Question

If a random variable \( X \) has the probability density function \[ f(x) = \begin{cases} \frac{5}{32}x^4 & \text{if } 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases} \] and if \( Y = X^2 \), then the expected value of \( Y \) is .................... (round off to one decimal place)

Hint:

To find the expectation of a function of a random variable, \(E[g(X)]\), you don't need to find the PDF of the new variable Y. You can directly use the formula \(E[g(X)] = \int g(x)f(x)dx\), which is usually much simpler.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the expected value of a function of a random variable, \( E[Y] = E[X^2] \). The expected value of any function \(g(X)\) of a continuous random variable \(X\) is found by integrating the product of \(g(X)\) and the probability density function \(f(x)\) over the entire range of \(X\).
Step 2: Key Formula or Approach:
The expected value of \(Y = g(X)\) is given by the formula: \[ E[Y] = E[g(X)] = \int_{-\infty}^{\infty} g(x) f(x) \,dx \] In this problem, \(g(X) = X^2\), and the PDF \(f(x)\) is non-zero only for \(0 \le x \le 2\). So the formula becomes: \[ E[Y] = E[X^2] = \int_{0}^{2} x^2 . \left(\frac{5}{32}x^4\right) \,dx \] Step 3: Detailed Calculation:
Set up the integral: \[ E[Y] = \int_{0}^{2} \frac{5}{32}x^6 \,dx \] Pull the constant out of the integral: \[ E[Y] = \frac{5}{32} \int_{0}^{2} x^6 \,dx \] Perform the integration: \[ E[Y] = \frac{5}{32} \left[ \frac{x^7}{7} \right]_{0}^{2} \] Evaluate the integral at the limits: \[ E[Y] = \frac{5}{32} \left( \frac{2^7}{7} - \frac{0^7}{7} \right) = \frac{5}{32} \left( \frac{128}{7} \right) \] Simplify the expression: \[ E[Y] = \frac{5 \times 128}{32 \times 7} \] Since \( 128 = 4 \times 32 \), we can cancel the 32s: \[ E[Y] = \frac{5 \times 4}{7} = \frac{20}{7} \] Convert to a decimal: \[ E[Y] \approx 2.85714 \] Rounding off to one decimal place: \[ E[Y] = 2.9 \] Step 4: Final Answer:
The expected value of Y is 2.9.