Question

Let \( X_1, X_2, \dots, X_{10} \) be i.i.d. \( N(0, 1) \) random variables. If \( T = X_1^2 + X_2^2 + \dots + X_{10}^2 \), then \( E\left( \frac{1}{T} \right) \) equals ..................

Hint:

The expectation of the reciprocal of a chi-squared random variable is \( \frac{1}{k-2} \), where \( k \) is the degrees of freedom. Ensure that \( k>2 \) for this formula to hold.

Answer 1

Correct Answer: 0.12

Step 1: Identify the distribution of \( T \).
Since \( X_1, X_2, \dots, X_{10} \) are i.i.d. standard normal random variables, each \( X_i^2 \) follows a chi-squared distribution with 1 degree of freedom. Therefore, \( T \) is the sum of 10 independent chi-squared random variables, which follows a chi-squared distribution with 10 degrees of freedom: \[ T \sim \chi^2(10). \]
Step 2: Use the expectation formula for a chi-squared random variable.
For a chi-squared random variable \( \chi^2_k \), the expectation of \( \frac{1}{T} \) is given by: \[ E\left( \frac{1}{T} \right) = \frac{1}{k-2} \quad \text{for} \quad k>2. \] In our case, \( T \sim \chi^2(10) \), so we have: \[ E\left( \frac{1}{T} \right) = \frac{1}{10 - 2} = \frac{1}{8}. \] Final Answer: \[ \boxed{\frac{1}{8}}. \]