e θ2 0 F1 (− y +1,θ1θ2 ) , y = K,−1,0,1,K( formula III)

−θ1 −θ 2 max{}0,− y y ~ P()Y = y = e ()θ1θ2 θ1 0 F1( y +1,θ1θ2 ) , y = K,−1,0,1,K( formula IV )

The advantages of the new formulas are: 1. Easier and more direct notation. Following the steps of deriving the Poisson difference distribution, it is more logical to use the regularized hypergeometric function instead of the Bessel function as follows.

Let X 1 ~Poisson(θ1 ) be independent of X 2 ~Poisson(θ 2 ) then

Y = X 1 − X 2 ~PD(θ1 ,θ 2 ) ∞ P()(Y = y = P X 1 − X 2 = y )= ∑ P()X 1 − X 2 = y X 2 = k k=0 ∞ = ∑ P ()()X 1 = y + k P X 2 = k k=max()− y,0 ∞ θ θ k −θ1 −θ2 y ()1 2 = e θ1 ∑ k =max()− y,0 k!()y + k ! ∞ θ θ k −θ1 −θ2 y ()1 2 P()Y = y = e θ1 ∑ y = K,−1,0,1,K with the convention that any k =0 k!()y + k ! term with negative factorial in the denominator is zero.

−θ1 −θ 2 y ~ P()Y = y = e θ1 0 F1 (; y +1;θ1θ2 ) y = K,−1,0,1,K

2. The special case when θ2 = 0 can be considered directly using formula II to get

the Poisson difference(θ1 ,0) ≡ Poisson(θ1 ).

Let Y ~PD(θ1 ,θ 2 ), and assume that θ2 = 0 then

−θ1 y ~ P(Y = y) = e θ1 0 F1(); y +1;0 , y = K,−1,0,1,K

5 ⎧ −θ1 y ⎧ 1 e θ1 , y = 0,1,2, ⎪ , y = 0,1,2,K ~ ⎪ K = ⎨ y! , since 0 F1 (); y +1;0 = ⎨ y! . ⎪ ⎪ ⎩ 0 , otherwise ⎩0 , otherwise This special case is not applicable when using the notation with the modified

Bessel function of the first kind since θ 2 appears in the denominator.

3. The special case when θ1 = 0 can be considered directly using formula III to get

the Poisson difference (0,θ2 ) ≡ ’negative’ Poisson(θ2 ).

Let Y ~PD(θ1 ,θ 2 ), and assume that θ1 = 0 then

−θ 2 − y ~ P(Y = y) = e θ2 0 F1();−y +1;0 , y = K,−1,0,1,K

−θ2 − y ⎧e θ 2 ⎧ 1 ⎪ , y = 0,−1,−2,K ~ ⎪ , y = 0,−1,−2,K = ⎨ ()− y ! since 0 F1 ();−y +1;0 = ⎨()− y ! . ⎪ ⎪ ⎩0 , otherwise ⎩0 , otherwise

This special case is not applicable using the notation with the modified Bessel

function of the first kind since θ1 appears in the numerator and a direct substitution will yield zero. 4. A more general formula for the probability mass function of the Poisson difference distribution for which formula II and III are special cases is as follows:

−θ1 −θ 2 max{}0,− y y ~ P()Y = y = e ()θ1θ2 θ1 0 F1(; y +1;θ1θ2 ) , y = K,−1,0,1,K(Formula 4)

4. Estimation

The Poisson difference distribution had been introduced more than 70 years ago. Till now only moment estimates are used in the literature and recently maximum likelihood estimates via EM algorithm were obtained by Karlis and Ntzoufras (2000) avoiding to maximize the likelihood directly. Karlis and Ntzoufras (2006) derived also Bayesian estimates and used the Bayesian approach for testing the equality of the two parameters of the Poisson difference distribution.

In this section, we focus on the estimation of the parameters θ1 and θ2 of the Poisson difference distribution. The maximum likelihood estimates are presented and are compared with the moment estimates via a Monte Carlo study. Asymptotic properties of the maximum likelihood estimates are exploited and confidence interval for each parameter is obtained for the first time. Likelihood ratio test for testing the equality of the two parameters is introduced.

4.1 The Method of Moments:

Let Z1,Z2 ,K,Zn be i.i.d. PD()θ1,θ2 , then 1 θˆ = ()S 2 + Z and (4.1) 1MM 2

6 1 θˆ = ()S 2 − Z , (4.2) 2MM 2 where Z is the sample mean and S 2 is the sample variance. The moment estimators are unbiased estimators. The moment estimates do not exist if S 2 − Z < 0 since in this case we would obtain negative estimates of θ1 or θ2 (Karlis and Ntzoufras (2000)). That is, moment estimates do not exist when the sample variance is less than the absolute value of the sample mean. In simulated samples or real data, cases like this happen usually when one of the parameters is very small compared to the other i.e. θi θ j ≥ 10 fori, j = 1,2. To solve this problem, a modification is done such that the negative estimate is set to zero since zero is the smallest possible value and the other estimate is set to equal the absolute value of the mean.

4.2 Maximum Likelihood Estimation

Let Z1,Z2 ,K,Zn be i.i.d. PD()θ1,θ2 . The likelihood function is given by z / 2 n n ⎡ ⎛ ⎞ i ⎤ −θ1 −θ 2 θ1 L = P()Zi = zi = ⎢e ⎜ ⎟ I z ()2 θ1θ2 ⎥ . ∏∏⎜θ ⎟ i i==11i ⎣⎢ ⎝ 2 ⎠ ⎦⎥

Using the differentiation formula for the modified Bessel function we differentiate the log- likelihood with respect to θ1 and θ2 as follows n zi z I z (2 θ1θ2 )+ I z +1(2 θ1θ2 ) ∂ ln L ∑ i θ n 2 θ θ i i = −n + i=1 + 2 ∑ 1 2 ∂θ1 2θ1 θ1θ2 i=1 I z ()2 θ1θ2 i n

