Question

If, \(f(x, y) = xe^{-x(y+1)}; x \ge 0, y \ge 0\), then \(E(Y|X = x)\) is

• A. \( \frac{1}{x} \)
• B. \( \frac{1}{x^2} + 5 \)
• C. \( \frac{1}{x^2} \)
• D. \( \frac{1}{x} + 3 \)

Hint:

After finding a conditional distribution, \(f(y|x)\), always check if it matches a standard distribution (like Normal, Exponential, Gamma, etc.). If it does, you can use the known formula for its mean to find the conditional expectation, which is much faster than computing the integral from scratch.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question asks for the conditional expectation of Y given X=x, denoted \(E(Y|X=x)\). To find this, we first need to determine the conditional probability density function (PDF) of Y given X=x, denoted \(f(y|x)\). Then, we can find the expectation of this conditional distribution.

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, \(f_X(x)\): \[ f_X(x) = \int_{0}^{\infty} xe^{-x(y+1)} dy = \int_{0}^{\infty} xe^{-xy-x} dy \] We can factor out the terms not involving y: \[ f_X(x) = xe^{-x} \int_{0}^{\infty} e^{-xy} dy \] The integral evaluates to: \[ \int_{0}^{\infty} e^{-xy} dy = \left[ -\frac{1}{x} e^{-xy} \right]_{y=0}^{y=\infty} = (0) - (-\frac{1}{x}e^0) = \frac{1}{x} \] So, the marginal PDF is: \[ f_X(x) = xe^{-x} \left(\frac{1}{x}\right) = e^{-x}, \quad \text{for } x \ge 0 \] (This means 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 the conditional distribution of Y given X=x is exponential 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 \( \frac{1}{x} \).