By: Neil E. Cotter s2

C O NCEPTU A L TOOLS By: Neil E. Cotter STATISTICS STUDENT'S OR t-DISTRIBUTION Derivation

TOOL: The random variable T, defined as follows, is the sampled data analogue of a standard normal (or gaussian) distribution. Because the value of T depends on S, however, T has a t-distribution that differs slightly from the standard normal (or gaussian) distribution. X - m T ә S / n

where n  number of data points, X , (which are independent and normally distributed i with mean  and variance ) n 1 X ә X  sample mean n е i i=1 n 2 1 2 S ә (X - X )  sample variance n - 1е i i=1 The probability density function of T is a t-distribution with  = n – 1 degrees of freedom:

- (n+1) /2 G((n +1)/2)ж t2 ц f (t) = з1 + ч T з n ч G(n /2) pn и ш

DERIV: The first step is to express T in terms of random variables with known

distributions: Z T = c / n

where X - m S2 Z ә and c2 ә n s / n s 2

C O NCEPTU A L TOOLS By: Neil E. Cotter STATISTICS STUDENT'S OR t-DISTRIBUTION Derivation (cont.)

As shown in other Conceptual Tools, Z has a standard normal (or gaussian) distribution, and 2 has a chi-squared distribution with  = n – 1 degrees of freedom:

1 - z2 /2 fZ (z) = e 2p 1 м x(n /2)- 1e- x /2 x > 0 п n /2 f (x) = 2 G(n /2) c 2 н п о0 x Ј 0 As shown in another tool, Z and 2 are also independent. This allows us to write the joint distribution of Z and 2 as the product of the respective

probability density functions for Z and 2:

м 1 - z2 /2 1 (n /2)- 1 - x /2 e x e x > 0 п 2p n /2 f (z,x) = н 2 G(n /2) п о0 x Ј 0 To find the probability density for T, we consider the cumulative distribution for T and take its derivative:

d d d Ҙ Ҙ fT (t) = FT (t) = P(T Ј t) = z f (z,x)dxdz т-Ҙ т =t dt dt dt x / n

NOTE: We treat the case of t > 0. We have t in the lower limit in this case because larger values of x give smaller values of t. For the

case of t < 0, we would integrate from –∞ to t. We move the derivative inside the outer integral and write the lower limit in terms of the x more clearly: й щ Ҙ к d Ҙ ъ fT (t) = f (z,x)dx dz т-Ҙ кdt т z2 ъ x=n лк t2 ыъ The chain rule yields an interpretation of the derivative of the integral:

d Ҙ d Ҙ dx 2 f (z,x)dx = 2 f (z,x)dx Ч dt тx=n z dx тx=n z dt t2 t2

or 2 d Ҙ - 2z f (z,x)dx = - f (z,x) z2 Чn dt т z2 x=n 3 x=n t2 t t2

Making this substitution, we have the following expression: 2 Ҙ 2 2z fT (t) = f (z,x) z Чn dz т-Ҙ x=n 3 t2 t Now we use the expression for f(z, x), assuming t > 0:

м жn z2 ц (n /2)- 1 - / 2 п 2 з 2 ч 1 - z2 /2 1 жnz ц з t ч 2 п e e и ш z > 0 f (z,x) nz = n /2 з 2 ч x= н 2p 2 G(n /2)з t ч t 2 п и ш п о0 z Ј 0

NOTE: We have the condition that z > 0 because the definition of T requires Z > 0 to achieve T > 0. (X and X are always

positive.) Incorporating the constraint on z into the lower limit of the outer integral

yields the following expression:

жn z2 ц (n /2)- 1 - з ч/ 2 1 2 1 жn z2 ц з 2 ч 2z2 f (t) = Ҙ e- z /2 e и t ш n dz T т0 n /2 з 2 ч 3 2p 2 G(n /2)и t ш t It is helpful to define a term for the constants:

1 1 (n /2)- 1 1 1 n /2 k = 2 n Чn = n 2p 2n /2G(n /2) p 2(n- 1) /2G(n /2)

Using this new term, we have the following expression:

жn z2 ц (n /2)- 1 - / 2 2 з 2 ч 2 Ҙ - z2 /2жz ц з t ч жz ц1 f (t) = k e з ч e и ш з ч dz T т0 з 2 ч з 2 ч t иt ш иt ш or

2ж n ц n /2 - z 1 + / 2 2 Ҙ з 2 ч жz ц 1 f (t) = k e и t ш з ч dz T т0 з 2 ч t иt ш We define the following variable that allows us to use a convenient change of variables:

n zt ә z 1+ t2

We have the following relationships for the change of variables:

dzt = dz n , z = 0 z = 0, z ® z ® 1+ Ю t Ҙ Ю t Ҙ t2

- 1 z2 z2 1 1 1 1 ж t2 ц = t = z2 = z2 = z2n- 1 1 + 2 n Ч 2 t 2 t 2 t з ч t 1+ t t + n n t v 2 1+ и ш t v

In terms of zt we have the following integral expression:

2 - n /2 Ҙ 2 n /2 - n /2ж t ц 1 dz f (t) = k e- zt /2 z2 (n) 1 + t T т0 ( t ) з ч n n t2 и ш 1+ n

2 - (n+1) /2 - (n+1) /2ж t ц Ҙ - z2 /2 2 n /2 f (t) = kn з1 + ч e t z dz T з n ч т0 ( t ) t и ш We change variables again in order to make the integrand have the form of a chi-squared distribution:

2 w ә zt We have the following relationships for the change of variables: dw dw = = dzt , zt = 0 Ю w = 0, zt ® Ҙ Ю w ® Ҙ 2zt 2 w

In terms of w we have the following integral expression:

2 - (n+1) /2 - (n+1) /2ж t ц Ҙ n /2 - w /2 dw f (t) = kn з1 + ч w e T з n ч т0 и ш 2 w By moving appropriate constants inside the integral, we obtain the integral of an entire chi-squared distribution with  + 1 degrees of freedom. In

other words the probability represented by the following integral must equal unity: 1 Ҙ w((n+1) / 2)- 1e- w / 2dw =1 т0 2(n+1) /2G((n +1)/2)

It follows that we have the following expression for the probability density function for T:

2 - (n+1) /2 (n- 1) /2 - (n+1) /2ж t ц f (t) = k2 G((n +1)/2)n з1 + ч T з n ч и ш Simplification of the constants yields the final result: - (n+1) /2 G((n +1)/2)ж t2 ц f (t) = з1 + ч T з n ч G(n /2) pn и ш