∑ zi n θ I z +1()2 θ1θ2 = −n + i=1 + 2 ∑ i , (4.3) θ1 θ1θ2 i=1 I z ()2 θ1θ2 i n zi z I z (2 θ1θ2 )+ I z +1(2 θ1θ2 ) ∂ ln L ∑ i θ n 2 θ θ i i = −n − i=1 + 1 ∑ 1 2 ∂θ2 2θ2 θ1θ2 i=1 I z ()2 θ1θ2 i n θ I z +1(2 θ1θ2 ) = −n + 1 ∑ i . (4.4) θ1θ2 i=1 I z ()2 θ1θ2 i ˆ ˆ The maximum likelihood estimators θ1MLE and θ2MLE are obtained by setting (4.3) and (4.4) to zero and solving the two nonlinear equations n ⎛ ˆ ˆ ⎞ ∑ zi n I ⎜2 θ θ ⎟ θˆ zi +1 1MLE 2MLE 0 = −n + i=1 + 2MLE ⎝ ⎠ and (4.5) ˆ ∑ θ ˆ ˆ i=1 ⎛ ˆ ˆ ⎞ 1MLE θ1MLEθ2MLE I ⎜2 θ θ ⎟ zi ⎝ 1MLE 2MLE ⎠

7 ⎛ ˆ ˆ ⎞ n I ⎜2 θ θ ⎟ θˆ zi +1 1MLE 2MLE 0 = −n + 1MLE ∑ ⎝ ⎠ . (4.6) ˆ ˆ i=1 ⎛ ˆ ˆ ⎞ θ1MLEθ2MLE I ⎜2 θ θ ⎟ zi ⎝ 1MLE 2MLE ⎠ ˆ ˆ Note that multiplying equation (4.5) by θ1MLE and equation (4.6) by θ2MLE and subtracting them we get n ˆ ˆ − nθ1MLE − nθ2MLE + ∑ zi = 0 i=1 ˆ ˆ θ1MLE = θ2MLE + z (4.7) Now, substituting equation (4.7) into equation (4.6) we obtain ⎛ ˆ ˆ ⎞ n I z +1⎜2 (θ2MLE + z)θ2MLE ⎟ ˆ i ()θ2MLE + z 0 = −n + ∑ ⎝ ⎠ . (4.8) ˆ ˆ i=1 ⎛ ˆ ˆ ⎞ ()θ2MLE + z θ2MLE I z ⎜2 ()θ2MLE + z θ2MLE ⎟ i ⎝ ⎠ ˆ ˆ Hence, we can find θ2MLE by solving the nonlinear equation (4.8) and then find θ1MLE using equation (4.7). ~ ∂ F (); x +1;θ ~ Using the identity, 0 1 = F (); x + 2;θ , maximum likelihood estimates could ∂θ 0 1 also be obtained using formulas II and III. ˆ Using formula II, one can find θ2MLE by solving the nonlinear equation n ~ ˆ ˆ 0 F1 (; xi + 2;(θ 2MLE + X )θ 2MLE ) 0 = −n + θˆ + X (4.9) ()2MLE ∑ ~ ˆ ˆ i=1 0 F1 (); xi +1;()θ 2MLE + X θ 2MLE ˆ ˆ and θ1MLE = θ2MLE + X . (4.10)

Remark:

All three formulas (I, II and III) gave identical maximum likelihood estimates when the relative difference between θ1 and θ 2 was not large (less than 10) when solving the nonlinear equation. But whenθ1 > 10θ2 , the nonlinear equations using formulas I or III were not as fast to converge as using formula II and were more willing to obtain negative estimate for θ2 than formula II. On the other side, for θ2 > 10θ1 , the nonlinear equations using formulas I or II were not as fast to converge as using formula III and were more willing to obtain negative estimates for θ1 than formula III. Hence, for maximum likelihood estimation, when the relative difference between θ1 and θ 2 is not large any formula can be used. If θ1 is much larger than θ 2 formula II gives better estimate. If θ 2 is much larger than θ1 formula III gives better estimate. This is also an advantage of using the new representation. It is possible (but very rare) that the maximum likelihood estimates result as negative values when the relative difference between the two estimates is very large a modification as stated in the method of moments is considered.

8 4.3 Asymptotic properties of the maximum likelihood estimates:

Tests and confidence intervals can be based on the fact that the maximum likelihood ˆ ˆ ˆ −1 estimator Θ = (θ1MLE ,θ2MLE ) is asymptotically normally distributed N2 (Θ, I ()Θ ) or more ˆ −1 accurately that n(Θ − Θ) is asymptotically N2 (0,nI (Θ)), where I()Θ is the Fisher information matrix with entries ⎛ − ∂ 2 log L()Θ ⎞ I Θ = E⎜ ⎟ , i, j = 1,2 . i, j () ⎜ ⎟ ⎝ ∂θ i ∂θ j ⎠ (4.11) Under mild regularity conditions, n−1 times the observed information matrix I(Θˆ ) is a consistent estimator of I()Θ / n . The observed information matrix using formula II is given by I I ˆ ⎛ 11 12 ⎞ I()Θ = ⎜ ⎟ where ⎝I21 I22 ⎠ ∂2 log L I11 = − 2 ∂θ1 ˆ θ1=θ1 n z ~ ~ ~ 2 ∑ i n F z +1,θˆ θˆ F z + 3,θˆ θˆ − F z + 2,θˆ θˆ = i=1 −θˆ2 0 1()()()i 1MLE 2MLE 0 1 i 1MLE 2MLE 0 1 i 1MLE 2MLE , 2 2MLE ∑ ~ 2 θˆ i=1 ˆ ˆ 1MLE 0 F1()zi +1,θ1MLEθ2MLE ∂2 log L I22 = − 2 ∂θ2 ˆ θ2 =θ2 ~ ~ ~ 2 n F z +1,θˆ θˆ F z + 3,θˆ θˆ − F z + 2,θˆ θˆ = −θˆ2 0 1()i 1MLE 2MLE 0 1( i 1MLE 2MLE ) 0 1( i 1MLE 2MLE ) and 1MLE ∑ ~ 2 i=1 ˆ ˆ 0 F1()zi +1,θ1MLEθ2MLE ∂2 log L I21 = I12 = − ∂θ1∂θ2 ˆ ˆ θ1 =θ1 ,θ 2 =θ 2 ~ n F z + 2,θˆ θˆ = − 0 1( i 1MLE 2MLE ) ∑ ~ ˆ ˆ i=1 0 F1()zi +1,θ1MLEθ2MLE ~ ~ ~ 2 n F z +1,θˆ θˆ F z + 3,θˆ θˆ − F z + 2,θˆ θˆ −θˆ θˆ 0 1( i 1MLE 2MLE )0 1( i 1MLE 2MLE ) 0 1( i 1MLE 2MLE ) . 1MLE 2MLE ∑ ~ 2 i=1 ˆ ˆ 0 F1()zi +1,θ1MLEθ2MLE

