Question

Let 𝑋₁, 𝑋₂, 𝑋₃ be 𝑖. 𝑖. 𝑑. random variables, each having the 𝑁(0, 1) distribution. Then, which of the following statements is/are TRUE?

• A. \(\frac{\sqrt2(X_1-X_2)}{\sqrt{(X_1+X_2)^2+2X^2_3}} ∼𝑡_1\)
• B. \(\frac{(X_1-X_2)^2}{\sqrt{(X_1+X_2)^2+2X^2_3}} ∼F_{1,2}\)
• C. \(E(\frac{X_1}{X^2_2+X^2_3})=0\)
• D. 𝑃(𝑋₁ < 𝑋₂ + 𝑋₃ ) =\(\frac{1}{3}\)

Answer 1

The Correct Option is C

The question involves independent and identically distributed (i.i.d.) random variables following a normal distribution, \(X_1, X_2, X_3 \sim \mathcal{N}(0, 1)\). Let's evaluate each given option:

1. \(\frac{\sqrt2(X_1-X_2)}{\sqrt{(X_1+X_2)^2+2X^2_3}} ∼ t_1\):
   In general, expressions involving a linear combination of normal variables divided by a function of their sum do not simply follow a \(t\)-distribution unless there is a specific form resembling a \(t\)-distribution. Here, there is no direct transformation that indicates a \(t_1\)-distribution. Therefore, this statement is likely false.
2. \(\frac{(X_1-X_2)^2}{\sqrt{(X_1+X_2)^2+2X^2_3}} ∼ F_{1,2}\):
   For an expression to follow an \(F\)-distribution, typically, the numerator would be a Chi-square distribution divided by its degrees of freedom, and the denominator should also have a similar form. Here, neither the numerator nor the denominator takes the classical form required for an \(F_{1,2}\) distribution. Thus, this statement is likely false.
3. \(E\left( \frac{X_1}{X^2_2+X^2_3} \right) = 0\):
   Explanation: The expression involves calculating the expectation of a ratio. Given that \(X_1\), \(X_2\), and \(X_3\) are symmetric random variables with mean 0, the expectation considers symmetry:
   • Since \(X_1\) is normally distributed with mean 0, and \((X_2^2 + X_3^2)\) is always positive, any odd function (like the one in the expression) around 0 results in 0.
4. \(P(X_1 < X_2 + X_3) = \frac{1}{3}\):
   This statement involves calculating the probability over a linear combination of i.i.d. normal variables. The distribution of \(X_2 + X_3\) is also normal, and determining the exact probability without extensive computation is non-trivial. Generally, it does not yield simple fractions like \(\frac{1}{3}\). Therefore, this is likely false.

Based on the evaluations above, the only true statement is: 

• \(E\left( \frac{X_1}{X^2_2+X^2_3} \right) = 0\)