Let's evaluate the given expression for the conditional expectation:

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

We will break this down by evaluating the two integrals separately.

Step 1: Evaluating the first integral

The first integral is:

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

Since $x$ is an odd function (i.e., $f(-x) = -f(x)$), and we are integrating over the symmetric interval $[-y, y]$, this integral evaluates to 0:

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

Step 2: Evaluating the second integral

The second integral is:

$\int_{-y}^{y} x^3 \, dx.$

Since $x^3$ is also an odd function, this integral also evaluates to 0 over the symmetric interval $[-y, y]$:

$\int_{-y}^{y} x^3 \, dx = 0.$

Step 3: Substituting the results back

Substitute the evaluated integrals into the original expression:

[ E[X | Y = y] = \frac{3}{4y^3} \left[ y^2 \cdot 0 - 0 \right] = 0. ]

Thus, the conditional expectation is:

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

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Would you like further clarification on any part of this solution? Here are five related questions to deepen your understanding:

1. What is the significance of odd functions when integrating over symmetric intervals?
2. How would this result change if the integrals involved even functions instead?
3. Could you apply the same reasoning to higher-order odd powers, such as $x^5$?
4. How does this conditional expectation relate to the independence of $X$ and $Y$?
5. Can you explain how to compute conditional expectations in general probability distributions?

Tip: When dealing with integrals over symmetric intervals, checking whether the function is odd or even can help simplify your calculations!