Question

Let \[ A = \left\{ (x, y) \in \mathbb{R}^2 : x^2 - \frac{1}{2\sqrt{\pi}}<y<x^2 + \frac{1}{2\sqrt{\pi}} \right\} \] and let the joint probability density function of \((X, Y)\) be \[ f(x, y) = \begin{cases} e^{-(x - 1)^2}, & (x, y) \in A,
0, & \text{otherwise.} \end{cases} \] Then, the covariance between the random variables \(X\) and \(Y\) is equal to .............

Hint:

For narrow uniform strips around a function \(y = g(x)\), \(E(Y|X = x) \approx g(x)\), which simplifies covariance calculations.

Answer 1

Correct Answer: 1

Step 1: Identify the support.
For every \(x\), \(y\) varies in a narrow band centered at \(y = x^2\). The width of this band is \(\frac{1}{\sqrt{\pi}}\), and \(f(x, y)\) does not depend on \(y\).
Step 2: Compute the marginal density of \(X\).
\[ f_X(x) = \int_{x^2 - \frac{1}{2\sqrt{\pi}}}^{x^2 + \frac{1}{2\sqrt{\pi}}} e^{-(x - 1)^2} \, dy = \frac{1}{\sqrt{\pi}} e^{-(x - 1)^2}. \]
Step 3: Compute conditional expectation.
Since \(y\) is uniformly distributed about \(x^2\), \[ E(Y|X = x) = x^2. \]
Step 4: Compute covariance.
\[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y]. \] Now, \[ E[Y] = E[E(Y|X)] = E[X^2], \] and \[ E[XY] = E[X E(Y|X)] = E[X^3]. \] Hence, \[ \text{Cov}(X, Y) = E[X^3] - E[X]E[X^2]. \]
Step 5: For \(X \sim N(1, \frac{1}{2})\), \[ E[X] = 1, \quad E[X^2] = 1 + \frac{1}{2} = \frac{3}{2}, \quad E[X^3] = 1^3 + 3(1)\left(\frac{1}{2}\right) = \frac{5}{2}. \] \[ \text{Cov}(X, Y) = \frac{5}{2} - (1)\left(\frac{3}{2}\right) = 1. \] Final Answer: \[ \boxed{1} \]