Convergence in probability with a maximum value

Question

Let X₁, X₂,...,X_n be i.i.d. continuous random variables of the uniform distribution $Unif(0, \beta$), $\beta>0$. Consider M_n = max_1≤i≤n X_i. Prove that M_n converges in probability to $\beta$.

I know how to prove a sample $\bar X$ converges in probability to an expected value $\mu$ with the Chebyshev's inequality $P(|\bar X-\mu|> {\epsilon})\le \frac{{\sigma}^2}{{\epsilon}^2}$ with (in this case) E(X_i) = ${\mu}=\frac{\beta}{2}$ and Var(X_i) = $\frac{\beta^2}{12}$, but the new concept of M_n = max_1≤i≤n X_i added to this confuses me a lot.

Can someone explain this to me? Thank you!

Answer 1

Recall that if $M_n \overset{p}{\to} \beta$, a constant, if and only if $M_n \overset{d}{\to}\beta$ (convergence in distribution).

Let us then consider the CDF of $M_n$ for each $n$.

As an exercise, prove that the CDF of $M_n$ is given by $$F_{M_n}(m) = \begin{cases} 0, & m \leq 0\\ \left(\dfrac{m}{\beta}\right)^n, & 0 < m < \beta \\ 1, & m \geq\beta \end{cases}$$ Then it is clear that $$\lim_{n \to \infty}F_{M_n}(m) = \begin{cases} 0, & m < \beta \\ 1, & m \geq \beta \end{cases}$$ which is identical in distribution to a random variable which is equal to $\beta$ with probaiblity $1$. Thus, $M_n \overset{d}{\to}\beta$, implying $M_n \overset{p}{\to} \beta$.