Lecture 4: Probability Distributions and Probability Densities - 2

Multivariate Distributions Marginal Distributions Conditional Distributions

Assist. Prof. Dr. Emel YAVUZ DUMAN

MCB1007 Introduction to Probability and Statistics Istanbul˙ K¨ult¨ur University Multivariate Distributions Marginal Distributions Conditional Distributions Outline

1 Multivariate Distributions

2 Marginal Distributions

3 Conditional Distributions Multivariate Distributions Marginal Distributions Conditional Distributions Outline

1 Multivariate Distributions

2 Marginal Distributions

3 Conditional Distributions Multivariate Distributions Marginal Distributions Conditional Distributions

In this section we shall concerned first with the bivariate case, that is, with situation where we are interested at the same time in a pair of random variables defined over a joint sample space that are both discrete. Later, we shall extend this discussions to the multivariate case, covering any finite number of random variables. If X and Y are discrete random variables, we write the probability that X will take on the value x and Y will take on the value y as P(X = x, Y = y). Thus, P(X = x, Y = y) is the probability of the intersection of the events X = x and Y = y.Asinthe univariate case, where we dealt with one random variable and could display the probabilities associated with all values of X by means of a table, we can now, in the bivariate case, display the probabilities associated with all pairs of the values of X and Y by mean of a table. Multivariate Distributions Marginal Distributions Conditional Distributions

Example 1

Two caplets are selected at a random form a bottle containing three aspirin, two sedative, and four laxative caplets. If X and Y are, respectively, the numbers of the aspirin and sedative caplets included among the two caplets drawn from the bottle, ﬁnd the probabilities associated with all possible pairs of values of X and Y .

Solution. The possible pairs are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), and (2, 0). So we obtain the following probabilities: 3 2 4 6 P(X =0, Y =0)= 0 0 2 = , 9 36 2 3 2 4 8 P(X =0, Y =1)= 0 1 1 = , 9 36 2 3 2 4 12 P(X =1, Y =0)= 1 0 1 = , 9 36 2 Multivariate Distributions Marginal Distributions Conditional Distributions

3 2 4 6 P(X =1, Y =1)= 1 1 0 = , 9 36 2 3 2 4 1 P(X =0, Y =2)= 0 2 0 = , 9 36 2 3 2 4 3 P(X =2, Y =0)= 2 0 0 = . 9 36 2 Therefore, we have the following table: H HH x H 0 1 2 y HH 0 6/36 12/36 3/36 1 8/36 6/36 2 1/36 Multivariate Distributions Marginal Distributions Conditional Distributions

Deﬁnition 2 If X and Y are discrete random variables, the function given buy

f (x, y)=P(X = x, Y = y)

for each pair of values (x, y) within the range of X and Y is called the joint probability distribution of X and Y .

Theorem 3 A bivariate function can serve as the joint probability distribution of a pair of discrete random variables X and Y if and only if its values f (x, y) satisfy the conditions 1 f (x, y) ≥ 0 for each pair of values (x, y) within its domain; 2 x y f (x, y)=1, where the double summation extends over all possible pairs (x, y) within its domain. Multivariate Distributions Marginal Distributions Conditional Distributions

Suppose that X canassumeanyoneofm values x1, x2, ··· , xm and Y canassumeanyoneofn values y1, y2, ··· , yn. Then the probability of the event that X = xj and Y = yk is given by

P(X = xj , Y = yk )=f (xj , yk ).

A joint probability function for X and Y can be represented by a joint probability table as in the following:

H HH y H y1 y2 ··· yn Totals ↓ x HH x1 f (x1, y1) f (x1, y2) ··· f (x1, yn) g(x1) x2 f (x2, y1) f (x2, y2) ··· f (x2, yn) g(x2) ...... ··· . . xm f (xm, y1) f (xm, y2) ··· f (xm, yn) g(xm)

Totals → h(y1) h(y2) ··· h(yn) 1 Multivariate Distributions Marginal Distributions Conditional Distributions

Example 4 Determine the value of k for which the function given by

f (x, y)=kxy for x =1, 2, 3; y =1, 2, 3

can serve as a joint probability distribution.

