# Joint probability- mean and variance? Pl refer image attached

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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# .

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