Joint probability- mean and variance? Pl refer image attached

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1 Answer
Dec 26, 2017

The answer is "(d) "3/4x_2 and 3/80x_2^2.

Explanation:

Part 1: Conditional Mean
By definition, "E"(X_1|X_2)=int_Ax_1*color(blue)(f(x_1|x_2))dx_1,

where

color(blue)(f(x_1|x_2))=(f(x_1, x_2))/color(red)(f_(X_2)(x_2)),

where

color(red)(f_(X_2)(x_2))=int_Af(x_1, x_2)dx_1.

In this case, since x_1 is bounded by 0 < x_1 < x_2, our integration interval is A = (0,x_2). Thus:

color(red)(f_(X_2)(x_2))=int_0^(x_2)21" "x_1^2" "x_2^3" "dx_1
color(white)(f_(X_2)(x_2))=21" "x_2^3int_0^(x_2)x_1^2" "dx_1
color(white)(f_(X_2)(x_2))=21" "x_2^3[1/3x_1^3]_(0)^(x_2)
color(white)(f_(X_2)(x_2))=7" "x_2^6

Thus,

color(blue)(f(x_1|x_2))=(f(x_1, x_2))/color(red)(f_(X_2)(x_2))

color(white)(f(x_1|x_2))=(21" "x_1^2" "x_2^3)/(7" "x_2^6)

color(white)(f(x_1|x_2))=(3" "x_1^2)/(x_2^3).

Finally,

"E"(X_1|X_2)=int_Ax_1*color(blue)(f(x_1|x_2))dx_1

color(white)("E"(X_1|X_2))=int_Ax_1*(3x_1^2)/x_2^3dx_1

Since we are once again integrating with respect to x_1, the integration interval is the same as before:

"E"(X_1|X_2)=int_0^(x_2)x_1*(3x_1^2)/x_2^3dx_1

color(white)("E"(X_1|X_2))=3/x_2^3 int_0^(x_2)x_1^3" "dx_1

color(white)("E"(X_1|X_2))=3/x_2^3 [x_1^4/4]_0^(x_2)

color(white)("E"(X_1|X_2))=3/(4x_2^3) [x_2^4]

color(white)("E"(X_1|X_2))=3/4x_2

Part 2: Conditional Variance

The conditional variance is

"Var"(X_1|X_2)="E"(X_1^2|X_2)-["E"(X_1|X_2)]^2

I'll leave the calculation of "E"(X_1^2|X_2) as an exercise. (Hint: just replace x_1 with x_1^2 in the formula for "E"(X_1|X_2).)

The result is:

"Var"(X_1|X_2)=(3x_2^2)/5-[(3x_2)/4]^2

color(white)("Var"(X_1|X_2))=(3x_2^2)/5-(9x_2^2)/16

color(white)("Var"(X_1|X_2))=3x_2^2[1/5-3/16]

color(white)("Var"(X_1|X_2))=3x_2^2[(16-15)/80]

color(white)("Var"(X_1|X_2))=3/80x_2^2.