Question

Let $X_1, X_2, X_3, X_4, X_5$ be a random sample from $N(0,1)$ distribution and let \[ W = \frac{X_1^2}{X_1^2 + X_2^2 + X_3^2 + X_4^2 + X_5^2}. \] Then $E(W)$ equals

• A. $\dfrac{1}{2}$
• B. $\dfrac{1}{3}$
• C. $\dfrac{1}{4}$
• D. $\dfrac{1}{5}$

Hint:

By symmetry, in a ratio of identical $\chi^2$ components, each has expected contribution equal to $\frac{1}{n}$.

Answer 1

The Correct Option is A

Step 1: Recognize the distribution.
Each $X_i^2 \sim \chi^2(1)$, so the denominator $\sum X_i^2 \sim \chi^2(5)$.

Step 2: Use symmetry of components.
Since all $X_i$ are i.i.d., \[ E\left(\frac{X_1^2}{\sum X_i^2}\right) = E\left(\frac{X_2^2}{\sum X_i^2}\right) = \cdots = E\left(\frac{X_5^2}{\sum X_i^2}\right). \] Adding all, \[ E\left(\frac{\sum X_i^2}{\sum X_i^2}\right) = 1 $\Rightarrow$ 5E(W) = 1 $\Rightarrow$ E(W) = \frac{1}{5}. \]

Step 3: Conclusion.
\[ \boxed{E(W) = \frac{1}{5}}. \]