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Good evening !

Let's say that $X_1,X_2,...,X_n$~$U[0,X]$

I want to prove or disprove that $P(|max(X_1,X_2,...,X_n)-X| \ge \varepsilon) \rightarrow 0$.

My intuition tell's me at start that this is true becouse when the $n$ will be bigger then maximum will be closer to X.

My first idea was to do this by inequality :

$$P(|max(X_1,X_2,...,X_n)-X|\ge \varepsilon) \le \frac{E(|max(X_1,X_2,...,X_n)-X|)}{\varepsilon}$$ and then to prove that $ E(|max(X_1,X_2,...,X_n)-X|) \rightarrow 0$ .

So $|max(X_1,X_2,...,X_n)-X|=X-max(X_1,X_2,...,X_n)$

$E(X-max(X_1,X_2,...,X_n))=X-E(max(X_1,X_2,...,X_n))$

Let's calculate $ E(max(X_1,X_2,...,X_n)), Y_n :=(max(X_1,X_2,...,X_n)$

$P(Y_n \le t)=P(max(X_1,X_2,...,X_n) \le t)=P(X_1 \le t, X_2\le t,...,X_n \le t)=P(X_1\le t)P(X_2\le t)...P(X_n \le t)=\frac{t^n}{X^n}$

So now compute cumulative distribution function :

$$ F_{Y_n}(t)= \begin{cases} 0, &for \;t\;\in\;(-\infty,0)\\ \frac{t^n}{X^n}&for\;t\in\;[0,X]\\ 1&for\;t\;\in\;(X,+\infty) \end{cases} $$

The density function :

$$D_{Y_n}(t)=\begin{cases} 0, &for \;t\;\in\;(-\infty,0) \cup (X,+\infty) \\ \frac{nt^{n-1}}{X^n}&for\;t\in\;[0,X]\\ \end{cases} $$

$E(Y_n)=\int_{0}^{X}t\cdot \frac{nt^{n-1}}{X^n}dt=X \cdot \frac{n}{n+1}$

And it goes to $X$ when $n \rightarrow \infty$. So $$\frac{E(max(X_1,X_2,...,X_n)-X)}{\varepsilon} \rightarrow 0 \Rightarrow P(|max(X_1,X_2,...,X_n)-X| \ge \varepsilon) \rightarrow 0$$.

Am i thinking correctly ?

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There is a much simpler way to do that:$${\Pr\Big\{|\max(X_1,X_2,...,X_n)-X| \ge \varepsilon\Big\}\\=\Pr\Big\{\max(X_1,X_2,...,X_n)-X \ge \varepsilon\Big\}\\+\Pr\Big\{\max(X_1,X_2,...,X_n)-X \le -\varepsilon\Big\}}$$also$$\Pr\Big\{\max(X_1,X_2,...,X_n)-X \ge \varepsilon\Big\}=0$$and$$\Pr\Big\{\max(X_1,X_2,...,X_n)-X \le -\varepsilon\Big\}=\left(1-{\varepsilon\over X}\right)^n$$therefore$$\Pr\Big\{|\max(X_1,X_2,...,X_n)-X| \ge \varepsilon\Big\}=\left(1-{\varepsilon\over X}\right)^n\to 0$$

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