The 95% confidence intervals for θ1 and θ2 are obtained by

ˆ I 22 ˆ I11 θ1 ±1.96 2 and θ 2 ±1.96 2 . (4.12) I11I 22 − I12 I11I 22 − I12

5.Testing:

The likelihood ratio test is a statistical test for making a decision between two hypotheses based on the value of this ratio.

5.1 Likelihood Ratio Test for equality of the parameters:

Let x1, x2 ,K, xn be the outcome of a random sample of size n with respect to the variable X . We consider the likelihood ratio test (LRT) for the null hypothesis

H0 : The data is drawn from PD()θ,θ against the alternative

H1 : The data is drawn from PD()θ1,θ2 . The LRT statistic is written as f x , x , , x ;θˆ λ = ( 1 2 K n ) , (5.1) n ˆ ˆ f ()x1, x2 ,K, xn ;θ1,θ2 ˆ where f (x1, x2 ,K, xn ;θ ) denotes the likelihood function of the sample under the null ˆ ˆ hypothesis calculated at maximum likelihood estimate of θ and f (x1, x2 ,K, xn ;θ1,θ2 ) denotes the likelihood function of the sample under the alternative hypothesis calculated at maximum likelihood estimates of θ1 and θ2 .

Under H0 the likelihood function is given by n n ∑ xi −2nθ i=1 ~ 2 f ()x1, x2 ,K, xn ;θ = e θ ∏ 0 F1()xi +1,θ . (5.2) i=1 The log-likelihood is given by n n ⎛ ⎞ ~ 2 ln f ()x1, x2 ,K, xn ;θ = −2nθ + ⎜∑ xi ⎟lnθ + ∑ln0F1()xi +1,θ . (5.3) ⎝ i=1 ⎠ i=1 The maximum likelihood estimate θˆ of θ is obtained by solving the nonlinear equation n x ~ ∂ ln f ()x , x , , x ;θ ∑ i n F ()x + 2,θ 2 1 2 K n = −2n + i=1 + 2θ 0 1 i = 0 . (5.4) ∑ ~ 2 ∂θ θ i=1 0 F1 ()xi +1,θ And hence, n n ˆ ∑ xi ~ ˆ −2nθ ˆi=1 ˆ2 f ()x1, x2 ,K, xn ;θ = e θ ∏ 0 F1 ()xi +1,θ . (5.5) i=1

Under H1 , n x n ˆ ˆ ∑ i ~ ˆ ˆ −nθ1−nθ2 ˆ i=1 ˆ ˆ f ()x1, x2 ,K, xn ;θ1,θ2 = e θ1 ∏ 0 F1 ()xi +1,θ1θ2 . (5.6) i=1

10 ⎛ ⎛ n ⎞ n ~ ⎞ ⎜ ˆ ˆ ˆ2 ⎟ − 2nθ + ⎜∑ xi ⎟lnθ + ∑ln0F1()xi +1,θ ⎜ ⎝ i=1 ⎠ i=1 ⎟ (5.7) − 2ln λn = −2⎜ ⎟ ⎛ ⎛ n ⎞ n ~ ⎞ ⎜− ⎜− nθˆ − nθˆ + ⎜ x ⎟lnθˆ + ln F x +1,θˆθˆ ⎟⎟ ⎜ ⎜ 1 2 ∑ i 1 ∑ 0 1()i 1 2 ⎟⎟ ⎝ ⎝ ⎝ i=1 ⎠ i=1 ⎠⎠

Under regularity condition for large values of n, − 2lnλn has chi-square distribution with 2 one degree of freedom. We reject H0 if − 2lnλn > χ1−α ,1 .

5.2 Likelihood Ratio Test for θ2 = 0 :

If the observed data were all nonnegative integer values even though they are differences, it is interesting to test if Poisson distribution can fits the data as well as the Poisson difference or not.

Let x1, x2 ,K, xn be the outcomes of a random sample of size n with respect to the variable X where all these outcomes are nonnegative integer values. We consider the likelihood ratio test (LRT) for the null hypothesis

H0 : The data is drawn from Poisson(θ1 ) (i.e. θ2 = 0 ) against the alternative

H1 : The data is drawn from PD()θ1,θ2 . The LRT statistic is written as f x , x , , x ;θˆ λ = ( 1 2 K n 01 ) , (5.8) n ˆ ˆ f ()x1, x2 ,K, xn ;θ1,θ2 ˆ where f (x1, x2 ,K, xn ;θ01 ) denotes the likelihood function of the sample under the null ˆ ˆ hypothesis calculated at maximum likelihood estimate of θ1 and f (x1, x2 ,K, xn ;θ1,θ2 ) denotes the likelihood function of the sample under the alternative hypothesis calculated at maximum likelihood estimates of θ1 and θ2 .

