I am studying probability theory on my own and am trying to work the following problem in the book:

Let $X_1, X_2, . . .$ be independent, $U(0, 1)$-distributed random variables, and let $N_m \in Po(m)$ be independent of $X_1, X_2, \dots$. Set $V_m = \max\{X_1, \dots ,X_{N_m}\}$ $(V_m = 0$ when $N_m = 0)$. Determine

(a) the distribution function of $V_m$,

(b) the moment generating function of $V_m$.

(c) Show that $E[V_m] \to 1$ as $m \to \infty$.

(d) Show that $m(1 − V_m)$ converges in distribution as $m \to \infty$, and determine the limit distribution.

I got stumped at the very first part when trying to find the distribution function of $V_m$. I know that the CDF of $X$ is given by

$F_X(x) = \begin{cases} 0, & x < 0 \\ x, & 0 \leq x \leq 1 \\ 1 & x > 1 \end{cases}$

This implies that the CDF of $V_m$ should be given by

$F_{V_m}(x) = \begin{cases} 0, & x < 0 \\ x^n, & 0 \leq x \leq 1 \\ 1 & x > 1 \end{cases}$ where $n \in Po(m)$

Then by differentiating $F_{V_m}(x)$, we get the density function $f_{V_m} = nx^{n-1}$. Then using conditional probability, we want to find the CDF of $V_m$ and I did the following to get there - $P(V_m = x) = \sum_{n=0}^{\infty} P(V_m = x|N_m=n) . f_N(n) = \sum_{n=1}^{\infty} nx^{n-1} e^{-m} \frac{m^n}{n!} = me^{m(x-1)}$

The problem is when I try to integrate this over x from 0 to 1,the answer is not 1 which implies that this might not be the correct pdf and hence integrating this may not give me the correct CDF. Incidentally, when integrating this over 1 to x, I get $ e^{m(x-1)} - e^{-m}$ which is not the correct answer.

I am not sure if I have made a big blunder in my understanding of something or a small goof up. I have looked at my calculations a few times to ensure that I have not made any silly error. I think once I am past this first part, I should be able to get to the rest. I would highly appreciate any help as I feel that this is essential for me to understand before I move forward with my course. Thanks to anyone ande

  • $\begingroup$ What is the subscript $m$ referring to? $\endgroup$
    – AdamO
    Commented Jul 16, 2014 at 17:43
  • $\begingroup$ The subscript $m$ does not refer to anything specific, it is just to show that $V_m$ is the maximum of $n$ uniform random variables where $n$ follows a poisson distribution with parameter $m$. Please let me know if this answers your question. Thanks AdamO. $\endgroup$
    – user42754
    Commented Jul 16, 2014 at 17:46
  • 1
    $\begingroup$ I suspect the limit in (c) is $m \to \infty$; that would fix the problem @whuber has pointed out. $\endgroup$
    – jbowman
    Commented Jul 16, 2014 at 19:59
  • 2
    $\begingroup$ Is it absolutely necessary to use $m$ (in a subscript) to denote the maximum and also to use $m$ as the parameter of the Poisson distribution? I mean there are more than $40$ unused upper-case and lower-case letters breathlessly awaiting a call from you to serve as symbols. $\endgroup$ Commented Jul 16, 2014 at 20:27
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    $\begingroup$ If your notation is copied from your textbook, then I will merely apply a review attributed to Dorothy Parker to your text: "This is not a book to be put down lightly; it should be thrown with great force." $\endgroup$ Commented Jul 16, 2014 at 20:56

2 Answers 2


The calculations in the question look correct, but care is needed because the distribution of $V_\mu$ is not continuous. (I will use $\mu$ instead of $m$ throughout.)

From first definitions we may find the distribution function (CDF) of $V_\mu$ is

$$F_\mu(x) = \Pr(V_\mu) \le x) = \sum_{n=0}^\infty x^n \Pr(N_\mu = n) = e^{\mu(x-1)}$$

provided $0 \le x \le 1$. For $x \gt 1$, $F_\mu(x) = 1$ of course. But for $x \lt 0$, necessarily $F_\mu(x) = 0$. Here is its graph when $\mu=1$ showing the jump at $x=0$:


The moment generating function, $\phi_\mu(t) = \mathbb{E}(\exp(t V_\mu))$, must be computed with similar care near zero. It can be obtained as a Lebesgue-Stieltjes integral,

$$\phi_\mu(t) = \int_\mathbb{R} e^{t x} dF_{\mu}(x)$$

via integration by parts as

$$\phi_\mu(t) = e^{t x} F_\mu(x) \vert_{-\infty}^1 - \int_0^1 t e^{t x} e^{\mu(x-1)}dx = e^t - t\frac{e^t - e^{-\mu}}{t+\mu}.$$

As a check, its McLaurin series begins

$$\phi_\mu(t) = 1 + \left(\frac{\mu-1+e^{-\mu}}{\mu}\right) t + \left(\frac{\mu^2 - 2\mu + 2 - 2e^{-\mu}}{\mu^2}\right)t^2/2 + \cdots$$

The constant term of $1$ shows the total probability mass is $1$. The next two terms will be useful in addressing the rest of the questions.

  • $\begingroup$ thank you very much for this. I highly appreciate your time and help. I knew I was missing something and needed to be pointed in the right direction. Thanks again for this. $\endgroup$
    – user42754
    Commented Jul 16, 2014 at 21:49

notice that since this continous you want to find CDF so going off what you had we have $$P(V_{m}<x)=\sum_{n=0}^{\infty}P(V_{m}<x|N_{m}=n)f_{N}(n)=\sum_{n=0}^{\infty}e^{-m}\frac{(xm)^{n}}{n!}=e^{-m}e^{xm}=e^{m(x-1)}$$

  • 1
    $\begingroup$ This answer is incorrect because the distribution is not continuous. It gives the wrong values when $x\lt 0$ or $x \gt 1$. $\endgroup$
    – whuber
    Commented Jul 16, 2014 at 21:34
  • $\begingroup$ I think that this is only when $0<x\leq 1$ $\endgroup$
    – Kamster
    Commented Jul 16, 2014 at 22:12
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    $\begingroup$ @user159813 Although it may seem like a triviality, failing to make that distinction explicit is the root cause of the difficulties expressed in the question. $\endgroup$
    – whuber
    Commented Jul 16, 2014 at 22:21

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