The joint density of X and Y is
f\left(x,y\right)=\frac{(y^2-x^2)}{8}e^{-y},\ \ \ \ 0<y<\infty,\ \ \ \ \ \ \ -y\le x\le y
Show that E\left[Y=y\right]=0


We are tasked with showing that \( E[X \mid Y = y] = 0 \), where the joint probability density function (PDF) of \( X \) and \( Y \) is given as:

\[
f(x, y) = \frac{(y^2 - x^2)}{8} e^{-y}, \quad 0 < y < \infty, \quad -y \leq x \leq y
\]

### Step 1: Conditional Expectation of \( X \) Given \( Y = y \)
The conditional expectation \( E[X \mid Y = y] \) is computed as:

\[
E[X \mid Y = y] = \frac{\int_{-y}^{y} x f_{X \mid Y}(x \mid y) \, dx}{\int_{-y}^{y} f_{X \mid Y}(x \mid y) \, dx}
\]

Where \( f_{X \mid Y}(x \mid y) \) is the conditional PDF of \( X \) given \( Y = y \), which can be derived as:

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

Here, \( f(x, y) \) is the joint PDF, and \( f_Y(y) \) is the marginal density of \( Y \).

### Step 2: Marginal Density of \( Y \)

The marginal density of \( Y \), \( f_Y(y) \), is obtained by integrating the joint density over the range of \( x \):

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

The integral of \( y^2 - x^2 \) over \( x \) simplifies as follows:

\[
\int_{-y}^{y} (y^2 - x^2) \, dx = 2 \int_{0}^{y} (y^2 - x^2) \, dx = 2 \left[ y^2 x - \frac{x^3}{3} \right]_0^y = 2 \left[ y^3 - \frac{y^3}{3} \right] = \frac{4y^3}{3}
\]

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

\[
f_Y(y) = \frac{1}{8} e^{-y} \cdot \frac{4y^3}{3} = \frac{y^3}{6} e^{-y}
\]

### Step 3: Conditional Expectation \( E[X \mid Y = y] \)

Now, we calculate the conditional expectation:

\[
E[X \mid Y = y] = \frac{\int_{-y}^{y} x f(x, y) \, dx}{f_Y(y)} = \frac{\int_{-y}^{y} x \cdot \frac{(y^2 - x^2)}{8} e^{-y} \, dx}{\frac{y^3}{6} e^{-y}}
\]

Since \( e^{-y} \) appears in both the numerator and denominator, it cancels out, giving:

\[
E[X \mid Y = y] = \frac{\int_{-y}^{y} x \cdot (y^2 - x^2) \, dx}{\frac{y^3}{6}}
\]

Now, simplify the integral in the numerator. Notice that the integrand is an odd function (a product of an odd function \( x \) and an even function \( y^2 - x^2 \)):

\[
\int_{-y}^{y} x (y^2 - x^2) \, dx = 0
\]

This is because the integral of an odd function over a symmetric interval (from \( -y \) to \( y \)) is zero. Therefore:

\[
E[X \mid Y = y] = 0
\]

### Conclusion:

We have shown that \( E[X \mid Y = y] = 0 \).

Would you like further clarification or details on any part of the process?

Here are 5 related questions you might consider:
1. How is the marginal distribution \( f_Y(y) \) derived in other cases?
2. Why does the symmetry of the function result in a zero integral?
3. What are some properties of joint and conditional distributions?
4. How does the concept of odd and even functions help in simplifying integrals?
5. Can we use similar steps to find \( E[Y \mid X = x] \)?

**Tip:** Identifying odd and even functions in integrals can often simplify your calculations, especially over symmetric limits.

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.

Explore the solution to finding the conditional expectation E[X | Y = y] = 0 given a joint density function f(x, y) = (y^2 - x^2)/8 e^{-y}. This problem covers key concepts such as marginal density, odd functions, and conditional expectation, suitable for undergraduate students.

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