If $X$ is standard Normal then find $\lim_{x\to0}P(X>x+\frac{a}{x}|X>x)$

Question

If $X$ is Standard Normal and $a>0$ is a constant then find $\displaystyle\lim_{x\to0}\mathbb{P}\left(X>x+\dfrac{a}{x}\Big|X>x\right)$.

This is an exercise from a book whose name I cannot immediately recall. I compiled problems from different books and am stuck at one of these. The answer to this is $\mathrm{e}^{-a}$.

However, I have some doubts regarding the validity of this answer. If $x\to0$, consider the negative subsequence of this sequence, and then $$\mathbb{P}\left(X>x+\dfrac{a}x\Big|X>x\right)=\dfrac{\mathbb{P}\left(X>x+\dfrac{a}x,X>x\right)}{\mathbb{P}(X>x)}=\dfrac{\mathbb{P}(X>x)}{\mathbb{P}(X>x)}=1,$$ so this subsequence always goes to $1$, hence the limit can't be $\mathrm{e}^{-a}$ which renders the question meaningless.

However, supposing I restrict the question to finding the limit as $x\to0^+$, then I get $$\dfrac{\mathbb{P}\left(X>x+\dfrac{a}x,X>x\right)}{\mathbb{P}(X>x)}=\dfrac{\mathbb{P}\left(X>x+\dfrac{a}x\right)}{\mathbb{P}(X>x)}$$.

Now as $x\to0^+$ the limit of this quantity is $\dfrac{\mathbb{P}\left[X>\displaystyle\lim_{x\to0}\left(x+\dfrac{a}x\right)\right]}{\mathbb{P}\left(X>\displaystyle\lim_{x\to0}x\right)}=0$ by continuity property of probability.

Hence I am not getting $\mathrm{e}^{-a}$. Whenever $x\to0$, the positive subsequence yields answer as $0$ and the negative subsequence yields answer as $1$.

Any help will be appreciated.

Answer 1

I agree with your analysis: the limit does not exist, since the right- and left-sided limits differ.

I suspect that either the original problem was erroneous, or perhaps you made a mistake when copying it.

Answer 2

I agree with you and Nate Eldredge that the limit doesn't exist. However, if we replace $0$ with $+\infty$, then the limit exists and equals $e^{-a}$.

Let $F$ denote the cdf of a standard normal, and let $\;f=F'$ be the pdf. Then $$\lim_{x \to \infty} \frac{P(X>x+\frac{a}{x})}{P(X>x)} = \lim_{x \to \infty} \frac{1-F(x+\frac{a}{x})}{1-F(x)}$$Now as $x \to \infty$, both numerator and denominator converge to $0$, so we can use L'hopital's rule to say that $$\lim_{x \to \infty} \frac{1-F(x+\frac{a}{x})}{1-F(x)} = \lim_{x \to \infty} \frac{-(1-\frac{a}{x^2})f(x+\frac{a}{x})}{-f(x)} = \lim_{x \to \infty} \frac{-(2\pi)^{-\frac{1}{2}}(1-\frac{a}{x^2})\;e^{-\frac{1}{2}(x+\frac{a}{x})^2}}{-(2\pi)^{-\frac{1}{2}}\;e^{-\frac{x^2}{2}}}=\lim_{x \to \infty} \bigg(1-\frac{a}{x^2}\bigg)e^{-\big(a+\frac{a^2}{2x^2}\big)}=(1-0)e^{-(a+0)}=e^{-a}$$ where in the second-to-last equality we used the fact that $\frac{1}{x^2} \to 0$ as $x \to \infty$