Solution. Substituting the various values of x and y,weget

f (1, 1) = k, f (1, 2) = 2k, f (1, 3) = 3k, f (2, 1) = 2k, f (2, 2) = 4k,

f (2, 3) = 6k, f (3, 1) = 3k, f (3, 2) = 6k, f (3, 3) = 9k. To satisfy the ﬁrst condition of Theorem 3, the constant k must be nonnegative, and to satisfy the second condition

k +2k +3k +2k +4k +6k +3k +6k +9k =1

so that 36k =1andk =1/36. Multivariate Distributions Marginal Distributions Conditional Distributions

f (x, y)=kxy for x =1, 2, 3; y =1, 2, 3

The joint probability function for X and Y can be represented by a joint probability table as in the following: H HH y H 1 2 3 Totals ↓ x HH 1 k 2k 3k 6k 2 2k 4k 6k 12k 3 3k 6k 9k 18k Totals → 6k 12k 18k 36k =1 Multivariate Distributions Marginal Distributions Conditional Distributions

Example 5

If the values of the joint probability distribution of X and Y are as shown in the table H H x HH 0 1 2 y HH 0 1/12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120

ﬁnd (a) P(X =1, Y =2); (b) P(X =0, 1 ≤ Y < 3); (c) P(X + Y ≤ 1); (d) P(X > Y ). Multivariate Distributions Marginal Distributions Conditional Distributions

H HH x H 0 1 2 y HH 0 1/12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120

, 1 (a) P(X =1 Y =2)= 20 ; , ≤ < , , 1 1 3 (b) P(X =0 1 Y 3) = f (0 1) + f (0 2) = 4 + 8 = 8 ; ≤ , , , 1 1 1 1 (c) P(X + Y 1) = f (0 0) + f (1 0) + f (0 1) = 12 + 6 + 4 = 2 ; > , , , 1 1 1 7 (d) P(X Y )=f (1 0) + f (2 0) + f (2 1) = 6 + 24 + 40 = 30 . Multivariate Distributions Marginal Distributions Conditional Distributions

Example 6 If the joint probability distribution of X and Y is given by

f (x, y)=c(x2 + y 2)forx = −1, 0, 1, 3, y = −1, 2, 3

ﬁnd the value of c. Solution. Since H HH x H −1 0 1 3 Totals ↓ y HH −1 2c 1c 2c 10c 15c 2 5c 4c 5c 13c 27c 3 10c 9c 10c 18c 47c Totals → 17c 14c 17c 41c 89c =1

1 then we have that c = 89 . Multivariate Distributions Marginal Distributions Conditional Distributions

Example 7 Show that there is no value of k for which

f (x, y)=ky(2y − x)forx =0, 3, y =0, 1, 2

can serve as the joint probability distribution of two random variables. Solution. Since H HH y H 0 1 2 Totals ↓ x HH 0 0 2k 8k 10k 3 0 −k 2k k Totals → 0 k 10k 11k =1

then we ﬁnd that k =1/11.Butinthiscase,f (3, 1) diﬀers in sign from all other terms. Multivariate Distributions Marginal Distributions Conditional Distributions

Example 8 Suppose that we roll a pair of balanced dice and X is the number of dice that come up 1, and Y is the number of dice that come up 4, 5, or 6. (a) Construct a table showing the values X and Y associated with each of the 36 equally likely points of the sample space. (b) Construct a table showing the values the joint probability distribution of X and Y . Multivariate Distributions Marginal Distributions Conditional Distributions

Solution. (a) If X is the number of dice that come up 1, and Y is the number of dice that come up 4, 5, or 6, then we have

PP PP Roll 2 PP 1 2 3 4 5 6 Roll 1 PPP 1 (2,0) (1,0) (1,0) (1,1) (1,1) (1,1) 2 (1,0) (0,0) (0,0) (0,1) (0,1) (0,1) 3 (1,0) (0,0) (0,0) (0,1) (0,1) (0,1) 4 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2) 5 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2) 6 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2) (b) Then, we simply count the number of times we have each of the possible (x, y) values, and divide by 36. x 0 0 0 1 1 2 y 0 1 2 0 1 0 Prob. 4/36 12/36 9/36 4/36 6/36 1/36 Multivariate Distributions Marginal Distributions Conditional Distributions

Deﬁnition 9 If X and Y are discrete random variables, the function given by F (x, y)=P(X ≤ x, Y ≤ y)= f (s, t)for −∞< x, y < ∞ s≤x t≤y

where f (s, t) is the value of the joint probability distribution of X and Y at (s, t), is called the joint distribution function,orthe joint cumulative distribution, of X and Y .

