Joint probability- mean and variance? Pl refer image attached
1 Answer
The answer is
Explanation:
Part 1: Conditional Mean
By definition,
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
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
"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
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 .