Step 1: Understanding the Concept:
The problem requires us to find the conditional expectation of Y given X=x, written as \(E(Y|X=x)\). To do this, we first determine the conditional probability density function (PDF) of Y given X=x, \(f(y|x)\), and then compute its expectation.
Step 2: Key Formula or Approach:
1. Find the marginal PDF of X: \(f_X(x) = \int_{-\infty}^{\infty} f(x,y) dy\).2. Find the conditional PDF of Y given X: \(f(y|x) = \frac{f(x,y)}{f_X(x)}\).3. Calculate the conditional expectation: \(E(Y|X=x) = \int_{-\infty}^{\infty} y . f(y|x) dy\).
Step 3: Detailed Explanation:
1. Find the marginal PDF of X, \(f_X(x)\): \[ f_X(x) = \int_{0}^{\infty} xe^{-x(y+1)} dy = \int_{0}^{\infty} xe^{-xy-x} dy \] Factor out terms not involving y: \[ f_X(x) = xe^{-x} \int_{0}^{\infty} e^{-xy} dy \] Evaluate the integral: \[ \int_{0}^{\infty} e^{-xy} dy = \left[ -\frac{1}{x} e^{-xy} \right]_{y=0}^{y=\infty} = (0) - (-\frac{1}{x}e^0) = \frac{1}{x} \] Thus, the marginal PDF is: \[ f_X(x) = xe^{-x} \left(\frac{1}{x}\right) = e^{-x}, \quad \text{for } x \ge 0 \] (X follows an exponential distribution with rate 1).2. Find the conditional PDF of Y given X, \(f(y|x)\): \[ f(y|x) = \frac{f(x,y)}{f_X(x)} = \frac{xe^{-x(y+1)}}{e^{-x}} = \frac{xe^{-xy}e^{-x}}{e^{-x}} = xe^{-xy}, \quad \text{for } y \ge 0 \] This is the PDF of an exponential distribution with rate parameter \(\lambda = x\).3. Calculate the conditional expectation, \(E(Y|X=x)\): The expected value of an exponential distribution with rate parameter \(\lambda\) is \(1/\lambda\). Since Y given X=x follows an exponential distribution with rate \(x\), its expected value is: \[ E(Y|X=x) = \frac{1}{x} \]
Step 4: Final Answer:
The conditional expectation \(E(Y|X = x)\) is \( \frac{1}{x} \).