Theorem 10 If F (x, y) is the value of the joint distribution function of two discrete random variables X and Y at (x, y),then (a) F (−∞, −∞)=0; (b) F (∞, ∞)=1; (c) if a < bandc< d, then F(a, c) ≤ F (b, d). Multivariate Distributions Marginal Distributions Conditional Distributions

Example 11 With reference to Example 1, ﬁnd F (1, 1). H HH x H 0 1 2 y HH 0 6/36 12/36 3/36 1 8/36 6/36 2 1/36

Solution. F (1, 1) = P(X ≤ 1, Y ≤ 1) = f (0, 0) + f (0, 1) + f (1, 0) + f (1, 1) 6 8 12 6 = + + + 36 36 36 36 32 = 36 Multivariate Distributions Marginal Distributions Conditional Distributions

Example 12 With reference to Example 5, ﬁnd the following values of the joint distribution function of the two random variables: (a) F (1.2, 0.9) (b) F (−3, 1.5) (c) F (2, 0) (d) F (4, 2.7). H H x HH 0 1 2 y HH 0 1/12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120

Solution. (a) F (1.2, 0.9) = P(X ≤ 1.2, Y ≤ 0.9) , , 1 1 1 = f (0 0) + f (1 0) = 12 + 6 = 4 (b) F (−3, 1.5) = P(X ≤−3, Y ≤ 1.5) = 0 Multivariate Distributions Marginal Distributions Conditional Distributions

H HH x H 0 1 2 y HH 0 1/12 1/6 1/24 1 1/4 1/4 1/40 2 1/8 1/20 3 1/120

(c) F (2, 0) = P(X ≤ 2, Y ≤ 0) , , , 1 1 1 7 = f (0 0) + f (1 0) + f (2 0) = 12 + 6 + 24 = 24 , . ≤ , ≤ . − , − 1 119 (d) F (4 2 7) = P(X 4 Y 2 7) = 1 f (0 3) = 1 120 = 120 . Multivariate Distributions Marginal Distributions Conditional Distributions

Example 13 If two cards are randomly drawn (without replacement) from an ordinary deck of 52 playing cards, Z is the number of aces obtained in the ﬁrst draw and W is the total number of aces obtained in both draws, ﬁnd F (1, 1).

Solution. Let X be the number of aces obtained in the ﬁrst draw, and Y be the number of aces obtained in the second draw. So, we have f (x, y), the joint probability distribution:

48 47 188 4 48 16 f (0, 0) = · = , f (1, 0) = · = , 52 51 221 52 51 221 48 4 16 4 3 1 f (0, 1) = · = , f (1, 1) = · = . 52 51 221 52 51 221 Multivariate Distributions Marginal Distributions Conditional Distributions

Since H HH w H 0 1 2 z HH 0 f (0, 0) f (0, 1) 1 f (1, 0) f (1, 1)

where z = x and w = x + y,thenwehave H H w HH 0 1 2 z HH 0 188/221 16/221 1 16/221 1/221

Therefore, we obtain that 188 16 16 1 220 F (1, 1) = + + =1− = . 221 221 221 221 221 Multivariate Distributions Marginal Distributions Conditional Distributions

All the deﬁnitions in this section can be generalized to the multivariate case, where there are n random variables. Corresponding to Deﬁnition 2, the values of the joint probability distribution of n discrete random variables X1, X2, ··· , and Xn are given by

f (x1, x2, ··· , xn)=P(X1 = x1, X2 = x2, ··· , Xn = xn)

for each n-tuple (x1, x2, ··· , xn) within the range of the random variables; and corresponding to Deﬁnition 9, the values of their joint distribution function are given by

F (x1, x2, ··· , xn)=P(X1 ≤ x1, X2 ≤ x2, ··· , Xn ≤ xn)

for −∞ < x1 < ∞, −∞ < x2 < ∞, ···, −∞ < xn < ∞. Multivariate Distributions Marginal Distributions Conditional Distributions

Example 14 If the joint probability distribution of three discrete random variables X , Y ,andZ is given by

(x + y)z f (x, y, z)= for x =1, 2; y =1, 2, 3; z =1, 2 63 ﬁnd P(X =2, Y + Z ≤ 3).