Under H0 the likelihood function is given by n n ∑ xi ˆ −nx i=1 f ()x1 , x2 ,K, xn ;θ 01 = e x / ∏ xi !. i=1 (5.9)

Under H1 , n x n ˆ ˆ ∑ i ~ ˆ ˆ −nθ1 −nθ 2 ˆ i=1 ˆ ˆ f ()x1, x2 ,K, xn;θ1,θ2 = e θ1 ∏ 0 F1 ()xi +1,θ1θ2 . (5.10) i=1 Therefore, ⎛ ⎛ n ⎞ n ⎛ ⎛ n ⎞ n ~ ⎞⎞ − 2ln λ = −2⎜− nX + ⎜ x ⎟ln X − ln x !−⎜− nθˆ − nθˆ + ⎜ x ⎟lnθˆ + ln F x +1,θˆ θˆ ⎟⎟. n ⎜ ∑ i ∑∑i ⎜ 1 2 ∑ i 1 0 1 ()i 1 2 ⎟⎟ ⎝ ⎝ i=1 ⎠ i==11⎝ ⎝ i=1 ⎠ i ⎠⎠ (5.11)

Under regularity condition for large values of n, − 2lnλn has chi-square distribution with 2 one degree of freedom. We reject H0 if − 2lnλn > χ1−α ,1 .

6. Simulation study:

The main objective of this section is to discuss some simulation results for computing the estimates of the parameters of PD(θ1,θ2 ) using the method of moments and the maximum likelihood method.

To generate one observation , Z , from PD(θ1,θ2 ) we generated one observation, X , from the Poisson distribution with parameter θ1 and an independent observation, Y , from the Poisson distribution with parameter θ2 and computed Z = X − Y . In this simulation study we used 1000 samples of size n=10, 20, 30, 50, 100, 150 and

200 and different values of θ1 and θ2 . We calculated the bias and used the relative mean square error (RMSE) as measures of the performance of the estimates in all the considered methods of estimation, where

r ˆ 1 ˆ 1. BIAS()θi = ∑()θ ji −θi (6.1) r j =1 1/ 2 r ˆ 1 ⎡1 ˆ 2 ⎤ 2. RMSE()θ i = ⎢ ∑()θ ji −θi ⎥ (6.2) θ i ⎣r j=1 ⎦ for i = 1,2 and r = 1000 . Tables 1-3 and graphs 1-4 illustrate some of the results.

In order to investigate the power of LRT for equality of the two parameters, the empirical power of the test was examined. The empirical power of the test is defined as the proportion of times the null hypothesis was rejected when the data actually were generated under the alternative hypothesis using 1000 replications. For each of a sample size n = 30, 50 and 100, the power of the test is computed under various choices for the parameters of the alternative distribution. We obtain the power at

θ1 = 0.1, 0.3, 0.5, 1, 3, 5, 10, 20 and 50 and θ2 = cθ1 for c = 0.1, 0.2, 0.3, 0.5, 1, 1.1, 1.2, 1.3 and 1.5. Note that c = 1 corresponds to the null hypothesis and the calculated values are the empirical type one error of the test. Table 4 shows the power of the test when the significance level is 5%, while Table 5 the power of the test when the significance level is 1%.

Discussion of the Simulation Results:

1. The maximum likelihood estimates are better than the moment estimates in terms of relative mean square error. Out of the 700 different cases considered in this ˆ ˆ ˆ simulation the RMSE of θ1MLE is less than θ1MM in 669 cases and RMSE of θ2MLE ˆ is less than θ2MM in 672 cases. 2. The RMSE differ substantially between the two methods of estimation when there is a large relative difference between the two parameters. In these cases the

12 maximum likelihood estimates are much better than the moment estimates in terms of relative mean square error as shown by the graphs 1-4. 3. In terms of bias, the method of moment is better than the maximum likelihood method when the relative difference between the two parameters is small or moderate since the moment estimates are unbiased. The maximum likelihood estimates are much better than the moment estimates in terms of the bias when the relative difference between the two parameters is large and the sample size is small, while the method of moment becomes better for large sample size. 4. When there is no large relative difference between the two parameters both methods are good. As well as, for large sample size, both methods can be used even when there is large relative difference .

5. When both θ1 and θ2 are large moment estimates and ML estimates are very close as expected since the distribution approaches normality. 6. As expected the RMSE always decreases as the sample size increases in both method of estimation. It has been noticed that the RMSE increases with the decreases of the parameter in both method. 7. The maximum likelihood estimators are frequently negatively biased and the bias decreases as the sample size increases. 8. The LRT test has lower performance when it is used to detect components that are very close, in other words the power of the test increase with the relative distance of the components.

9. Considering the value of c fixed, the power increases as the values of θ1 and θ2 increase. 10. When we increase the sample size, the power improves as expected. 11. At c=1, the type one error is smaller or around 0.05 in the 5% level of significance table and is smaller or around 0.01 in the 1% level of significance table.

7. Applications

7.1 Application to the Saudi Stock Exchange Data:

The data has been downloaded from the Saudi Stock Exchange and further filtered. Trading in Saudi Basic Industry (SABIC) and Arabian Shield from the Saudi stock exchange (TASI) were recorded at June 30, 2007, every minute. The Saudi Stock Exchange opens at 11.00 am and closes at 3.30 pm. Missing minutes have been added with a zero price change. The first and final 15 minutes of the trading day, were deleted from the data. The reason for this is that we only focus on studying the price formation during ordinary trading. The minimum amount a price can move is SAR 0.25 in Saudi stock i.e. the tick size is 0.25. The price change is therefore characterized by discrete jumps. The data consists of the difference in price every minute as number of ticks=(close price-open price)*4. Note that, our considered data could take both positive and negative integer values. In Figure 5 and 6, the price change at every minute is illustrated in terms of number of ticks for SABIC and Arabian Shield. Descriptive statistics of the considered data are presented in Table 6.

