Let $X_i \sim U[0,1]$ are independent. Let $Y_n=\max\{ X_1,X_2,...,X_n\}$. And let $Z_n=\frac{Y_n}{1+nY_n}$.

Does $Z_n$ converge in probability to $0$?

My attempt: $P(\frac{Y_n}{1+nY_n}\le \epsilon)=P(Y_n \le \epsilon(1+nY_n))=P(Y_n(1-\epsilon \cdot n)\le \epsilon)=...$

If $\epsilon$ is very large the..

$...=P(Y_n \ge \frac{\epsilon}{1-\epsilon \cdot n})$

And $\lim_{n --> \infty}P(Y_n \ge \frac{\epsilon}{1-\epsilon \cdot n})=1$

But If $\epsilon$ is very small then:

$...=P(Y_n \le \frac{\epsilon}{1-\epsilon \cdot n})$

And $\lim_{n --> \infty}P(Y_n \le \frac{\epsilon}{1-\epsilon \cdot n})=0$

So, $Z_n$ is not converge in probability.

What with almost sure convergence?

  • $\begingroup$ If something doesn't converge in probability then it doesn't converge almost surely, as a.s convergence implies convergence in probability. $\endgroup$
    – fGDu94
    Jan 23 '20 at 22:18

For any $y\in [0,1]$, $$ 0\le\frac{y}{1+ny}\le \frac{1}{1+n}. $$

  • $\begingroup$ Is it hint for almost sure convergence? $\endgroup$
    – pawelK
    Jan 23 '20 at 20:31
  • 1
    $\begingroup$ It is a hint for the a.s. convergence. $\endgroup$
    – d.k.o.
    Jan 23 '20 at 20:32

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