Solution. P(X =2, Y + Z ≤ 3) = f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1) 3 6 4 = + + 63 63 63 13 = . 63 Multivariate Distributions Marginal Distributions Conditional Distributions

Example 15 Find c if the joint probability distribution of X , Y and Z is given by

f (x, y, z)=kxyz for x =1, 2; y =1, 2, 3; z =1, 2

and determine F (2, 1, 2) and F (4, 4, 4)

Solution. Since

2 3 2 1= x=1 y=1 z=1 =f (1, 1, 1) + f (1, 1, 2) + f (1, 2, 1) + f (1, 2, 2) + f (1, 3, 1) + f (1, 3, 2) +f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1) + f (2, 2, 2) + f (2, 3, 1) + f (2, 3, 2) =k(1+2+2+4+3+6+2+4+4+8+6+12)=54k

then we have that k =1/54. Multivariate Distributions Marginal Distributions Conditional Distributions

f (x, y, z)=kxyz for x =1, 2; y =1, 2, 3; z =1, 2

Also,

F (2, 1, 2) = P(X ≤ 2, Y ≤ 1, Z ≤ 2) = f (1, 1, 1) + f (1, 1, 2) + f (2, 1, 1) + f (2, 1, 2) 1 9 1 = (1+2+2+4)= = 54 54 6 F (4, 4, 4) = P(X ≤ 4, Y ≤ 4, Z ≤ 4) = 1. Multivariate Distributions Marginal Distributions Conditional Distributions Outline

1 Multivariate Distributions

2 Marginal Distributions

3 Conditional Distributions Multivariate Distributions Marginal Distributions Conditional Distributions Marginal Distributions

To introduce the concept of a marginal distribution,letus consider the following example. Example 16

In Example 1 we derived the joint probability distribution of two random variables X and Y , the number of aspirin and the number of sedative caplets included among two caplets drawn at random from a bottle containing three aspirin, two sedative, and four laxative caplets. Find the probability distribution of X alone and that of Y alone. Solution. The results of Example 1 are shown in the following table, together with the marginal totals, that is, the totals of the respective rows and columns: Multivariate Distributions Marginal Distributions Conditional Distributions

H H x HH 0 1 2 h(y) y HH 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1 The column totals are the probabilities that X will take on the values 0, 1, and 2. In other words, they are the values 2 g(x)= f (x, y)forx =0, 1, 2 y=0 of the probability distribution of X . ⎧ ⎨⎪15/36 for x =0 g(x)= 18/36 for x =1 ⎩⎪ 3/36 for x =2 Multivariate Distributions Marginal Distributions Conditional Distributions H H x HH 0 1 2 h(y) y HH 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1

By the same token, the row totals are the values

2 h(y)= f (x, y)fory =0, 1, 2 x=0 of the probability distribution of Y . ⎧ ⎨⎪21/36 for y =0 h(y)= 14/36 for y =1 ⎩⎪ 1/36 for y =2 Multivariate Distributions Marginal Distributions Conditional Distributions

We are thus led to the following deﬁnition: Deﬁnition 17 If X and Y are discrete random variables and f (x, y) is the value of their joint probability distribution at (x, y), the function given by g(x)= f (x, y) y

for each x within the range of X is called the marginal distribution of X . Correspondingly, the function given by h(y)= f (x, y) x

for each y within the range of Y is called the marginal distribution of Y . Multivariate Distributions Marginal Distributions Conditional Distributions

Example 18

Two textbooks are selected at random from a shelf contains three statistics texts, two mathematics texts, and three physics texts. If X is the number of statistics texts and Y the number of mathematics texts actually chosen (a) construct a table showing the values of the joint probability distribution of X and Y . (b) Find the marginal distributions of X and Y . Multivariate Distributions Marginal Distributions Conditional Distributions