2 2

1 0

-1 0

Sabic -2 A rabian Shield -3 -1

-4

-2 -5

1 24 48 72 96 120 144 168 192 216 240 1 24 48 72 96 120 144 168 192 216 240 minut e minut e Figure 5: Plot of the price change every minute for Figure 6: Plot of the price change every minute for SABIC Arabian Shield

Table 6: Descriptive Statistics for SABIC and Arabian Shield Variable Sample size Mean Standard Minimu Maximum Deviation m SABIC 240 -0.0833 0.6479 -2 2 Arabian Shield 240 -0.1042 1.0276 -5 2

In order to test if our samples are random samples we conduct the runs test on every sample. (Runs tests test whether or not the data order is random. No assumptions are made about population distribution parameters.) For SABIC the p-value= 0.852 and for Arabian Shield the p-value=0.123. Since both P- values are greater than 0.05, our samples can be considered as random samples. The numbers of ticks of price change take values on the integer numbers. An appropriate distribution to fit these samples could be the Poisson difference distribution.

Maximum likelihood and moment estimates of θ1 and θ2 are obtained using methods discussed in the previous section and illustrated in the next table.

Table 7: Estimation result for SABIC and Arabian Shield

ˆ ˆ ˆ ˆ Stock θ1MLE θ2MLE θ1MM θ2MM SABIC 0.1681 0.2514 0.1682 0.2516 Arabian Siheld 0.451 0.5551 0.4737 0.5779

The Pearson Chi-square test is performed to both samples to test if Poisson difference distribution gives good fit to the data. The null hypothesis is that the sample comes from PD distribution and the alternative hypothesis is that sample does not come from PD distribution. For SABIC the p-value= 0.449862, which implies that PD(0.168,0.251) fits the data well. For Arabian Shield the p-value= 0.137931, which implies that PD(0.451,0.5551) fits the data well.

The 95% confidence intervals for θ1 and θ2 are calculated for SABIC and Arabian Shield.

14 SABIC

The 95% confidence intervals for θ1 is (0.1088, 0.2273) and 95% CI for θ 2 is (0.1818, 0.3211).

Arabian Shield

The 95% confidence intervals for θ1 is (0.3355, 0.5664) and 95% CI for θ 2 is (0.4327, 0.6776).

In both cases, SABIC and Arabian Shield the two confidence intervals overlapped indicating that θ1 and θ2 could be equal. The likelihood ratio test for equality of means is conducted and the statistic for SABIC=3.979 and for Arabian Shield=2.582. The 2 tabulated value to compare with is χ1,0.95 = 3.84146 . For SABIC, we reject the hypothesis that θ1 and θ2 are equal. While for Arabian Shield we fail to reject the hypothesis that θ1 and θ2 are equal confirming the overlapping in the confidence ˆ intervals. The maximum likelihood estimate of θ is θMLE = 0.507185. Hence PD(0.507,0.507) gives a good fit for these data. Bar charts of the relative frequency, PD distribution estimated using the method of moments and the maximum likelihood method for the two stocks are ploted in the figures 7 and 8. In figure 8 PD distribution with equal parameters estimated using maximum likelihood is also plotted. Clearly, from the graphs that PD fits well both data and no significance difference is found from the two methods of estimation for SABIC while this is not the case for Arabian shield.

0.8

0.7

0.6

0.5 PD MLE 0.4 PD MM Saudi basic industry 0.3

0.2

0.1

0 -5 -4 -3 -2 -1 0 1 2 3 4 5

Figure 7: SABIC and fitted distributions

15 0.5 0.45 0.4 0.35 0.3 PD MLE PD MM 0.25 Arabian Shield 0.2 PD MLE Equal means 0.15 0.1 0.05 0 -5 -4 -3 -2 -1 0 1 2 3 4 5

Figure 8: Arabian Shield and fitted distributions

7.2 Application to Nursery Intensive Care Unit Data:

The numbers of occupied beds of the NICU (nursery intensive care unit) in Dallah hospital at Riyadh, Saudi Arabia from 12/12/2005 to 22/03/2006 are time dependent but after taking difference of every two consecutive days the resulting data is a random sample of size 100. This data represents the change in number of beds during 24 hours in NICU. Descriptive statistics of the original and the differenced data are as follows:

Table 8: number of occupied beds and the difference

Variable Sample size Mean Standard Deviation Minimum Maximum No of occupied Beds 100 9.861 2.250 5 17 Difference 101 -0.0300 1.654 -4 4

Figures 9 and 10 illustrate the plots of the number of occupied beds and the difference of every two consecutive days, respectively. The runs test was applied to both the original and the differenced data. The result of the test was that the number of occupied beds is not a random sample (p-value=0), while after differencing the resulting data is a random sample (p-value=0.979).

4 19 3 17 2 15 1

13 0

11 -1

-2 9

no of occupied beds in NICU in beds occupied of no -3 7 difference no of in occupied beds -4 5 -5 1 10 20 30 40 50 60 70 80 90 100 1 10 20 30 40 50 60 70 80 90 100 day Index

Figure 9: number of occupied beds in NICU Figure 10: difference of number of occupied beds in NICU

The changes in number of occupied beds during 24 hours in NICU take values on the integer numbers. A good candidate to fit this sample could be the Poisson difference distribution. The estimation result are summarized in table 9. Figure 11 represents the fitted Poisson difference distributions using both methods with the relative frequency of the data.

Table 9 estimation result for NICU ˆ ˆ ˆ ˆ θ1MLE θ2MLE θ1MM θ2MM 1.34236 1.37236 1.35323 1.38323

The Pearson Chisquare test is performed to test if Poisson difference distribution gives good fit to the data. The p-value=0.407497, which implies that PD(1.342, 1.372) fits the data well.

The 95% CI for θ1 is (0.9089, 1.7758) and the 95% CI for θ 2 is (0.93761, 1.8071). The likelihood ratio test for equality of the parameters is conducted. The 2 statistic = 0.0331479 < χ1,0.95 = 3.84146 which implies that a Poisson difference with equal parameters fits the data well. The maximum likelihood estimate of θ is given by θˆ = 1.3578 .

