Answer to Question #260561 in Statistics and Probability for Anisha

Question #260561

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?


1
Expert's answer
2021-11-15T17:54:24-0500

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"and "\\mu"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""\\sum" x P (X = x)

x

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

= 0

"\\mu""\\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\\ \u03b4x\u03b4y}" , where "\\delta"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} \n \n\u200b" = 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.



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