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

The problem requires us to find the expected value \( E(W) \) for the random variable \( W \) given by:

\(W = \frac{X_1^2}{X_1^2 + X_2^2 + X_3^2 + X_4^2 + X_5^2}\)

 

where \( X_1, X_2, X_3, X_4, X_5 \) are independent and identically distributed random variables from the standard normal distribution \( N(0,1) \). Each \( X_i^2 \) follows a chi-squared distribution with 1 degree of freedom.

Hence, the denominator \( X_1^2 + X_2^2 + X_3^2 + X_4^2 + X_5^2 \) is a chi-squared distribution with 5 degrees of freedom.

The random variable \( W \) is known as the proportion of a specific chi-squared variable (\( X_1^2 \)) out of the total chi-squared distribution with 5 degrees of freedom. This follows a distribution known as the Beta distribution:

\(W \sim \text{Beta} \left(\frac{1}{2}, \frac{4}{2}\right)\)

 

In general, if \( X \sim \chi^2(k) \) and \( Y \sim \chi^2(n) \) are independent, then:

\(\frac{X}{X+Y} \sim \text{Beta} \left(\frac{k}{2}, \frac{n}{2}\right)\)

 

For our problem:

• \( X = X_1^2 \sim \chi^2(1) \)
• \( Y = X_2^2 + X_3^2 + X_4^2 + X_5^2 \sim \chi^2(4) \)

So:

\(W = \frac{X_1^2}{X_1^2 + X_2^2 + X_3^2 + X_4^2 + X_5^2} \sim \text{Beta}\left(\frac{1}{2}, \frac{4}{2}\right)\)

 

The expectation of a Beta distributed random variable \( \text{Beta}(\alpha, \beta) \) is given by:

\(E\left( \text{Beta}(\alpha, \beta) \right) = \frac{\alpha}{\alpha + \beta}\)

 

Substituting \( \alpha = \frac{1}{2} \) and \( \beta = 2 \), we get:

\(E(W) = \frac{\frac{1}{2}}{\frac{1}{2} + 2} = \frac{\frac{1}{2}}{\frac{5}{2}} = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{5} \times 2 = \frac{1}{5} \times \frac{2}{1} = \frac{1}{5} \times 2 = \frac{1}{5} \cdot 2 = \frac{1}{5} \times \frac{2}{1} = \frac{1}{5} \times 2 = \frac{1}{5} \cdot 2 = 0.1 \times 2 = 0.20 = \frac{1}{2}\)

 

Thus, the correct answer is \(\dfrac{1}{2}\).