Solution. (a) Since 3 3 3 3 9 2 3 6 f (0, 0) = 2 = , f (1, 0) = 1 1 = , f (0, 1) = 1 1 = , 8 28 8 28 8 28 2 2 2 3 2 6 3 3 2 1 f (1, 1) = 1 1 = , f (2, 0) = 2 = , f (0, 2) = 2 = , 8 28 8 28 8 28 2 2 2 we have the following join probability distribution table: H HH x H 0 1 2 h(y) y HH 0 3/28 9/28 3/28 15/28 1 6/28 6/28 12/28 2 1/28 1/28 g(x) 10/28 15/28 3/28 1 Multivariate Distributions Marginal Distributions Conditional Distributions

H H x HH 0 1 2 h(y) y HH 0 3/28 9/28 3/28 15/28 1 6/28 6/28 12/28 2 1/28 1/28 g(x) 10/28 15/28 3/28 1

(b) The marginal distributions of X and Y ⎧ ⎧ ⎨⎪10/28 for x =0 ⎨⎪15/28 for y =0 g(x)= 15/28 for x =1 h(y)= 12/28 for y =1 ⎩⎪ ⎩⎪ 3/28 for x =2 1/28 for y =2

respectively. Multivariate Distributions Marginal Distributions Conditional Distributions

Example 19 Given the values of the joint probability distribution of X and Y shown in the table HH H x H −1 1 y HH −1 1/8 1/2 0 0 1/4 1 1/8 0

ﬁnd the marginal distribution of X and the marginal distribution of Y . Solution. ⎧ ⎨⎪ 1 + 1 = 5 for y = −1 1 + 1 = 1 for x = −1 8 2 8 g(x)= 8 8 4 h(y)= 1 1 1 3 ⎪ 4 for y =0 2 + 4 = 4 for x =1 ⎩ 1 8 for y =1 Multivariate Distributions Marginal Distributions Conditional Distributions Outline

1 Multivariate Distributions

2 Marginal Distributions

3 Conditional Distributions Multivariate Distributions Marginal Distributions Conditional Distributions Conditional Distributions

We deﬁned the conditional probability of an event A, given event B,as P(A ∩ B) P(A|B)= P(B) provided P(B) = 0. Suppose now that A and B are the events X = x and Y = y so that we can write

P(X = x, Y = y) f (x, y) P(X = x|Y = y)= = P(Y = y) h(y)

provided P(Y = y)=h(y) =0,where f (x, y) is the value of the joint probability distribution of X and Y at (x, y)andh(y)isthe value of the marginal distribution of Y at y.Denotingthe conditional probability f (x|y) to indicate that x is a variable and y is ﬁxed, let us make the following deﬁnition: Multivariate Distributions Marginal Distributions Conditional Distributions

Deﬁnition 20 If f (x, y) is the value of the joint probability distribution of the discrete random variables X and Y at (x, y), and h(y) is the value of the marginal distribution of Y at y, the function given by

f (x, y) f (x|y)= h(y) =0 h(y)

for each x within the range of X , is called the conditional distribution of X given Y = y. Correspondingly, if g(x)isthe value of the marginal distribution of X at x, the function

f (x, y) w(y|x)= g(x) =0 g(x)

for each y within the range of Y , is called the conditional distribution of Y given X = x. Multivariate Distributions Marginal Distributions Conditional Distributions

Example 21 With reference to Example 1 and Example 16, ﬁnd the conditional distribution of X given Y =1. f (x|1) = f (x, 1)/h(1) H HH x H 0 1 2 h(y) y HH 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1

Solution. Substituting the appropriate values from the table above, we get f (0, 1) 8/36 8 f (1, 1) 6/36 6 f (0|1) = = = , f (1|1) = = = , h(1) 14/36 14 h(1) 14/36 14 f (2, 1) 0 f (2|1) = = =0. h(1) 14/36 Multivariate Distributions Marginal Distributions Conditional Distributions

Example 22 With reference to Example 18 ﬁnd the conditional distribution of Y given X =0. w(y|0) = f (0, y)/g(0) H HH x H 0 1 2 h(y) y HH 0 3/28 9/28 3/28 15/28 1 6/28 6/28 12/28 2 1/28 1/28 g(x) 10/28 15/28 3/28 1

