Let X and Y be independent random variables each with density f(x) = 1 /2theta &for- theta<x< theta\\ 0&otherwise

If the Var(XY) = 64 , then what is the value of theta ?

Let's solve for the value of \(\theta\) given that \(X\) and \(Y\) are independent random variables with a given density function, and the variance of their product, \(XY\), is 64.

### Step-by-Step Solution:

1. **Understand the Density Function**:

   The density function \(f(x)\) for both \(X\) and \(Y\) is given by:
   \[
   f(x) = 
   \begin{cases} 
   \frac{1}{2\theta} & \text{for } -\theta < x < \theta, \\
   0 & \text{otherwise}.
   \end{cases}
   \]

   This is the probability density function (PDF) of a uniform distribution on the interval \((- \theta, \theta)\). Therefore, \(X\) and \(Y\) are uniformly distributed on this interval.

2. **Find the Variance of \(X\) and \(Y\)**:

   For a uniform random variable \(U \sim \text{Uniform}(-a, a)\), the variance is given by:
   \[
   \text{Var}(U) = \frac{(b-a)^2}{12}.
   \]

   Here, \(a = -\theta\) and \(b = \theta\), so:
   \[
   \text{Var}(X) = \text{Var}(Y) = \frac{(\theta - (-\theta))^2}{12} = \frac{(2\theta)^2}{12} = \frac{4\theta^2}{12} = \frac{\theta^2}{3}.
   \]

3. **Calculate the Variance of \(XY\)**:

   Since \(X\) and \(Y\) are independent, the variance of their product \(XY\) is given by:
   \[
   \text{Var}(XY) = \text{E}[X^2]\text{E}[Y^2] - (\text{E}[XY])^2.
   \]

   Since \(X\) and \(Y\) are independent, \(\text{E}[XY] = \text{E}[X]\text{E}[Y]\).

   The mean of a uniform random variable on \((- \theta, \theta)\) is 0, so:
   \[
   \text{E}[X] = \text{E}[Y] = 0 \implies \text{E}[XY] = 0.
   \]

   Thus, the variance simplifies to:
   \[
   \text{Var}(XY) = \text{E}[X^2] \text{E}[Y^2].
   \]

   The expectation \(\text{E}[X^2]\) for \(X \sim \text{Uniform}(-\theta, \theta)\) is:
   \[
   \text{E}[X^2] = \frac{\theta^2}{3}.
   \]

   Similarly, \(\text{E}[Y^2] = \frac{\theta^2}{3}\).

   Therefore:
   \[
   \text{Var}(XY) = \left( \frac{\theta^2}{3} \right) \left( \frac{\theta^2}{3} \right) = \frac{\theta^4}{9}.
   \]

4. **Set Up the Equation for the Given Variance**:

   Given that \(\text{Var}(XY) = 64\), we have:
   \[
   \frac{\theta^4}{9} = 64.
   \]

5. **Solve for \(\theta\)**:

   Multiply both sides by 9:
   \[
   \theta^4 = 576.
   \]

   Take the fourth root of both sides:
   \[
   \theta = \sqrt[4]{576}.
   \]

   Simplify the expression:
   \[
   \theta = \sqrt{24}.
   \]

   Since \(\theta > 0\), we take the positive root:
   \[
   \theta = 2\sqrt{6}.
   \]

### Final Answer:

\[
\theta = 2\sqrt{6}.
\]

Would you like any further details or have any questions?

### Relative Questions:

1. What is the expectation of the product of two independent random variables?
2. How do you calculate the variance of a random variable uniformly distributed over an interval?
3. How does independence between two random variables affect their joint moments?
4. What are the properties of a uniform distribution?
5. How does the variance of the product of two random variables differ if they are not independent?

**Tip**: The variance of a product of two independent random variables depends on both their variances and their expected values. If the expected values are zero, the variance of the product simplifies significantly.

Learn how to determine the value of theta from the variance of independent random variables X and Y with a uniform distribution. Explore the detailed step-by-step solution and formula application for theta calculation.

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