The joint density of X and Y is f(x, y) = (y2 − x2) 8 e−y, 0 < y < ∞, −y x y Show that E[X|Y = y] = 0.

To prove that \( E[X | Y = y] = 0 \), we need to calculate the conditional expectation of \( X \) given \( Y = y \). Here's a step-by-step breakdown of the process:

### 1. **Given Information**:

The joint probability density function of \( X \) and \( Y \) is given as:
\[
f(x, y) = (y^2 - x^2) \cdot 8 e^{-y}, \quad 0 < y < \infty, \, -y \leq x \leq y.
\]

The marginal density of \( Y \), denoted as \( f_Y(y) \), is required to compute the conditional density of \( X \) given \( Y = y \).

### 2. **Find the Marginal Density \( f_Y(y) \)**:

To obtain the marginal density \( f_Y(y) \), we integrate the joint density \( f(x, y) \) with respect to \( x \):

\[
f_Y(y) = \int_{-y}^{y} f(x, y) \, dx.
\]

Substitute \( f(x, y) = (y^2 - x^2) \cdot 8 e^{-y} \):

\[
f_Y(y) = 8 e^{-y} \int_{-y}^{y} (y^2 - x^2) \, dx.
\]

Now, compute the integral:

\[
\int_{-y}^{y} (y^2 - x^2) \, dx = 2 \int_0^{y} (y^2 - x^2) \, dx.
\]

This can be solved as:

\[
2 \int_0^y (y^2 - x^2) \, dx = 2 \left[ y^2x - \frac{x^3}{3} \right]_0^y = 2 \left( y^3 - \frac{y^3}{3} \right) = 2 \cdot \frac{2y^3}{3} = \frac{4y^3}{3}.
\]

Thus, the marginal density of \( Y \) becomes:

\[
f_Y(y) = 8 e^{-y} \cdot \frac{4y^3}{3} = \frac{32y^3}{3} e^{-y}.
\]

### 3. **Find the Conditional Density \( f_{X|Y}(x | y) \)**:

The conditional density of \( X \) given \( Y = y \) is defined as:

\[
f_{X|Y}(x | y) = \frac{f(x, y)}{f_Y(y)}.
\]

Substitute the expressions for \( f(x, y) \) and \( f_Y(y) \):

\[
f_{X|Y}(x | y) = \frac{(y^2 - x^2) \cdot 8 e^{-y}}{\frac{32y^3}{3} e^{-y}} = \frac{3(y^2 - x^2)}{4y^3}.
\]

Thus, the conditional density of \( X \) given \( Y = y \) is:

\[
f_{X|Y}(x | y) = \frac{3(y^2 - x^2)}{4y^3}, \quad -y \leq x \leq y.
\]

### 4. **Calculate \( E[X | Y = y] \)**:

The conditional expectation \( E[X | Y = y] \) is given by:

\[
E[X | Y = y] = \int_{-y}^{y} x \cdot f_{X|Y}(x | y) \, dx.
\]

Substitute \( f_{X|Y}(x | y) = \frac{3(y^2 - x^2)}{4y^3} \):

\[
E[X | Y = y] = \int_{-y}^{y} x \cdot \frac{3(y^2 - x^2)}{4y^3} \, dx.
\]

Now, separate the terms and compute the integral:

\[
E[X | Y = y] = \frac{3}{4y^3} \int_{-y}^{y} x (y^2 - x^2) \, dx.
\]

This integral can be split into two parts:

\[
E[X | Y = y] = \frac{3}{4y^3} \left[ y^2 \int_{-y}^{y} x \, dx - \int_{-y}^{y} x^3 \, dx \right].
\]

First, compute each of these integrals:

- \( \int_{-y}^{y} x \, dx = 0 \) (since it's an odd function integrated over a symmetric interval),
- \( \int_{-y}^{y} x^3 \, dx = 0 \) (again, because it's an odd function).

Thus, both integrals are zero, and we get:

\[
E[X | Y = y] = 0.
\]

### 5. **Conclusion**:

Therefore, we have shown that:

\[
E[X | Y = y] = 0.
\]

Would you like further details or clarification on any part of this solution?

Here are 5 questions to further explore:
1. How would we approach finding \( E[Y] \) from the given joint density?
2. What is the intuition behind conditional expectation being zero in this case?
3. Can we extend this result to higher moments, such as \( E[X^2 | Y = y] \)?
4. How does the symmetry of the joint density function play a role in this problem?
5. How would you compute the variance of \( X \) given \( Y = y \)?

**Tip**: Recognizing symmetries in probability density functions can often simplify expectation and variance calculations.

The joint density of X and Y is f(x, y) = (y^2 − x^2) 8 e^{-y}, 0 < y < ∞, −y ≤ x ≤ y. Show that E[X|Y = y] = 0.

Learn how to prove that E[X | Y = y] = 0 for a given joint density function f(x, y) = (y^2 - x^2) 8 e^{-y}. This solution involves calculating the conditional expectation by finding the marginal and conditional density functions, integrating with respect to x, and utilizing properties of symmetric functions. Suitable for advanced undergraduate students in probability and statistics.

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