More than one variable

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Chapter More than one variable.1 Bivariate discrete distributions Suppose that the r.v. s X and Y are discrete and take on the values x j and y j, j 1, respectively. Then the joint p.d.f. of X and Y, to be denoted by f X,Y, is defined by: f X,Y (x j,y j ) P(X x j,y y j ) and f X,Y (x,y) 0 when (x,y) (x j,y j ) (i.e., at least one of x or y is not equal to x j or y j, respectively). The marginal distribution of X is defined by the probability function P(X x i ) j P(X x i,y y j ). P(Y y j ) i P(X x i,y y j ). Note that P(X x i ) 0 and i P(X x i) 1. The mean and variance of X can be defined in the usual way. The conditional distribution of X given Y y j is defined by the probability function P(X x i Y y j ) P(X x i,y y j ). P(Y y j ) The conditional mean of X given Y y j is defined by E[X Y y j ] i x i P(X Y y j ). and similarly for the variance: V ar[x Y y j ] E(X 2 Y y j ) (E(X Y y j )) 2. Although the E[X Y y j ] depends on the particular values of Y, it turns out that its average does not, and, indeed, is the same as the E[X]. More precisely, it holds: E[E(X Y )] E[X] and E[E(Y X)] E[Y ]. 1

CHAPTER. MORE THAN ONE VARIABLE 2 That is, the expectation of the conditional expectation of X is equal to its expectation, and likewise for Y. The covariance of X and Y is defined by where cov[x,y ] E[(X E[X])(Y E[Y ])] E[XY ] E[X]E[Y ] E[XY ] i x i y j P(X x i,y y j ). j The result obtained next provides the range of values of the covariance of two r.v.s; it is also referred to as a version of the CauchySchwarz inequality. Theorem.1 Cauchy Schwarz inequality 1. Consider the r.v.s X and Y with E[X] E[Y ] 0 and V ar[x] V ar[y ] 1. Then always 1 E[XY ] 1, and E[XY ] 1 if and only if P(X Y ) 1, and E[XY ] 1 if and only if P(X Y ) 1. 2. For any r.v.s X and Y with finite expectations and positive variances σx 2 and σy 2, it always holds: σ Xσ Y Cov(X,Y ) σ X σ Y, and Cov(X,Y ) σ X σ Y if and only if P[Y E[Y ]+ σ Y σ X (X E[X])] 1, Cov(X,Y ) σ X σ Y if and only if P[Y E[Y ] σ Y σ X (X EX)] 1. The correlation coefficient between X and Y is defined by ( )( ) corr[x,y ] E[ X E[X] Y E[Y ] σ X σ Y ] Cov[X,Y ] E[XY ] E[X]E[Y ] σ X σ Y σ X σ Y. The correlation always lies between 1 and +1. Example.1 Let X and Y be two r.v.s with finite expectations and equal (finite) variances, and set U X + Y and V X Y. Calculate if r.v.s U and V are correlated. E[UV ] E[(X + Y )(X Y )] E(X 2 Y 2 ) E[X 2 ] E[Y 2 ] E[U]E[V ] [E(X + Y )][E(X Y )] (E[X] + E[Y ])(E[X] E[Y ]) (E[X]) 2 (E[Y ]) 2 Cov(U,V ) E[UV ] E[U]E[V ] (E[X 2 ] E[X] 2 ) (E[Y 2 ] E[Y ] 2 ) V ar(x) V ar(y ) 0 U and V are uncorrelated. For two r.v.s X and Y with finite expectations, and (positive) standard deviations σ X and σ Y, it holds:

