Let X be N~(0,1). Show that: $$ \mathcal{P} (X > t + \frac{a}{t} | X>t) \underbrace{\longmapsto}_{t \longmapsto \infty} e^{-a} $$
I used firstly the memoryless property and then the markov inequality: $$ \mathcal{P} (X > t + \frac{a}{t} | X>t) = P(X > \frac{a}{t}) $$ by Markov and applying the exponential function since increasing and continuos: $$ \mathcal{P}(e^X > e^{\frac{a}{t}}) \leq \frac{E(e^x)}{e^{\frac{a}{t}}} = \frac{e^{\frac{1}{2}}}{e^{\frac{a}{t}}} $$
Thus, it's not the result that I needed when t goes to infinity. Can anyone spot my mistake?
Exactly, it suffices to verify the assertion by definition. Let $\Phi(x)$ be the c.d.f. of standard normal distribution. $$ \begin{align} P(X > t + \frac at | X > t) &= \frac{P(X > t + \frac at, X > t)}{P(X > t)} \\ &=\frac{P(X > t + \frac at)}{P(X > t)} \\ &= \frac{1 - \Phi(t + a/t) }{1 - \Phi(t)} \quad\text{ apply L'Hôpital's rule}\\ &\to \frac{-\Phi^\prime(t+a/t)(1 -a/t^2)}{-\Phi^\prime(t)} \\ &\to \frac{e^{-\frac12(t+a/t)^2}}{e^{-\frac12t^2}} \\ &\to e^{-a} \end{align} $$