# Convergence of running maximum of uniform random variables [closed]

Let $X_1, X_2, ... X_n$ be an IID sequence of IID random variables that have a uniform distribution $(0,1)$.

Let Max$(n) =$ max$(X_k:1\le k \le n)$, where $n\in \mathbb N$.

How do I show that Max$(n)$ converges to $1$ almost surely?

## closed as off-topic by Did, Adam Hughes, William, Kirill, CookieJul 22 '14 at 7:29

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Trivially we observe that the maximum of the list can only increase as we add more variables to it (for any given instance of the list). $$\newcommand{\Max}{\operatorname{Max}} \forall n, 0\leq \Max(n)\leq 1, \Max(n+1)\geq \Max(n)$$

However, as each $X_i$ is a random variable, so to is $\Max(n)$. So we need to examine the behaviour of the probability distribution of this.

\begin{align}\Pr(\Max(n) > m) & = 1-\Pr(\bigcap\limits_{i=1}^n (X_i\leq m)) & \forall m \in [0, 1) \\ ~ & = 1-\prod_{i=1}^n \Pr(X_i\leq m) & \text{by independence} \\ ~ & = 1- m^n & \text{since } \forall i\in\{1..n\}, X_i\sim\mathcal{U}(0,1) \\ ~ & ~ \\ \therefore \lim_{n\to\infty}\Pr(Max(n)\geq m) & = 1 & \forall m\in [0, 1) \end{align}

Thus by squeezing, the probability that the maximum is 1 approaches certainty as $n$ positively approaches infinitude.

Let $Y_n = \max \{ X_1,\dots,X_n \}$. Then $Y_{n+1} = \max \{ Y_n,X_{n+1} \}$.

Consider $E[Y_{n+1} | Y_n]$. Note that $Y_{n+1} - Y_n \geq 0$, so

$$0 \leq E[Y_{n+1} - Y_n | Y_n] = E[Y_{n+1} | Y_n] - Y_n$$

So $Y_n$ is a submartingale. It is also nonnegative, so it converges a.s. by Doob's martingale convergence theorem. This does not state what the limit actually is.

To prove that the limit is $1$, suppose it were $y<1$. Then for all but finitely many $n$ we would have $X_n \in \left [0,\frac{y+1}{2} \right ]$. But the probability that any particular $k$ of the $X_n$ are there is $\left ( \frac{y+1}{2} \right )^k$, which tends to zero as $k \to \infty$. Thus the probability of convergence to something other than $1$ is zero, and the probability of not converging at all is zero, so the probability of converging to $1$ is $1$.