Joint probability- mean and variance? Pl refer image attached

Question

Grace Agate · Accepted Answer

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

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#.

Emery Adams · Answer

undefined To calculate the mean of joint probability distribution:

$$ E(X) = \sum_{i=1}^{n} \sum_{j=1}^{m} x_{ij} P(X = x_{ij}) $$

To calculate the variance of joint probability distribution:

$$ Var(X) = \sum_{i=1}^{n} \sum_{j=1}^{m} (x_{ij} - E(X))^2 P(X = x_{ij}) $$

Where:
- $ E(X) $ is the mean of the joint probability distribution.
- $ Var(X) $ is the variance of the joint probability distribution.
- $ x_{ij} $ are the possible values of the random variable $ X $.
- $ P(X = x_{ij}) $ is the probability of the random variable $ X $ taking on the value $ x_{ij} $.
- $ n $ is the number of possible values for the random variable $ X $.
- $ m $ is the number of possible values for the random variable $ Y $.