# Maximum of exponential random variables is bigger than maximum of normal random variables almost surely as $n$ tends to infinity.

Let $X_1,X_2,\cdots,X_n$ be iid Gaussian random variables with mean $0$ and variance $1$. Let $Y_1,Y_2,\cdots,Y_n$ be iid exponential random variables with mean $1$. Prove that $$\lim_{n\to\infty}P\left(\max{(Y_1,Y_2,\cdots,Y_n)}\ge \max{(X_1,X_2,\cdots,X_n)}\right)=1$$

I let $W_n=\max{(X_1,X_2,\cdots,X_n)}$ and $Z_n=\max{(Y_1,Y_2,\cdots,Y_n)}$.As $X_i$'s and $Y_i$'s are iid, it is easy to get the distribution of $W$ and $Z$ as $F_W(w)=\left(F_X(w)\right)^n$ and $F_Z(z)=\left(F_Y(z)\right)^n$. I computed $$P(Z_n\ge W_n)=\int_{-\infty}^{\infty}F_W(w)f_Z(w)dw=\int_0^{\infty}\left(\Phi(w)\right)^nn(1-e^{-w})^{n-1}e^{-w}dw$$ where $\Phi$ is the CDF of standard normal distribution. I am not sure how to estimate the resulting integral as $n\to\infty$, although I could verify that it converges to 1 by using numerical computation.

Also, I believe that there should be a more clever and elegant way to deal with this problem, but I don't know how. Thanks for any hint in advance.

The question does not explain if $(X_n)$ should be assumed independent of $(Y_n)$, and in fact this hypothesis is not necessary for the desired result to hold.

To see why, note that for every $x$, $$P(\max(Y_1,\ldots,Y_n)<\max(X_1,\ldots,X_n))\leqslant P(\max(X_1,\ldots,X_n)>x)+P(\max(Y_1,\ldots,Y_n)<x)$$ with $$P(\max(X_1,\ldots,X_n)>x)=1-(1-P(X_1>x))^n$$ and $$P(\max(Y_1,\ldots,Y_n)<x)=P(Y_1<x)^n$$ Thus, if one can choose some sequence $(x_n)$ such that the convergences $$(1-P(X_1>x_n))^n\to1\qquad\&\qquad P(Y_1<x_n)^n\to0$$ hold simultaneously, we are done. To do so, recall that, for every $x>1$, $$P(X_1>x)\leqslant e^{-x^2/2}\qquad\&\qquad P(Y_1>x)=e^{-x}$$ thus, $$(1-P(X_1>x_n))^n\geqslant(1-e^{-x_n^2/2})^n\qquad\&\qquad P(Y_1<x_n)^n=(1-e^{-x_n})^n$$ hence the two convergences we need are simultaneously realized as soon as $x_n\to\infty$ with $$ne^{-x_n^2/2}\to0\qquad\&\qquad ne^{-x_n}\to\infty$$ that is, $$e^{x_n}\ll n\ll e^{x_n^2/2}$$ or, equivalently, $$x_n^2-2\log n\to\infty\qquad\&\qquad x_n-\log n\to-\infty$$ which happens, say, if $$x_n\sim\tfrac12\log n$$ Thus, as desired, $$P(\max(Y_1,\ldots,Y_n)<\max(X_1,\ldots,X_n))\to0$$

• I am still a little bit confused about the very first inequality. If we have $\omega\in\{max(Y_1\cdots Y_n)<max(X_1\cdots X_n)\}$, does this implies that $\omega\in\{max(Y_1\cdots Y_n)<x\}$ or $\{max(X_1\cdots X_n)>x\}$ for any fixed $x$? Nov 22, 2016 at 2:39
• Also, why $P(X_1\ge x)\le e^{-x^2/2}$ is true? Thanks. Nov 22, 2016 at 2:50
• I am not sure of the amount of time you spent trying to solve the two questions you raise in your comments, it was probably not much... but anyway, yes, for every random variables $U$, $V$ and every real number $w$, $$[V<x]\cap[x<U]\subseteq[V<U]$$ hence $$[U<V]\subseteq[V>x]\cup[U<x]$$ Second, regarding the inequality $$P(X_1>x)\leqslant e^{-x^2/2}$$ yes it holds, and what did you try to check it?
– Did
Nov 22, 2016 at 8:16
• *every real number $x$...
– Did
Nov 22, 2016 at 12:37
• I am sorry that I didn't think too much about those two questions I raised in comments yesterday. Now I figure it out. For the second one, let $g(x)=1-F(x)-e^{-x^2/2}$ we can find its upper bound for $x\ge0$. Thanks so much for help. Nov 22, 2016 at 16:26