Question

Joint probability- mean and variance? Pl refer image attached

Answer 1

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`.