Difference in number of beds in NICU and fitted distributions

0.35

0.3

0.25 NICU 0.2 PDMM

P(x) 0.15 PDMLE PD equal parameters 0.1

0.05

0 -7-6-5-4-3-2-1012345678 x

Figure 11: Relative frequency of occupied beds of NICU and fitted distributions

Interpretation: If a is positive, then P()X = a represent the probability that the number of occupied beds in NICU increase by a during 24 hours, and if a is negative it is the probability that the number of occupied beds in NICU decrease by a during hours, while a zero value of a gives the probability that the number of occupied beds remain the same during 24 hours.

17 Table 1: Estimation result when θ1 = 0.1 and θ 2 = 0.1,0.5,1,5,10,20,40, 100

θ1 θ 2 n BIASθ1MLE BIASθ2MLE RMSE θ1MLE RMSEθ2MLE BIAS θ1MM BIASθ2MM RMSE θ1MM RMSE θ2MM 0.1 0.1 10 -0.01318 -0.00538 1.292956 1.032222 -0.00021 0.007589 1.121226 1.157216 0.1 0.1 30 0.001254 -0.00175 0.607665 0.590262 0.001867 -0.00113 0.637939 0.620011 0.1 0.1 50 -0.00095 0.000432 0.477679 0.476944 -0.00068 0.000699 0.499026 0.505239 0.1 0.5 10 0.004554 -4.6E-05 1.34112 0.494076 0.008511 0.003911 1.801878 0.556341 0.1 0.5 30 -0.00535 -0.00245 0.703185 0.272131 -0.00813 -0.00523 0.908524 0.298558 0.1 0.5 50 -0.00409 0.000312 0.590546 0.208273 -0.00411 0.000286 0.772062 0.226391 0.1 1 10 -0.00023 0.000574 1.720408 0.346265 0.071756 0.072556 2.496205 0.427179 0.1 1 30 -0.00043 -0.00606 0.97797 0.199335 0.019552 0.013918 1.290641 0.22228 0.1 1 50 -0.00391 -0.00175 0.723831 0.149405 0.010211 0.012371 1.087228 0.176764 0.1 5 10 -0.08654 -0.06414 1.382104 0.144153 0.346533 0.368933 8.910697 0.235896 0.1 5 30 -0.07852 -0.07819 1.260665 0.083531 0.221034 0.221368 5.515384 0.13896 0.1 5 50 -0.06459 -0.04975 1.194913 0.068403 0.144352 0.159192 3.907787 0.106189 0.1 10 10 -0.09708 -0.05988 0.991578 0.101255 0.733356 0.770556 17.42675 0.205457 0.1 10 30 -0.09421 -0.09418 0.981888 0.058813 0.498315 0.498348 10.94255 0.12442 0.1 10 50 -0.09194 -0.07262 0.975239 0.046796 0.346387 0.365707 7.768611 0.093466 0.1 20 10 -0.09843 -0.12323 0.993664 0.072969 1.874689 1.849889 37.96931 0.208872 0.1 20 30 -0.09641 -0.09741 0.989546 0.040707 0.996414 0.995414 20.59289 0.11228 0.1 20 50 -0.09537 -0.08659 0.985593 0.03192 0.750696 0.759476 15.17944 0.080758 0.1 40 10 -0.09921 -0.13041 0.996822 0.051418 3.807911 3.776711 76.17532 0.200861 0.1 40 30 -0.09777 -0.09741 0.99314 0.028593 2.035603 2.03597 41.04851 0.107703 0.1 40 50 -0.09824 -0.08546 0.992922 0.022435 1.561344 1.574124 30.34522 0.077858 0.1 100 10 -0.099736 -0.151436 0.999046 0.032428 9.538211 9.4865111 187.9559 0.192701 0.1 100 30 -0.099574 -0.100108 0.998505 0.018047 5.181139 5.1806057 102.7201 0.104915 0.1 100 50 -0.099269 -0.077309 0.997578 0.014143 4.007431 4.029391 76.30223 0.076853

18 Table 2: Estimation result when θ1 = 3 and θ 2 = 0.1,0.5,1,5,10,20,40, 100 θ1 θ2 n BIASθ1MLE BIASθ2MLE RMSE θ1MLE RMSEθ2MLE BIAS θ1MM BIASθ2MM RMSE θ1MM RMSE θ2MM 3 0.1 10 -0.04 -0.0363 0.19678 2.476532 0.243878 0.247578 0.285589 6.215183 3 0.1 30 -0.03523 -0.03223 0.11635 1.601173 0.103909 0.106909 0.150275 3.113259 3 0.1 50 -0.02591 -0.03017 0.091817 1.361756 0.071904 0.067644 0.115182 2.388226 3 0.5 10 -0.11002 -0.11312 0.251362 1.185636 0.008789 0.005689 0.329483 1.755105 3 0.5 30 -0.05699 -0.04819 0.155813 0.761915 -0.00636 0.002441 0.17914 0.922178 3 0.5 50 -0.04041 -0.04709 0.125363 0.612523 -0.01238 -0.01906 0.141203 0.711609 3 1 10 -0.15926 -0.17196 0.314625 0.877726 0.003678 -0.00902 0.356776 1.002042 3 1 30 -0.06605 -0.05435 0.186557 0.505613 -0.0101 0.001603 0.197708 0.537326 3 1 50 -0.04357 -0.05009 0.146065 0.386746 -0.01382 -0.02034 0.152505 0.41042 3 5 10 -0.36627 -0.39117 0.60179 0.375197 0.006111 -0.01879 0.64498 0.404338 3 5 30 -0.14994 -0.12007 0.354225 0.217081 -0.02189 0.007979 0.355858 0.220883 3 5 50 -0.08335 -0.09005 0.261821 0.161954 -0.01368 -0.02038 0.267901 0.165491 3 10 10 -0.53944 -0.57604 0.893654 0.2858 0.0221 -0.0145 1.026357 0.323568 3 10 30 -0.28508 -0.24178 0.556272 0.171646 -0.04656 -0.00326 0.568225 0.176295 3 10 50 -0.1359 -0.1421 0.414424 0.129982 -0.0005 -0.0067 0.425559 0.133148 3 20 10 -0.68591 -0.74061 1.052776 0.171392 0.814289 0.759589 1.538699 0.243851 3 20 30 -0.35268 -0.29381 0.802618 0.123681 0.087179 0.146046 0.901271 0.140041 3 20 50 -0.17222 -0.18214 0.686815 0.107032 0.079154 0.069234 0.717855 0.111514 3 40 10 -2.04151 -2.11971 0.931211 0.08661 2.465933 2.387733 2.710573 0.210919 3 40 30 -1.20482 -1.12195 0.884336 0.071742 0.794782 0.877648 1.477545 0.114006 3 40 50 -0.99586 -1.0106 0.839517 0.066912 0.563998 0.549258 1.209934 0.093328 3 100 10 -2.59423 -2.71403 0.971201 0.043913 7.994667 7.874867 6.401175 0.195413 3 100 30 -2.36323 -2.22713 0.941309 0.033053 3.499068 3.635168 3.257403 0.099246 3 100 50 -2.24481 -2.25893 0.929114 0.031138 2.81066 2.79654 2.662162 0.081033

