Answer to Question #338597 in Statistics and Probability for Ken

Question

Let X, Y, Z have joint probability distribution function

f(x , y , z) = { k(x^2+ yz), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1

0, elsewhere

Find

(i) the value of k.

(ii) the conditional expectation of Z given X= x,Y= y

Answer 1

The following condition must be satisfied: $\int_0^1\int_0^1\int_0^1f(x,y,z)dxdydz=1$.

Compute the left side of the latter: $\int_0^1\int_0^1\int_0^1f(x,y,z)dxdydz=\int_0^1\int_0^1\int_0^1k(x^2+yz)dxdydz=k\int_0^1\int_0^1(\frac13x^3+xyz)|_0^1dydz=k\int_0^1\int_0^1(\frac13+yz)dydz=k\int_0^1|(\frac13y+\frac12y^2z)|_0^1dz=k\int_0^1(\frac13+\frac12z)dz=k(\frac13z+\frac14z^2)|_0^1=k(\frac13+\frac14)=\frac{7}{12}k.$

(i) From the latter we find that: $k=\frac{12}{7}$.

(ii) The aim is to find: $E(Z|X=x,Y=y)$. Using the definition of conditional expectation we get: $E(Z|X=x,Y=y)=\int_{0}^{1}zf_{Z|X,Y}(z|x,y)dz=\frac{1}{f_{X,Y}(x,y)}\int_{0}^{1}zf(z,x,y)dz$.

$f_{X,Y}(x,y)=\int_{0}^{1}k(x^2+yz)dz=k(x^2z+\frac12yz^2)|_0^1=k(x^2+\frac12y).$

$\int_{0}^{1}zf(z,x,y)dz=\int_{0}^{1}k(x^2z+yz^2)dz=k(\frac12x^2+\frac13y)$

We get: $E(Z|X=x,Y=y)=\frac{\frac12x^2+\frac13y}{x^2+\frac12y}$.

Answer: (i) $k=\frac{12}{7}$ (ii) $E(Z|X=x,Y=y)=\frac{\frac12x^2+\frac13y}{x^2+\frac12y}$