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I have a sequence of iid r.v. $(X_n,n\geq 1), X_i\sim \mathcal{E}(1)$.

I am studying properties of the random variable $T_n = \max({X_1,\ldots,X_n})$. Computing the cdf I get $$\mathbb{P}\left(\{T_n\leq t\}\right)= \prod_{i=1}^n\mathbb{P}(\{X_i\leq t\}) = F_X(t)^n = (1-e^{-t})^n$$

I was asked to study the convergence in distribution of $T_n-\log(n)$ I get that $$ \begin{split} \mathbb{P}(\{T_n-\log n\leq t\}) &=\mathbb{P}(\{T_n\leq t+\log n\}) \\ &= \left(1-e^{-t-\log n}\right)^n \\ &= \left(1+\frac{-e^{-t}}{n}\right)^n \to e^{((-e)^{-t})}, \quad n\to +\infty \end{split}$$

But I was also asked to compute the expectation of $T_n$. Now, I know that since the $X_i$ are positive then the $T_n$ will also be positive and thus I can compute $$\mathbb{E}(T_n) = \int_{0}^{+\infty}\mathbb{P}(\{T_n>t\})dt$$

But $$\int_{0}^{+\infty}\mathbb{P}(\{T_n>t\})dt = \int_{0}^{+\infty}1-(1-e^{-t})^n dt = x\big|^{+\infty}_0 - \int_{0}^{+\infty}(1-e^{-t})^n dt$$ and I tried, for instance, to solve the last integral by substitution $y = (1-e^{-t})$ but I don't get something "easy" to compute, and wolfram alpha also does not provide me an answer. Am I on the wrong track?

gt6989b
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user1868607
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2 Answers2

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Expand $(1-e^{-t})^{n}$ using Binomial theorem. you will be able to compute the integral easily.

Kavi Rama Murthy
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  • I tried, different times, but still I get stuck with the binomial coefficient that I don't Know how to get rid of. Do you have any hint? Thanks… @Kavi Rama Murthy – user1868607 Jun 27 '18 at 07:25
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Use 1-a^n = (1-a)(1+a+a^2+...+a^(n-1)) , change of variables from exp(-t) --> x and then integrate to get required answer as 1+1/2+1/3+...+1/n (which varies as log n)

Srivatsa
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