19 Table 3: Estimation result when θ1 = 20 and θ 2 = 0.1,0.5,1,5,10,20,40, 100 θ1 θ2 n BIASθ1MLE BIASθ2MLE RMSE θ1MLE RMSEθ2MLE BIAS θ1MM BIASθ2MM RMSE θ1MM RMSE θ2MM 20 0.1 10 -0.10421 -0.09821 0.071036 0.994113 1.870156 1.876156 0.211625 39.41315 20 0.1 30 -0.09571 -0.09648 0.042675 0.990861 0.964733 0.963967 0.104989 19.30099 20 0.1 50 -0.083 -0.09572 0.031964 0.98242 0.726688 0.713968 0.081143 14.86756 20 0.5 10 -0.46559 -0.46639 0.074859 0.978388 1.6849 1.6841 0.2135 8.007935 20 0.5 30 -0.41145 -0.40642 0.04942 0.946161 0.771648 0.776682 0.106884 3.942413 20 0.5 50 -0.36795 -0.38309 0.039426 0.921474 0.537333 0.522193 0.084079 3.093743 20 1 10 -0.78468 -0.79508 0.084748 0.956369 1.470678 1.460278 0.217115 4.090912 20 1 30 -0.62563 -0.6177 0.062705 0.901952 0.574832 0.582766 0.112717 2.098218 20 1 50 -0.50544 -0.52042 0.053941 0.878119 0.361774 0.346794 0.089447 1.669592 20 5 10 -0.87752 -0.90012 0.254507 0.995728 0.0209 -0.0017 0.303863 1.205384 20 5 30 -0.40616 -0.38006 0.151666 0.591397 -0.01518 0.010922 0.159795 0.627692 20 5 50 -0.31743 -0.33259 0.125275 0.489582 -0.06843 -0.08359 0.127258 0.497595 20 10 10 -2.28165 -2.30295 0.347421 0.698201 -0.47397 -0.49527 0.340446 0.683717 20 10 30 -0.54147 -0.48843 0.206997 0.413896 0.087491 0.140524 0.199993 0.39952 20 10 50 -0.26585 -0.26725 0.149505 0.298936 0.029506 0.028106 0.149887 0.299151 20 20 10 -2.48508 -2.51788 0.427011 0.431674 -0.60854 -0.64134 0.453715 0.458402 20 20 30 -0.4914 -0.42457 0.256362 0.256967 0.157414 0.224247 0.263921 0.264899 20 20 50 -0.372 -0.37444 0.196582 0.197804 0.009829 0.007389 0.199615 0.200815 20 40 10 -3.32938 -3.41208 0.70145 0.350294 0.439578 0.356878 0.712148 0.354942 20 40 30 -1.12493 -1.10773 0.403581 0.203392 0.100084 0.117284 0.387297 0.195578 20 40 50 -0.59344 -0.60972 0.319339 0.160506 0.155238 0.138958 0.308743 0.155333 20 100 10 -2.48622 -2.59042 1.04886 0.210463 0.705311 0.601111 1.400466 0.279987 20 100 30 -1.19129 -1.16739 0.705659 0.142139 0.334136 0.358036 0.770405 0.155002 20 100 50 -0.8106 -0.8495 0.577485 0.116146 0.330816 0.291916 0.607913 0.122086

bias of (1=0.3, 2=0.3) bias of 1 (1=0.3, 2=3)

0.008 0.2

0.004 0.15

0 0.1 0 50 100 150 200 250 bias MLE bias MLE -0.004 0.05 bias bias bias bias bias MM bias MM

-0.008 0 0 50 100 150 200 250

-0.012 -0.05

-0.016 -0.1 sample size sample size

(a) (b)

Figure 1: (a) Bias of θ1 using MM and ML versus sample size when θ1 = 0.3, θ 2 = 0.3

(b) Bias of θ1 using MM and ML versus sample size when θ1 = 0.3, θ 2 = 3

bias of theta1 (1=1, 2=1) bias of 1 (1=1, 2=20)

0.02 2

0.01 1.5 0 0 50 100 150 200 250 -0.01 1

-0.02 bias MLE bias MLE 0.5 bias bias -0.03 bias MM bias bias MM

-0.04 0 0 50 100 150 200 250 -0.05 -0.5 -0.06

-0.07 -1 sample size sample size

(a) (b)

Figure 2: (a) Bias of θ1 using MM and ML versus sample size when θ1 = 1, θ 2 = 1

(b) Bias of θ1 using MM and ML versus sample size when θ1 = 1, θ 2 = 20

21 RMSE of (1=0.3, 2=0.3) RMSE of 1 (1=0.3, 2=3)

0.8 2.5

0.7 2 0.6

0.5 1.5 RMSE MLE RMSE MLE 0.4 RMSE RMSE RMSE MM RMSE MM 1 0.3

0.2 0.5 0.1

0 0 0 50 100 150 200 250 0 50 100 150 200 250 sample size sample size

(a) (b)

