Given iid random variables $X_1, \dots, X_n$ with common density: $$ f(x) = 1\{ x > 0 \} \cdot \frac{1}{(x+1)^2} $$ it is supposed to be the case that $\frac{\max_i X_i}{n}$ converges in distribution.

Question: How can these random variables converge in distribution if the pointwise limit of their cumulative distribution functions isn't even an increasing function?

Details: I thought this was going to work out nicely and that the sequence would converge in distribution cleanly to an exponential random variable.

The CDF of each $X_i$ is $1\{x >0 \} (1 - \frac{1}{1+x})$, so since the $X_i$ are iid, the CDF of their maximum should just be the product of their CDF's, i.e. $1\{x >0 \} \left(1 - \frac{1}{1+x}\right)^n$.

Presumably, $P(\frac{\max_i X_i}{n} \le x) = P(\max_i X_i \le nx)$, so that we get for the CDF's of the relevant sequence of random variables:

$$P\left(\frac{\max_i X_i}{n} \le x\right) = \begin{cases}0 & x \le 0 \\ \left(1-\frac{1}{1+nx}\right)^n & x > 0 \end{cases}$$

Taking the limit as $n \to \infty$, this approaches $1\{x > 0 \} e^{-x}$, which isn't even an increasing function, and isn't the CDF of an exponential, $1\{x > 0 \}(1 - e^{-x})$.

Is the claim I am trying to prove wrong? Are my calculations wrong? Or is my understanding of convergence in distribution wrong? Or something else?

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    $\begingroup$ Please check the limit of $\left(1-\frac{1}{1+nx}\right)^n$. $\endgroup$ – Viktor Oct 29 '17 at 6:30
  • $\begingroup$ @Viktor The above equals $$\left( \frac{nx}{nx+1} \right)^n \,,$$ while $$e^{-x} = \lim\left(\frac{nx-1}{nx}\right)^n$$ so since $$\lim\left( \frac{nx}{nx+1} \right) = \lim \left( \frac{nx-1}{nx} \right) \,,$$ my guess is that $$\lim\left( \frac{nx}{nx+1} \right)^n = e^{-x}$$ although now that I think about it again it doesn't seem that I am able to verify it directly. I had tried earlier to expand the binomial to somehow get something easier to work with but embarrassingly was not able to. To apply L'Hospital's Rule one has to differentiate w.r.t. $n$ which doesn't lead to any simplification. $\endgroup$ – Chill2Macht Oct 29 '17 at 15:55
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    $\begingroup$ Your guess is not right! What is the value of $\lim_{n\to\infty}\left(1-\frac{1}{nx}\right)^{n}$? $\endgroup$ – Grassie Oct 29 '17 at 17:58
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    $\begingroup$ You should obtain $e^{-1/x}$ for the limit. One way to see this is that $$\log\left(1-\frac{1}{1+nx}\right)^n=n\log\left(1-\frac{1}{1+nx}\right)\approx -n\frac{1}{1+nx}\approx -\frac{1}{x}.$$ $\endgroup$ – whuber Oct 29 '17 at 17:58
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    $\begingroup$ Yes--those conclusions are correct and they seem to be the point of this exercise. BTW, if you write the logarithm in the form $$\frac{\log\left(1-\frac{1}{1+nx}\right)}{n^{-1}},$$ then L'Hopital does the trick. $\endgroup$ – whuber Oct 29 '17 at 20:38

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