Let $X_1,X_2,\cdots ,X_j \cdots $ be i.i.d. $\mathcal N(0, 1)$ random variables. Show that for any $a > 0$, $$\lim_{n\to \infty}P\left( \sum_{i=1}^n X_i^2\leq a\right)=0$$ It is clear that $\sum_{i=1}^n X_i^2 \sim \chi ^2_n$; then, I can't proceed. Please help.
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Define $Y_n = \sum_{i=1} X_n^2$. As you mentioned, $Y_n$ is distributed according to chi-squared with density: $$p(y) = C(n)x^{n/2-1}e^{-x/2}$$ where $C(n)$ is some normalization constant (defined only for $y \ge 0$). The probability of $Y_n$ being larger than $a$ is (as a function of $n$): $$\int_0^a C(n) x^{n/2-1} e^{-x/2}$$ where $C(n) = 2^{n/2} \Gamma(n/2)$. This is strictly smaller than $$\int_0^a C(n) x^{n/2-1} = C(n) \frac{2}{n} a^{n/2}$$ because $e^{-z} < 1$ for $z > 0$. This means that we need to show that $$D(n) = \frac{2}{n 2^{n/2} \Gamma(n/2)} a^{n/2} \le \frac{d^n}{\Gamma(n/2)}$$ goes to 0 as $n$ goes to infinity and $d = \sqrt{a/2}$. Then, $\Gamma(n/2)$ is always larger than $\Gamma(m/2)$ where $m$ is either $n$ if $n$ is even or $n-1$ if $n$ is odd (i.e. $m$ is always an even number smaller than $n$). Therefore, $\Gamma(n/2) \le \Gamma(m/2) = (m/2-1)!$. So we get $$D(n) \le \frac{d^n}{(m/2-1)!}$$ which clearly goes to 0 at the speed of light. |
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The stated result has nothing to do with the normal distribution and is fully general. It can also be proven using only the most basic properties of cumulative distribution functions. In particular, it is unnecessary to appeal to (relatively) "high-powered" theorems like the Law of Large Numbers. Proposition: Let $X_i \sim F$ be iid with any distribution other than $\delta_0$, a point-mass at zero. Then, for each $a \geq 0$, $$ \lim_{n\to\infty} \mathbb P\Big(\sum_{i=1}^n X_i^2 \leq a\Big) = 0 \>. $$ Define $Y_i = X_i^2$ and denote the distribution of $Y_i$ by $G$. If $F$ is of unbounded support, then so is $G$. In this case, everything is completely straightforward once we have the following lemma. Lemma: If $Y_i \geq 0$ are iid, then $\mathbb P(\sum_{i=1}^n Y_i \leq a) \leq G^n(a)$.
Now, if $G$ is of unbounded support, then $G(a) < 1$ for all $a \geq 0$, but then $G^n(a) \to 0$, so invoking the lemma, we are done. Extending to the case of bounded support is not much more difficult. Suppose there exists $B > 0$ such that $G(a) = 1$ for all $a \geq B$ and $G(a) < 1$ for $a < B$. The case where $a < B$ is already handled by the argument above. For $a > B$, there is a fixed $N := N(a) = [1+(a/B)]$ such that $$ G_N(a) := \mathbb P\Big(\sum_{i=1}^N Y_i \leq a\Big) < 1 \>. $$ (Why?) But, then this reduces to the previous case since NB The intuition in the bounded case is that once we add enough terms, the support of the distribution of the sum will eventually catch up to, and overtake, $a$. Once that happens, we find ourselves in the previous case. Epilogue The specific case of the normal distribution falls under the category of $F$ (hence, $G$) with unbounded support. So, we only need the first part of the answer (which requires no calculation whatsoever) to establish the result in the question statement. |
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We note that since each $X_i$ has mean 0 and variance 1, $(\sum_i X_i^2)/n$ converges to 1 a.s. But if $$ \lim_{n\to\infty} \mathbb P\Big( \sum_i X_i^2 < a \Big) > 0 \,, $$ then with positive probability $$ \frac{1}{n}\sum_i X_i^2 $$ goes to 0, which is a contradiction because if the sum have a positive probability of staying finite then dividing by n we have $(\sum_i X_i^2)/n$ converging to 0 with positive probability contradicting convergence to 1 almost surely. |
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