Figure 3: (a) RMSE of θ1 using MM and ML versus sample size when θ1 = 0.3, θ 2 = 0.3

(b) RMSE of θ1 using MM and ML versus sample size when θ1 = 0.3, θ 2 = 3

RMSE of 1 (1=1, 2=1) RMSE of 1 (1=1, 2=20)

0.6 4.5

0.5 4

3.5 0.4 3

0.3 MSE MLE 2.5 RMSE MLE MSE MSE

MSE MM MSE 2 RMSE MM 0.2 1.5

0.1 1

0.5 0 0 0 50 100 150 200 250 0 50 100 150 200 250 sample size sample size

(a) (b)

Figure 4: (a) RMSE of θ1 using MM and ML versus sample size when θ1 = 1, θ 2 = 1

(b) RMSE of θ1 using MM and ML versus sample size when θ1 = 1, θ 2 = 20

22 Table 4: Power of the LRT of equal parameters when the significance level is 5% θ1 n c=0.1 c=0.2 c=0.3 c=0.5 c=1 c=1.1 c=1.2 c=1.3 c=1.5 30 0.498 0.373 0.314 0.179 0.064 0.053 0.037 0.039 0.049 0.1 50 0.645 0.463 0.324 0.152 0.033 0.05 0.046 0.063 0.1 100 0.868 0.702 0.535 0.262 0.047 0.046 0.063 0.1 0.171 30 0.83 0.653 0.467 0.233 0.036 0.064 0.076 0.095 0.147 0.3 50 0.96 0.853 0.689 0.339 0.052 0.056 0.079 0.125 0.239 100 1 0.994 0.949 0.645 0.041 0.043 0.092 0.206 0.42 30 0.958 0.853 0.688 0.337 0.048 0.058 0.088 0.127 0.247 0.5 50 1 0.983 0.892 0.542 0.053 0.066 0.099 0.165 0.347 100 1 1 0.998 0.86 0.039 0.067 0.136 0.307 0.635 30 0.999 0.992 0.931 0.612 0.064 0.076 0.131 0.21 0.411 1 50 1 1 0.996 0.827 0.046 0.081 0.156 0.283 0.598 100 1 1 1 0.991 0.045 0.112 0.282 0.531 0.904 30 1 1 0.999 0.965 0.056 0.115 0.261 0.474 0.837 3 50 1 1 1 0.999 0.055 0.136 0.372 0.674 0.98 100 1 1 1 1 0.041 0.219 0.63 0.939 1 30 1 1 1 1 0.06 0.13 0.368 0.664 0.972 5 50 1 1 1 1 0.061 0.189 0.536 0.885 1 100 1 1 1 1 0.045 0.334 0.835 0.993 1 30 1 1 1 1 0.064 0.215 0.639 0.911 1 10 50 1 1 1 1 0.057 0.308 0.844 0.991 1 100 1 1 1 1 0.043 0.563 0.994 1 1 30 1 1 1 1 0.063 0.373 0.914 0.999 1 20 50 1 1 1 1 0.06 0.58 0.987 1 1 100 1 1 1 1 0.042 0.855 1 1 1 30 1 1 1 1 0.057 0.769 0.999 1 1 50 50 1 1 1 1 0.061 0.923 1 1 1 100 1 1 1 1 0.041 1 1 1 1

23 Table 5: Power of the LRT of equal parameters when the significance level is 1% θ1 n c=0.1 c=0.2 c=0.3 c=0.5 c=1 c=1.1 c=1.2 c=1.3 c=1.5 30 0.091 0.059 0.045 0.031 0.007 0.005 0.001 0.004 0.006 0.1 50 0.38 0.248 0.162 0.061 0.009 0.013 0.016 0.015 0.029 100 0.711 0.472 0.302 0.119 0.009 0.014 0.018 0.024 0.065 30 0.605 0.382 0.235 0.082 0.006 0.008 0.011 0.025 0.052 0.3 50 0.877 0.647 0.42 0.153 0.011 0.011 0.018 0.033 0.082 100 1 0.962 0.807 0.387 0.006 0.009 0.022 0.068 0.215 30 0.854 0.625 0.425 0.149 0.013 0.012 0.022 0.04 0.082 0.5 50 0.989 0.908 0.707 0.3 0.01 0.015 0.026 0.046 0.152 100 1 1 0.979 0.658 0.01 0.012 0.038 0.135 0.388 30 0.996 0.937 0.797 0.335 0.017 0.026 0.054 0.086 0.212 1 50 1 1 0.97 0.625 0.01 0.02 0.054 0.115 0.345 100 1 1 1 0.943 0.012 0.026 0.099 0.293 0.732 30 1 1 0.997 0.885 0.019 0.028 0.123 0.242 0.634 3 50 1 1 1 0.984 0.013 0.042 0.17 0.429 0.896 100 1 1 1 1 0.008 0.078 0.378 0.817 0.998 30 1 1 1 0.985 0.02 0.046 0.166 0.41 0.867 5 50 1 1 1 1 0.011 0.068 0.304 0.69 0.99 100 1 1 1 1 0.006 0.133 0.661 0.964 1 30 1 1 1 1 0.025 0.083 0.371 0.761 0.998 10 50 1 1 1 1 0.014 0.128 0.654 0.966 1 100 1 1 1 1 0.006 0.333 0.946 1 1 30 1 1 1 1 0.023 0.161 0.746 0.983 1 20 50 1 1 1 1 0.01 0.328 0.949 1 1 100 1 1 1 1 0.005 0.685 1 1 1 30 1 1 1 1 0.019 0.492 0.993 1 1 50 50 1 1 1 1 0.014 0.786 1 1 1 100 1 1 1 1 0.008 0.993 1 1 1

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