CHAPTER. MORE THAN ONE VARIABLE 3 and V ar(x + Y ) σ 2 X + σ 2 Y + 2Cov(X,Y ) V ar(x + Y ) σ 2 X + σ 2 Y if X and Y are uncorrelated. Proof V ar(x + Y ) E[(X + Y ) E(X + Y )] 2 E[(X E[X]) + E(Y E[Y ])] 2 E(X E[X]) 2 + E(Y E[Y ]) 2 + 2E[(X E[X])(Y E[Y ])] σ 2 X + σ 2 Y + 2Cov(X,Y ). Random variables X and Y are said to be independent if P(X x i,y y j ) P(X x i )P(Y y j ). If X and Y are independent then Cov[X,Y ] 0. The converse is NOT true. There exist many pairs of random variables with Cov[X,Y ] 0 which are not independent. Example.2 A fair dice is thrown three times. The result of first throw is scored as X 1 1 if the dice shows 5 or and X 1 0 otherwise; X 2 and X 3 are scored likewise for the second and third throws. Let Y 1 X 1 + X 2 and Y 2 X 1 X 3. Show that P(Y 1 0,Y 2 1) 4. Calculate the remaining probabilities in the bivariate distribution of the pair (Y 1,Y 2 ) and display the joint probabilities in an appropriate table. 1. Find the marginal probability distributions of Y 1 and Y 2. 2. Calculate the means and variances of Y 1 and Y 2. 3. Calculate the covariance of Y 1 and Y 2. 4. Find the conditional distribution of Y 1 given Y 2 0. 5. Find the conditional mean of Y 1 given Y 2 0.

CHAPTER. MORE THAN ONE VARIABLE 4 P(X 1 1) P({5, }) 1, P(X 3 2 1) 1, P(X 3 3) 1 3 For Y 1 to be 0, X 1 and X 2 must be 0. Then Y 2 to be 1, X 3 must be 1. P(Y 1 0,Y 2 1) P(X 1 0)P(X 2 0)P(X 3 1) 2 2 1 4 3 3 3 Y 1 0 1 2 4 2-1 8 1 Y 2 0 4 2 1 0 12 0 12 3 1 1. Marginal probability distribution of Y 1 : y 1 0 1 2 12 12 3 P(Y 1 y 1 ) Marginal probability distribution of Y 2 : 2. y 2-1 0-1 P(Y 2 y 2 ) E[Y 1 ] 0 12 + 1 12 + 2 3 2 3 E[Y1 2 ] 0 2 12 + 12 12 + 22 3 8 9 V ar[y 1 ] E[Y 2 1 ] (E[Y 1 ]) 2 8 9 ( 2 3 ) 2 4 9 E[Y 2 ] 1 + 0 + 1 0 E[Y 2 2 ] ( 1) 2 + 02 + 12 4 9 V ar[y 2 ] E[Y 2 2 ] (E[Y 2 ]) 2 4 9 3. Cov[Y 1,Y 2 ] E[Y 1 Y 2 ] E[Y 1 ]E[Y 2 ] E[Y 1 Y 2 ] 1 ( 1) 2 +1 1 4 + 2 1 2 2 9

CHAPTER. MORE THAN ONE VARIABLE 5 4. P(Y 1 0 Y 2 0) P(Y 1 0 Y 2 0) P(Y 2 0) P(Y 1 1 Y 2 0) P(Y 1 1 Y 2 0) P(Y 2 0) P(Y 1 2 Y 2 0) P(Y 1 2 Y 2 0) P(Y 2 0) 8 1 8, 1 5. E[Y 1 Y 2 0] 1 + 2 1 8 Exercises Exercise.1 The random variables X and Y have a joint probability function given by { c(x f(x,y) 2 y + x) x-2,-1,0,1,2 y1,2,3 0 otherwise Determine the value of c. Find P(X > 0) and P(X + Y 0) Find the marginal distributions of X and Y. Find E[X] and V ar[x]. Find E[Y ] and V ar[y ]. Find the conditional distribution of X given Y 1 and E[X Y 1]. Find the probability function for Z X + Y and show that E[Z] E[X] + E[Y ] Find Cov[X,Y ] and show that V ar[z] V ar[x] + V ar[y ] + 2Cov[X,Y ]. Find the correlation between X and Y. Are X and Y independent? Table for the joined probabilities:

