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Letting $N$ be a standard normal random variable and $X_n$ such that for $x\geq 0$: $$P(X_n \geq n+x) = P(N \geq n+x \vert N \geq n)$$. I'm asked to find deterministic $a_n$ such that $Y_n = a_n(X_n-n)$ converges in distribution to some finite positive random variable $Y$.

First of all we can have the distribution of $X_n - n$: Notice from the formula given, $X_n - n \geq 0$ a.s., and for $x \geq 0 $, $\displaystyle{P(X_n-n \geq x) = \frac{P(N \geq n+x)}{P(N \geq n)}}$. By Gaussian variable has no atoms, $X_n - n$ has no atoms and thus $$P(X_n-n \leq x) = 1-P(X_n-n \geq x) = \frac{P(n \leq N \leq n+x)}{P(N \geq N)}$$

I'm thinking of trying to make the CDF $F_n(y) = P(Y_n \leq y)$ converges to some other CDF. Because we want the distributional limit to be positive, then $a_n >0$. Then I found $$ P(Y_n \leq y) = P(a_n(X_n-n)\leq y) = P(X_n-x \leq \frac{y}{a_n}) = \frac{\int_{n}^{n+\frac{y}{a_n}}e^{-\frac{x^2}{2}}dx}{\int_{n}^{\infty}e^{-\frac{x^2}{2}}dx}$$, and to make it more convenient to guess such $a_n$, I further used weighted mean value theorem to get:$$F_n(y) = P(Y_n \leq y) = \frac{e^{\frac{-\epsilon_n^2}{2}}\frac{y}{a_n}}{\int_{n}^{\infty}e^{-\frac{x^2}{2}}dx} $$ where $\epsilon_n \in (n,n+\frac{y}{a_n})$. It seems that we should choose $a_n$ such that it cancels the denominator and yet should have a limit. But I'm stuck here. Any help or hint would be appreciated.

Stack_Underflow
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1 Answers1

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Fix some $t\in \mathbb R$. If $t\leq 0$, then as you noticed $P(a_n (X_n-n)<t) = 0$. Assume next that $t>0$ and note that $$P(a_n (X_n-n)<t) = 1-\frac{1-\Phi(n+\frac t{a_n})}{1-\Phi(n)},$$ where $\Phi$ denotes the cdf of the standard normal. By a classical result, letting $\phi$ denote the pdf of the standard normal, the following estimate holds as $x\to \infty$: $$\frac{1-\Phi(x)}{\phi(x)} = \frac 1x - \frac 1{x^3} + O\Big(\frac 1{x^5}\Big), $$ hence $1-\Phi(x) = \frac{\phi(x)}{x} + o(\frac{\phi(x)}{x})$, which rewrites $1-\Phi(x)\sim \frac{\phi(x)}{x}$.

Therefore $$\frac{1-\Phi(n+\frac t{a_n})}{1-\Phi(n)}\sim_{n\to\infty}\frac{n}{n+\frac t{a_n}} \exp\Big(-\frac 12 t(2\frac n{a_n} + \frac t{a_n^2})\Big).$$

This suggests letting $a_n=n$, which yields $$\lim_n \frac{1-\Phi(n+\frac t{n})}{1-\Phi(n)} = \exp(-t).$$

Thus for every $t\in \mathbb R$, $$P(n (X_n-n)<t)\to (1-\exp(-t))1_{t\geq 0}.$$

Let $Y$ denote a random variable with exponential distribution of parameter $1$. We have established that $P(n (X_n-n)<t)\to F_Y(t)$. This is enough to conclude that $n (X_n-n)$ converges in distribution to $Y$.

Note the technicality that we dealt with $P(n (X_n-n)<t)$ instead of the usual $P(n (X_n-n)\leq t)$. This makes no difference for convergence since $P(Z_n<t) = 1-P(-Z_n\leq -t)$, thus we have convergence in distribution of $-Z_n$ to $-Y$, and hence of $Z_n$ to $Y$ by the continuous mapping theorem.

Gabriel Romon
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