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Let $\left\{X_n\right\}$ be identically, independently exponentially distributed random variables with parameter/mean $1$. Let $M_n$ be the max of the $X_i$'s up to $n$. Find a sequence $\left\{a_n\right\}$ of real numbers such that $P[M_n-a_n\leq x]$ converges to a non-degenerate distribution function $G(x)$ as $n\rightarrow\infty$ for every $x\in\mathbb{R}$ such that $G(x)$ is continuous.

I am using A Probability Path. If this is a well covered subject, can someone direct me to the section that is relevant to the answer?

I know that $P[M_n-a_n\leq X]=\left(1-e^{-(x+a_n)} \right)^n$. I've been trying to manipulate the $a_n$ such that this starts to look like another CDF, but I run into contradictions... so this bashing approach doesn't seem to work. I don't know what concepts this problem is trying to demonstrate. The use of $M$ makes this seem like a Martingale, but I know it's not one since $M_n$ is monotonically nondecreasing.

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Also, if I wanted to evaluate the moment generating function $\int e^{tx}G(dx)$, I note that $G(dx)=e^{-e^{-x}}e^{-x}dx$. I get $\int_0^\infty e^{(t-1)x-e^{-x}}dx$. It's not further simplifiable. Is this correct? – sam.t Dec 11 '11 at 1:22
up vote 2 down vote accepted

We have that $\mathbb{P} (X_i \leq x) = 1 - \exp(-\lambda x)$ and $M_n = \displaystyle \max_{1 \leq i \leq n} X_i$. $$\mathbb{P} (M_n \leq x) = \mathbb{P} (X_i \leq x \text{ for all X}_i \in \{1,2,\ldots,n\}).$$ Since $X_i$'s are independent, we get $$\mathbb{P} (X_i \leq x \text{ for all X}_i \in \{1,2,\ldots,n\}) = \prod_{k=1}^{n} \mathbb{P} (X_i \leq x) = \left( 1 - \exp(-\lambda x) \right)^n.$$ Hence, we get $$F_{M_n}(x) = \mathbb{P} (M_n \leq x) = \mathbb{P} (X_i \leq x \text{ for all X}_i \in \{1,2,\ldots,n\}) = \left( 1 - \exp(-\lambda x) \right)^n.$$ $$F_{M_n - a_n}(x) = \mathbb{P} (M_n - a_n \leq x) = \mathbb{P} (M_n \leq x + a_n) = \left( 1 - \exp(-\lambda (x+a_n)) \right)^n.$$ Choose $a_n = \frac{\log(n)}{\lambda}$. We then get $$ \begin{align*} F_{M_n - a_n}(x) & = \left( 1 - \exp \left(-\lambda \left(x+ \frac{\log(n)}{\lambda} \right) \right) \right)^n \\ & = \left( 1 - \exp(-\lambda x) \exp(- \log(n)) \right)^n \\ & = \left(1 - \frac{\exp(-\lambda x)}{n} \right)^n \\ \lim_{n \rightarrow \infty} F_{M_n - a_n}(x) & = \lim_{n \rightarrow \infty} \left(1 - \frac{\exp(-\lambda x)}{n} \right)^n \\ & = \exp(-\exp(-\lambda x)) \end{align*}$$ Note that if we let $F_{M_{\infty}}(x) = \exp(-\exp(-\lambda x))$, we have $F_{M_{\infty}}(x)$ to be monotone non-decreasing and right continuous. Furthermore, we also have $\displaystyle \lim_{x \rightarrow -\infty} F(x) = 0$ and $\displaystyle \lim_{t \rightarrow \infty} F(x) = 1$. Hence, $F_{M_{\infty}}(x) = \exp(-\exp(-\lambda x))$ is a valid distribution function. Hence, choosing $a_n = \log(n)$, we find that the random variables $M_n$ converge in distribution to the random variable $M_{\infty}$ whose distribution is $F_{M_{\infty}}(x) = \exp(-\exp(-\lambda x))$.

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Oh... I actually did end up with $a_n=\log(n)$, but I overlooked that this was a legitimate CDF. Thanks! – sam.t Dec 11 '11 at 1:16

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