Compute $P(M_n<c)$ with $M_n=\max\{X_1,...,X_n\}$ and $X_i's$ are i.i.d

Question

$X_1,X_2...$ are i.i.d real-valued random variable defined on a probability space $(\Omega,\mathbb{F},P)$ and $M_n:=\max\{X_1,...,X_n\}$. $X_1$ have the distribution function $F$ with $F(x)<1$ for every $x$

I want to explore whether $P(M_n\rightarrow \infty)=1$ (my intuition says "yes it is").

I figured if I can compute $P(M_n<c)$ for every $c>0$ then I might be able to show the desired results by looking at the limits.

I need help to compute $P(M_n<c)$

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EDIT: By help froms David's comment I can show that $$P(M_n<c)= P(\max(X_1,...X_n)<c)= P(X_1<c,...,X_n<c)\\ =P(X_1<c)*...*P(X_n<c) \text{ (because they are independant)}\\ =F(c)^n\rightarrow0 \text{ as } n \rightarrow\infty.\\ P(M_n<c)\rightarrow 0$$Right?

My conclusion:

Now since $P(M_n<c)\rightarrow 0$, $M_n$ is exploding for larger $n$ and hence the $P(M_n\rightarrow \infty)=1$.

Is my conclusion sufficient or do I need to tell the same story with "math"?

Answer 1

Your analysis is mixed up. As $c\to \infty$, $P(X_k\le c)=F(c)\to 1$. Therefore for fixed $n$, $P(M_n\le c)=F^n(c)\to 1$.

What happens as $n\to \infty$ may be a little tricky, but that is not part of your original question.