To show that $E[X \mid Y = y] = 0$, we need to calculate the conditional expectation of $X$ given $Y = y$.

Step 1: Determine the Joint Density Function

The joint density function of $X$ and $Y$ is given by:

$f(x, y) = (y^2 - x^2) \cdot 8 e^{-y}, \quad 0 < y < \infty, \, -y < x < y.$

Step 2: Find the Marginal Density of $Y$

To find the marginal density $f_Y(y)$, integrate the joint density over all possible values of $X$:

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

Simplifying the expression inside the integral:

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

Next, let's compute the integral:

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

This is a standard integral that can be solved as:

[ \int_{-y}^{y} (y^2 - x^2) , dx = \left[y^2 x - \frac{x^3}{3}\right]_{-y}^{y} = \left[y^3 - \frac{y^3}{3}\right] - \left[-y^3 + \frac{(-y)^3}{3}\right]. ]

$= \left(y^3 - \frac{y^3}{3}\right) - \left(-y^3 + \frac{y^3}{3}\right) = \frac{4y^3}{3}.$

So, the marginal density of $Y$ is:

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

Step 3: Find the Conditional Density of $X \mid Y = y$

The conditional density of $X$ given $Y = y$ is given by:

$f_{X \mid Y}(x \mid y) = \frac{f(x, y)}{f_Y(y)} = \frac{(y^2 - x^2) \cdot 8 e^{-y}}{\frac{32 y^3}{3} e^{-y}}.$

Simplifying:

$f_{X \mid Y}(x \mid y) = \frac{(y^2 - x^2) \cdot 8}{\frac{32 y^3}{3}} = \frac{3 (y^2 - x^2)}{4 y^3}.$

$f_{X \mid Y}(x \mid y) = \frac{3 (y^2 - x^2)}{4 y^3}, \quad -y < x < y.$

Step 4: Calculate $E[X \mid Y = y]$

To find $E[X \mid Y = y]$, we use:

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

Simplifying the integral:

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

Notice that the integrand is an odd function of $x$ because it is $x \times (y^2 - x^2)$, and we are integrating it over a symmetric interval $[-y, y]$. Therefore, the integral of an odd function over a symmetric interval is zero:

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

Thus:

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

This confirms that:

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

Would you like any further details or have any questions about this proof?

Related Questions:

1. How do you find the variance of $X \mid Y = y$?
2. What is the significance of the expectation being zero in terms of symmetry?
3. How can we verify the marginal density of $Y$ is correctly normalized?
4. What is the joint distribution of $X$ and $Y$ if it were to represent a physical phenomenon?
5. How would the results change if the bounds on $X$ were different?

Tip: When dealing with joint density functions, always ensure that the function is normalized correctly to make valid probability statements!