We are tasked with showing that the conditional expectation $E[X | Y = y] = 0$ given the joint density $f(x, y) = \frac{y^2 - x^2}{8} e^{-y}$, where $0 < y < \infty$ and $-y < x < y$.

1. Understanding the Problem

The conditional expectation $E[X | Y = y]$ is defined as:

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

where $f_{X|Y}(x | y)$ is the conditional density of $X$ given $Y = y$, obtained by normalizing the joint density $f(x, y)$ over $x$, for a fixed $y$.

We know from the joint density that $X$ is supported on $-y < x < y$. So, the conditional expectation becomes:

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

2. Finding the Conditional Density $f_{X|Y}(x | y)$

The conditional density $f_{X|Y}(x | y)$ is given by:

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

where $f_Y(y)$ is the marginal density of $Y$.

a. Marginal Density $f_Y(y)$

To find $f_Y(y)$, we integrate the joint density over $x$:

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

This simplifies as follows:

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

$\int_{-y}^{y} (y^2 - x^2) \, dx = 2 \int_0^{y} (y^2 - x^2) \, dx$

The integral evaluates to:

$2 \left[ y^2 x - \frac{x^3}{3} \right]_0^y = 2 \left( y^3 - \frac{y^3}{3} \right) = 2 \times \frac{2y^3}{3} = \frac{4y^3}{3}$

Thus, the marginal density becomes:

$f_Y(y) = \frac{e^{-y}}{8} \times \frac{4y^3}{3} = \frac{y^3 e^{-y}}{6}$

b. Conditional Density $f_{X|Y}(x | y)$

Now, we can write the conditional density as:

$f_{X|Y}(x | y) = \frac{f(x, y)}{f_Y(y)} = \frac{\frac{y^2 - x^2}{8} e^{-y}}{\frac{y^3 e^{-y}}{6}} = \frac{3(y^2 - x^2)}{4y^3}$

3. Computing the Conditional Expectation

Now that we have the conditional density, we can compute the conditional expectation $E[X | Y = y]$:

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

This integral breaks into two parts:

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

Now, we compute the integral $\int_{-y}^{y} x(y^2 - x^2) \, dx$. Since the integrand is an odd function (i.e., $x \cdot (y^2 - x^2)$ is odd in $x$), the integral over the symmetric interval $[-y, y]$ is zero:

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

Thus:

$E[X | Y = y] = 0$

Conclusion

We have shown that the conditional expectation $E[X | Y = y] = 0$.

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Would you like further details or explanations?

Here are 5 related questions you might explore:

1. How would we compute $E[Y]$ for the given joint density?
2. What is the variance of $X$ given $Y = y$?
3. How can we generalize this result for different joint distributions?
4. What role does the symmetry of the density function play in this result?
5. How does the marginal density $f_Y(y)$ change if the limits for $x$ change?

Tip: When working with joint densities, it's crucial to normalize the conditional densities properly to compute expectations effectively.