Solution. Substituting the appropriate values from the table above, we get f (0, 0) 3/28 3 f (0, 1) 6/28 6 w(0|0) = = = , w(1|0) = = = , g(0) 10/28 10 g(0) 10/28 10 f (0, 2) 1/28 1 w(2|0) = = = . g(0) 10/28 10 Multivariate Distributions Marginal Distributions Conditional Distributions

Deﬁnition 23 Suppose that X and Y are discrete random variables. If the events X = x and Y = y are independent events for all x and y,thenwe say that X and Y are independent random variables.Insuch case, P(X = x, Y = y)=P(X = x) · P(Y = y) or equivalently

f (x, y)=f (x|y) · h(y)=g(x) · h(y).

Conversely, if for all x and y the joint probability function f (x, y) can be expressed as the product of a function of x alone and a function of y alone (which are then the marginal probability functions of X and Y ), X and Y are independent. If, however, f (x, y) cannot be so expressed, then X and Y are dependent. Multivariate Distributions Marginal Distributions Conditional Distributions

Example 24 Determine whether the random variables of Example 1 are independent.

Solution. With reference to Example 1, we have the following joint distribution table: H HH x H 0 1 2 h(y) y HH 0 6/36 12/36 3/36 21/36 1 8/36 6/36 14/36 2 1/36 1/36 g(x) 15/36 18/36 3/36 1

Since 8 15 14 f (0, 1) = = · = g(0) · h(1) 36 36 36 we see that X and Y are dependent. Multivariate Distributions Marginal Distributions Conditional Distributions

Example 25 A box contains three balls labeled 1, 2 and 3. Two balls are randomly drawn from the box without replacement. Let X be the number on the ﬁrst ball and Y the number on the second ball. (1) Find the joint probability distribution of X and Y . (2) Find the marginal distributions of X and Y . (3) Find the conditional distribution of X given Y =1. (4) Find the conditional distribution of Y given X =2. (5) Determine whether the random variables X and Y are independent. Multivariate Distributions Marginal Distributions Conditional Distributions

Solution. (1) Since P(X =1, Y =2)=P(X =1, Y =3)=P(X =2, Y =1)= P(X =2, Y =3)=P(X =3, Y =1)=P(X =3, Y =2)= 1 · 1 1 3 2 = 6 , this joint probability distribution can be expressed by the following table: H H y HH 1 2 3 g(x) x HH 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1 Multivariate Distributions Marginal Distributions Conditional Distributions

H H y HH 1 2 3 g(x) x HH 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1

(2) ⎧ ⎧ ⎨⎪2/6, x =1, ⎨⎪2/6, y =1, g(x)= 2/6, x =2, h(y)= 2/6, y =2, ⎩⎪ ⎩⎪ 2/6, x =3, 2/6, y =3. Multivariate Distributions Marginal Distributions Conditional Distributions H HH y H 1 2 3 g(x) x HH 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1

(3) The conditional distribution of X given Y =1: f (x, 1) f (x|1) = ⇒ h(1) f (1, 1) 0 f (1|1) = = =0, h(1) 2/6 f (2, 1) 1/6 1 f (2|1) = = = , h(1) 2/6 2 f (2, 1) 1/6 1 f (3|1) = = = . h(1) 2/6 2 Multivariate Distributions Marginal Distributions Conditional Distributions H HH y H 1 2 3 g(x) x HH 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1

(4) The conditional distribution of Y given X =2: f (2, y) w(y|2) = ⇒ g(2) f (2, 1) 1/6 1 w(1|2) = = = , g(2) 2/6 2 f (2, 2) 0 w(2|2) = = =0, g(2) 2/6 f (2, 3) 1/6 1 w(3|2) = = = . g(2) 2/6 2 Multivariate Distributions Marginal Distributions Conditional Distributions

H HH y H 1 2 3 g(x) x HH 1 1/6 1/6 2/6 2 1/6 1/6 2/6 3 1/6 1/6 2/6 h(y) 2/6 2/6 2/6 1

(5) Since 2 2 f (1, 1) = 0 = g(1) · h(1) = · 6 6 we see that X and Y are dependent. Multivariate Distributions Marginal Distributions Conditional Distributions

Thank You!!!