I have homework about convergence.

Question: Let $X_1,...,X_n$ be i.i.d. exponential(1) random variables. Let $X_{(n)}$ = max $X_i$. Find a sequence $a_n$ so that $X_{(n)} - a_n$ converges in distribution.

Answer so far :

$P(|X_{(n)} - \lambda e^{-\lambda}| \geq \epsilon)$ = $P(X_n \leq \lambda e^{-\lambda} - \epsilon)$ = $(\lambda e^{-\lambda} - \epsilon)^n$ $\rightarrow$ 0

I take $\epsilon$ = t/n,

$P(X_n \leq \lambda e^{-\lambda} - \frac{t}{n})$ = $(\lambda e^{-\lambda} - \frac{t}{n})^n \rightarrow e^{-\lambda t}$

$P(n(\lambda e^{-\lambda} - X_n) \leq t) \rightarrow 1 - e^{-\lambda t}$

$n(\lambda e^{-\lambda} - X_n) = X_n - a_n$

$a_n = n \lambda e^{- \lambda}$

Actually I am a bit confused about the sequence $a_n$. Your advanced feedback are really appreciated.

  • 2
    $\begingroup$ If $X_{(n)}-a_n$ is to converge in distribution, then--at the very least--the series of expectations $\mu_n = \mathbb{E}\left(X_{(n)}-a_n\right)$ of these random variables must converge. That reduces the question to one about a series of numbers. In your working out, the appearance of undefined symbols "$\lambda$" and "$t$" is surprising: they are superfluous. The $\mu_n$ are functions of $n$ alone. $\endgroup$ – whuber Sep 26 '16 at 13:48
  • $\begingroup$ Could you please guide me what should I do? $\endgroup$ – Jyanto Sep 26 '16 at 14:14
  • 2
    $\begingroup$ I thought I just did! Since you're dealing with problems in the convergence of random variables, which is a relatively advanced topic, it's fair to presume you know how to work with expectations and that you can compute them for simple distributions like the exponential. If you're uncomfortable with either of those, then you would be well served to refine those skills before you go any further. $\endgroup$ – whuber Sep 26 '16 at 14:19
  • $\begingroup$ I'm sorry. So the $a_n$ is series of numbers with function of $n$ only? $\endgroup$ – Jyanto Sep 26 '16 at 16:02
  • 1
    $\begingroup$ That's the intended meaning. If the $a_n$ were random variables, instead of just numbers, you could just set $a_n = X_{(n)}$: the differences are constantly the zero random variable and they trivially converge. $\endgroup$ – whuber Sep 26 '16 at 16:26

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