Limit of chi square distribution.

Question

Question in my exercise is written as $$\lim_{n \to \infty}\bigg(\dfrac{1}{2^{\frac{n}{2}}\Gamma{\frac{n}{2}}}\int_{n+\sqrt{2n}}^{\infty}e^{\frac{-t}{2}}t^{\frac{n}{2}-1}dt\bigg)$$ equals :

$(A)=.5$

$(B)=0$

$(C)=.0228$

$(D)=.1587$

As sample size increase chi square approaches normal distribution.(I am not sure if i wrote this statement correct so please correct me and give me little explanation on that). Using this fact i calculated $P(X>n-\sqrt{2n}) = \Phi(-1) = .1587$. Did i do everything correct using this intuition ?

Answer 1

There is at least one mistake in your main equation: $x$ should be $n.$ Perhaps also, in view of @Henry's Comment and my explanation below, the lower limit of the integral was meant to be $n + \sqrt{2n}.$

As it stands, the expression of which you are taking the limit amounts to $P_n = 1- F_n(n-\sqrt{2n}),$ or $$P_n = P\left[X_n > E(X_n) - SD(X_n)\right],$$ where $X_n \sim \mathsf{Chisq(n)}.$

The probability that $Z \sim \mathsf{Norm}(0,1)$ has $$P\left[Z > E(Z) - SD(Z)\right] = P(Z > -1) = 1 - P(Z \le -1) = P(Z \le 1) = 0.8413,$$ where the last step is by symmetry. In R statistical software, where pnorm (without 2nd and 3rd arguments) is $\Phi(\cdot),$ one obtains:

1 - pnorm(-1)
## 0.8413447
pnorm(1)
## 0.8413447

Thus according to the CLT (and the continuity of the CDF), $\lim_{n \rightarrow \infty}P_n = 0.8413.$ A simple computation in R, where pchisq denotes a CDF, indicates the speed of this convergence:

n = c(1:10, 25, 30, 50, 75, 100, 500, 1000)
p = 1 - pchisq(n-sqrt(2*n), n)

cbind(n, p)             # 17 x 2 matrix brackets [] show row indexes                
         n         p
 [1,]    1 1.0000000
 [2,]    2 1.0000000
 [3,]    3 0.9076624
 [4,]    4 0.8827564
 [5,]    5 0.8711126
 [6,]    6 0.8644286
 [7,]    7 0.8601196
 [8,]    8 0.8571235
 [9,]    9 0.8549262
[10,]   10 0.8532494
[11,]   25 0.8453927
[12,]   30 0.8446442
[13,]   50 0.8432274
[14,]   75 0.8425623
[15,]  100 0.8422420
[16,]  500 0.8415137
[17,] 1000 0.8414281

Below is a graph of standardized chi-squared density curves for degrees of freedom 10, 15, 25, 50, 100, 500, and 1000 (in various rainbow colors) along with the standard normal density (heavy black). At the resolution of the graph, the purple curve for 1000 DF is hardly distinguishable from the black curve.

Note: Because none of your suggested answers seems correct, I'm wondering if the lower limit of the integral in your initial expression should have been $n + \sqrt{2n}$ (or the limits should have been $-\infty$ to $n - \sqrt{2n}).$ Notice that $1 - .1587 = .8413.$

Answer 2

Remark. The question is changed after posting this answer. The lower bound of the integral was $n-\sqrt[] {2n}$ first. The idea remains the same.

Your idea to use the Central Limit Theorem is good. First note that the sum of chi distributed random variables is again chi distributed. Let $U_i\sim \chi_1^2$ i.i.d. then $\sum_{i=1}^n U_i\sim \chi_n^2 $. What you want to find is: \begin{align} \lim_{n\to\infty} \mathbb P\left(\sum_{i=1}^n U_i>n-\sqrt[]{2n}\right) = \lim_{n\to\infty} \frac{1}{2^{n/2}\Gamma(n/2)}\int^\infty_{n-\sqrt[]{2n}}e^{-t/2}t^{\frac{n}{2}-1}\,dt \end{align} The LHS is: \begin{align} \lim_{n\to \infty}\mathbb P\left(\frac{\sum_{i=1}^n U_i-n}{\sqrt[]{2n}}>-1\right)&=\lim_{n\to \infty}\mathbb P\left(\frac{\sum_{i=1}^n U_i-n\mathbb E[U_1]}{\sqrt[]{n\operatorname{Var}(U_1)}}>-1\right)\\&\stackrel{CLT}{=} 1-\Phi(-1) \end{align} where $\Phi(\cdot)$ is the CDF of standard normal distribution. But $1-\Phi(-1)=\Phi(1)\approx 0.8413447$ which is not one of the options. Strange..

We conclude anyway that:

\begin{align} \lim_{n\to\infty} \frac{1}{2^{n/2}\Gamma(n/2)}\int^\infty_{n-\sqrt[]{2n}}e^{-t/2}t^{\frac{n}{2}-1}\,dt=\Phi(1)\approx 0.8413447 \end{align}