Answer to Question #262869 in Statistics and Probability for John

Question

Suppose X and Y are random variables with P(X = 1) = P(X = −1) = 1 2 ; P(Y = 1) = P(Y = −1) = 1 2 . Let c = P(X = 1 and Y = 1) (a) Determine the joint distribution of X and Y, Cov(X,Y), and r(X,Y). (b) For what value(s) of c are X and Y independent? For what value(s) of c are X and Y 100% correlated?

Answer 1

a) i) The joint distribution of X and Y, P (x, y) is given by;

x = 1 x = -1 Marginal

y = 1 c 1/2-c 1/2

y = -1 1/2-c c 1/2

Marginal 1/2 1/2 1

ii) Cov (X , Y) = E (XY) - $\mu$_x $\mu$_y, where $\mu$_x and $\mu$_y are the means of x and y respectively.

E (XY) = $\sum\sum$ x y P (X=x , Y = y)

x y

= (1)(1)c + (1)(-1)(1/2 - c) + (-1)(1)(1/2-c) + (-1)(-1)(c)

= c + (-1/2+c) + (-1/2+c) + c

= c - 1/2 + c -1/2 + c + c

= 4c - 1

$\mu$_x = $\sum$ x P (X = x)

x

= 1(1/2) + (-1)(1/2)

= 0

$\mu$_y = $\sum$ y P (Y = y)

y

= 1(1/2) + (-1)(1/2)

= 0

=> Cov (X , Y) = (4c - 1) - (0)(0)

= 4c -1

iii) The correlation coefficient is given by;

r (X , Y) = ${Cov (X , Y)\over\ δxδy}$ , where $\delta$_x and $\delta$_y are the standard deviations of X and Y respectively.

Var (X) = E(X²) - [E(X)]²

E(X²) = $\sum$ x² P (X = x)

x

= (1)²(1/2) + (-1)²(1/2)

= 1/4

=> Var (X) = 1/4 - 0²

=> $\delta$_x = $\sqrt{1/4}$ = 1/2

Var (Y) = E(Y²) - [E(Y)]²

E(Y²) = $\sum$ y² P (Y = y)

y

= (1)²(1/2) + (-1)²(1/2)

= 1/4

=> Var (Y) = 1/4 - 0²

=> $\delta$y = $\sqrt{1/4}$ = 1/2

r (X , Y) = ${4c -1\over\ 1/2 1/2}$ = ${4c - 1\over\ 1/4} ​$ = 4 (4c - 1)

b) i) we know that c = P ( X = 1 and Y = 1). For X and Y to be independent, c = P (X = 1) P (Y = 1).

P (X = 1) = 1/2 , P (Y = 1) = 1/2

=> c = 1/2 x 1/2 = 1/4

Therefore, if c = 1/4, then X and Y are independent.

ii) X and Y are said to be 100% correlated if r (x , y) = 1 or r (x , y) = -1.

r (x , y) = 16c - 4 = 1

16c - 4 = 1

16c = 5

c = 5/16

or

r(x , y) = 16c -4 = -1

16c = 3

c = 3/16

Therefore c = 5/16 or 3/16.