Question

For n = 1, 2, 3, …, let the joint moment generating function of (X, Y_n) be
\(M_{X,Y_n}(t_1,t_2)=e^{\frac{t^2_1}{2}(1-2t_2)^{-\frac{n}{2}}}, t_1 \in \R,t_2 \lt \frac{1}{2}.\)
If \(T_n=\frac{\sqrt{n}X}{\sqrt{Y_n}},n \ge1,\) then which one of the following statements is true ?

• A. The minimum value of n for which Var(T_n) is finite is 2
• B. \(E(T^3_{10})=10\)
• C. \(Var(X+Y_4=7)\)
• D. \(\lim\limits_{n \rightarrow \infin}P(|T_n|>3)=1-\frac{\sqrt2}{\sqrt{\pi}}\int^3_0e^{-\frac{t^2}{2}}dt\)

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the given moment generating function (MGF) of the joint distribution of \((X, Y_n)\) and determine the distribution and behavior of \(T_n = \frac{\sqrt{n}X}{\sqrt{Y_n}}\).

Step 1: Understand the Moment Generating Function

The given MGF is:

\(M_{X,Y_n}(t_1,t_2)=e^{\frac{t_1^2}{2}(1-2t_2)^{-\frac{n}{2}}},\) where \( t_1 \in \mathbb{R}, t_2 < \frac{1}{2} \).

This MGF suggests that \(X\) is a standard normal random variable since for a standard normal distribution, the MGF is given by \( e^{\frac{t^2}{2}} \).

The term \((1-2t_2)^{-\frac{n}{2}}\) indicates a scaling transformation affecting the variance of \(Y_n\).

Step 2: Distribution of \( T_n \)

Considering \( T_n = \frac{\sqrt{n}X}{\sqrt{Y_n}} \), and noting the asymptotic behavior as \( n \to \infty \), we'll analyze the limit behavior.

As \( n \to \infty \), the Law of Large Numbers suggests that \( Y_n/n \to 1 \) by the Central Limit Theorem, assuming \( Y_n \) represents a sum of random variables (normal distribution consideration from the MGF's form).

The form \( \frac{\sqrt{n}X}{\sqrt{Y_n}} \) simplifies under the law of large numbers to a standard normal distribution asymptotically.

Step 3: Analyzing the Limit

We need to find the behavior of \( \lim\limits_{n \rightarrow \infty} P(|T_n| > 3) \). As established, \( T_n \) tends to a standard normal distribution for large \( n \).

The probability \( P(|T_n| > 3) \) for a standard normal distribution \( N(0, 1) \) is:

P(|Z| > 3) = 2P(Z > 3) = 2\(\left(1 - P(Z \leq 3)\right)\)

For a standard normal variable, this equals:

\(1 - \frac{\sqrt{2}}{\sqrt{\pi}} \int^3_0 e^{-\frac{t^2}{2}} \, dt\), following the computation of a Gaussian tail probability.

Conclusion:

The correct statement is:

\(\lim\limits_{n \rightarrow \infty} P(|T_n| > 3) = 1 - \frac{\sqrt{2}}{\sqrt{\pi}} \int^3_0 e^{-\frac{t^2}{2}} \, dt \)

This result aligns with the properties of the normal distribution and the asymptotic behavior of \( T_n \) as \( n \) increases.