Question

Let $X_{i} \sim N(0,1), i=1,2, \dots$ be independent random variables. If $T_{n}=\sum_{i=1}^{n} X_{i}^{2}$, then $\lim_{n \rightarrow \infty} P(T_{n} > n+2 \sqrt{2 n})$ is equal to

• A. $\Phi(2)$
• B. $\Phi(-2)$
• C. $1/2$
• D. $1-\Phi(-2)$

Hint:

For large $n$, $\chi^2_n$ is essentially normal. The term $n + k\sqrt{2n}$ represents the mean plus $k$ standard deviations. The probability of being above that is $1-\Phi(k)$.

Answer 1

The Correct Option is B

We analyze the distribution of $T_n$, which is a sum of squared standard normal variables.

Step 1: \color{redIdentify properties of $T_n$
$T_n = \sum X_i^2$ follows a Chi-square distribution with $n$ degrees of freedom ($\chi^2_n$).
$E(T_n) = n$.
$Var(T_n) = 2n$.

Step 2: \color{redApply Central Limit Theorem to $T_n$
As $n \to \infty$, $T_n$ is approximately $N(n, 2n)$.
Standardizing $T_n$: $Z_n = \frac{T_n - n}{\sqrt{2n}} \xrightarrow{d} N(0,1)$.

Step 3: \color{redEvaluate the target probability
$P(T_n > n + 2\sqrt{2n}) = P(T_n - n > 2\sqrt{2n})$.
Divide by $\sqrt{2n}$ on both sides:
$P\left(\frac{T_n - n}{\sqrt{2n}} > 2\right)$.

Step 4: \color{redFind the limit
The limit is $P(Z > 2) = 1 - \Phi(2)$.
Using symmetry: $1 - \Phi(2) = \Phi(-2)$.
Thus, the value is $\Phi(-2)$.