Question

The random variable \(Y \sim U(0, X)\), where the marginal density of \(X\) isv

\( f_X(x) = \begin{cases} 2x & \text{for } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases} \)

Then \(E(Y)\) is:

• A. $2/3$
• B. $4/3$
• C. $1/2$
• D. $1/3$

Hint:

The Law of Iterated Expectations ($E(Y) = E[E(Y|X)]$) is much faster than finding the joint density $f(x,y)$ and performing a double integral. Always look for hierarchical structures in probability problems.

Answer 1

The Correct Option is D

We can find the expected value of $Y$ using the Law of Iterated Expectations, which states $E(Y) = E[E(Y|X)]$.

Step 1: \color{redFind the Conditional Expectation E(Y|X)
We are given that $Y|X$ follows a Uniform distribution $U(0, X)$
For a uniform distribution $U(a, b)$, the mean is $(a+b)/2$.
Thus, $E(Y|X) = \frac{0 + X}{2} = \frac{X}{2}$.

Step 2: \color{redApply the Law of Iterated Expectations
$E(Y) = E[E(Y|X)] = E[\frac{X}{2}] = \frac{1}{2} E(X)$.

Step 3: \color{redCalculate E(X)
Using the provided marginal density $f_X(x) = 2x$ for $0 < x < 1$
$E(X) = \int_{0}^{1} x \cdot f_X(x) dx$
$E(X) = \int_{0}^{1} x \cdot (2x) dx = \int_{0}^{1} 2x^2 dx$
$E(X) = [ \frac{2x^3}{3} ]_{0}^{1} = \frac{2}{3}$.

Step 4: \color{redFind the Final Result for E(Y)
$E(Y) = \frac{1}{2} E(X) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$.
Thus, $E(Y) = 1/3$.