CHAPTER. MORE THAN ONE VARIABLE X -2-1 0 1 2 1 2c 0 0 2c c 10c Y 2 c c 0 3c 10c 20c 3 10c 2c 0 4c 14c 30c 18c 3c 0 9c 30c 0c Since the sum of probabilities must add to one, c 1. 0 P(X > 0) 39 0 P(X + Y 0) P(X 2,Y 2) + P(X 1,Y 1) 1 10 Marginal distributions for X: x -2-1 0 1 2 P(X x) 18/0 3/0 0 9/0 30/0 Marginal distributions for Y : y 1 2 3 P(Y y) 10/0 20/0 30/0 E[X] 2 18/0 1 3/0 + 0 0 + 1 9/0 + 2 30/0 30/0 1/2 E[X 2 ] ( 2) 2 18/0 + ( 1) 2 3/0 + 0 0 + 1 2 9/0 + 2 2 30/0 3.4 V ar[x] E[X 2 ] E[X] 2 3.4 0.5 2 3. E[Y ] 1 1/ + 2 1/3 + 3 1/2 14/ 7/3 E[Y 2 ] 1 2 1/ + 2 2 1/3 + 3 2 1/2 3/.0 V ar[y ].0 (7/3) 2 5/9 P(X 2 Y 1) 0.2, P(X 1 Y 1) 0, P(X 0 Y 1) 0, P(X 1 Y 1) 0.2, P(X 2 Y 1) 0. E[X Y 1] 2 0.2 1 0 + 0 0 + 1 0.2 + 2 0. 1 Z X + Y z -1 0 1 2 3 4 5 P(Z z) ( 2/0 /0 11/0 4/0 9/0 14/0 14/0 E[Z] 1 0 1 2 + 1 11 + 2 4 + 3 9 + 4 14 + 5 14) 170 2 5 1 + 2 21 E[X] + E[Y ] 3 0 E[X,Y ] 2 1 2/0 2 2 /0 2 3 10/0 1 2 1/0 1 3 2/0 + 1 1 2/0+1 2 3/0+1 3 4/0+2 1 /0+2 2 10/0+2 3 14/0 1 Cov[X,Y ] E[X,Y ] E[X]E[Y ] 1/ E[Z 2 ] 1 84 (1 2 + 1 11 + 4 4 + 9 9 + 1 14 + 25 14) 0 0 V ar[z] 84 ( 170 3.3722 0 0 V ar[x] + V ar[y ] + 2Cov[X,Y ] 3. + 5/9 2 1/ 3.3722 corr[x,y ] Cov[X,Y ] 1/ 0.12 V ar[x]v ar[y ] 3. 5/9

CHAPTER. MORE THAN ONE VARIABLE 7 X and Y are independent: P(X 1,Y 1) P(X 1)P(Y 1). Exercise.2 The following experiment is carried out. Three fair coins are tossed. Any coins showing heads are removed and the remaining coins are tossed. Let X be the number of heads on the first toss and Y the number of heads on the second toss. Note that if X 3 then Y 0. Find the joint probability function and marginal distributions of X and Y. We have that P(Y y,x x) P(Y y X x)p(x x). Suppose X 0, this has a probability 0.5 3. Then Y X 0 has a Binomial distribution with parameters n 3 and p 0.5. Similarly Y X 1 has a Binomial distribution with parameters n 2 and p 0.5. In this way we see we can produce a table of the joint probabilities: X 0 1 2 3 0 1/4 /4 12/4 8/4 /4 Y 1 5/4 12/4 12/4 0 /4 2 3/4 /4 0 0 9/4 3 1/4 0 0 0 1/4 1/8 3/8 3/8 1/8 1 Marginal distribution for X: x 0 1 2 3 P(X x) 1/8 3/8 3/8 1/8 Marginal distribution for Y : y 0 1 2 3 P(Y y) /4 /4 9/4 1/4

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