Question

Let \((X,Y)\) be a random vector with the joint probability density function

• A. The marginal distribution of \(X\) is \(Exp(1)\)
• B. \(E(Y|X=4)=0.25\)
• C. \(E(XY)=1.5\)
• D. \(E(Y)\) is not finite

Hint:

For joint densities, first find the marginal density, then use \(f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)}\) to identify the conditional distribution.

Answer 1

The Correct Option is A, B, D

Step 1: Find the marginal density of \(X\).
\[ f_X(x)=\int_{0}^{\infty}xe^{-x(y+1)}\,dy \] \[ =x e^{-x}\int_{0}^{\infty}e^{-xy}\,dy \] \[ =x e^{-x}\cdot \frac{1}{x} \] \[ =e^{-x},\qquad x>0 \] Thus,
\[ X\sim Exp(1) \] Hence, option (A) is true.

Step 2: Find the conditional distribution of \(Y|X=x\).
\[ f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)} \] \[ =\frac{xe^{-x(y+1)}}{e^{-x}} \] \[ =xe^{-xy},\qquad y>0 \] So,
\[ Y|X=x\sim Exp(x) \] Therefore,
\[ E(Y|X=x)=\frac{1}{x} \] At \(x=4\),
\[ E(Y|X=4)=\frac14=0.25 \] Hence, option (B) is true.

Step 3: Find \(E(XY)\).
Using conditional expectation,
\[ E(XY)=E\{X E(Y|X)\} \] Since
\[ E(Y|X)=\frac{1}{X}, \] we get
\[ E(XY)=E\left(X\cdot \frac{1}{X}\right) \] \[ =E(1)=1 \] So,
\[ E(XY)\neq 1.5 \] Hence, option (C) is false.

Step 4: Check whether \(E(Y)\) is finite.
\[ E(Y)=E\{E(Y|X)\} \] \[ =E\left(\frac{1}{X}\right) \] Since \(X\sim Exp(1)\),
\[ E\left(\frac{1}{X}\right)=\int_{0}^{\infty}\frac{1}{x}e^{-x}\,dx \] The integral diverges near \(x=0\).
Therefore,
\[ E(Y)=\infty \] Hence, option (D) is true.

Step 5: Final conclusion.
The true statements are
\[ \boxed{(A),\,(B)\text{ and }(D)} \]