Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$.

Question

Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$. Is this solution enough?... I cannot understand the solution provided in the below mentioned link: https://www.stat.cmu.edu/~larry/=stat705/Lecture4.pdf

Answer 1

Yes, it is enough. And as @noam.szyfer suggested, this does need independence.

I think you are okay with the first equality, since $X_i$-s cannot exceed 1.

For the second, note that $X_{(n)}\leq 1-\epsilon$ means that the maximal one of the first $n$ trials is smaller than $(1-\epsilon)$, therefore every one of the first $n$ trials should be smaller than that value. And they are independent, so the propability of $\{X_{(n)}\leq 1-\epsilon\}$ is the product of all $n$ probabilities of $\{X_k\leq 1-\epsilon\}$.

Hope this helps.

Answer 2

Yes @noam.szyfer, you need independence. In order for $X_{(n)}$ to converge to 1 in probability, we need to show that $\forall \epsilon>0, P(|X_{(n)}-1|<\epsilon)=1$. This is given via the following: \begin{equation} P(|X_{(n)}-1|<\epsilon)\\=P(|1-X_{(n)}|<\epsilon)\\ =P(1-X_{(n)}<\epsilon)\\ =1-P(1-X_{(n)}\geq \epsilon)=1-P(1-\epsilon\geq X_{(n)})\\ \end{equation} However, note that $X_{(n)}\leq 1-\epsilon$ if and only if $X_i\leq 1-\epsilon$ for all i. This implies that \begin{equation} 1-P(1-\epsilon\geq X_{(n)})=1-P(1-\epsilon\geq X_1,1-\epsilon\geq X_2,..)\\ =1-\prod_{i=1}^\infty P(1-\epsilon\geq X_i)\,\,(\text{By Independence})\\ =1-(1-\epsilon)^n \end{equation} The final term limits to one as $1-\epsilon<1$ for any $\epsilon>0$ and hence $\lim_{n\rightarrow \infty}(1-\epsilon)^n=0$.

Answer 3

As the different commentors here agree we also need the $X_i$ to be independent for this to work. Here is the same proof as in the PDF you linked, with additional explanations.

We need to show that $\Bbb{P}(|X_{(n)} - 1|\geq\varepsilon)$ converges to 0 as $n\to \infty$. (This is the definition of convergence in probability). So let $\varepsilon>0$, then we have:

$$\Bbb{P}(|X_{(n)} - 1|\geq\varepsilon) = \Bbb{P}(1-X_{(n)}\geq\varepsilon) = \Bbb{P}(X_{(n)}\leq1-\varepsilon)$$

For the first equality we used that $X_{(n)}-1$ is non-positive, since $X_{(n)}$ is bounded by 1.

Now comes the crucial step: What is the probability that the maximum of all the random variables $X_i$ is smaller than some value? It's the same as saying: every $X_i$ must be smaller than this value (since then also the maximum of all of them are smaller. If one was bigger, then also the maximum would be bigger!). So in that sense, the following two events are the same:

$$\Bbb{P}(X_{(n)}\leq1-\varepsilon) = \Bbb{P}(X_1 \leq 1-\varepsilon, ...,X_n \leq 1-\varepsilon).$$

Finally, we use the independence assumption to get

$$\Bbb{P}(X_1 \leq 1-\varepsilon, ...,X_n \leq 1-\varepsilon) = \prod_{i=1}^{n}\Bbb{P}(X_i < 1-\varepsilon) = (1-\varepsilon)^n.$$

We see that the last term tends towards 0 as $n$ tends towards infinity.