Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $n$ independent geometric random variables $X_n$, each with probability parameter $p$ (and thus expectation $E\left(X_n\right) = \frac{1}{p}$), what is $$E_n = E\left(\max_{i \in 1 .. n}X_n\right)$$

If we instead look at a continuous-time analogue, e.g. exponential random variables $Y_n$ with rate parameter $\lambda$, this is simple: $$E\left(\max_{i \in 1 .. n}Y_n\right) = \sum_{i=1}^n\frac{1}{i\lambda}$$

(I think this is right... that's the time for the first plus the time for the second plus ... plus the time for the last.)

However, I can't find something similarly nice for the discrete-time case.

What I have done is to construct a Markov chain modelling the number of the $X_n$ that haven't yet "hit". (i.e. at each time interval, perform a binomial trial on the number of $X_n$ remaining to see which "hit", and then move to the number that didn't "hit".) This gives $$E_n = 1 + \sum_{i=0}^n \left(\begin{matrix}n\\i\end{matrix}\right)p^{n-i}(1-p)^iE_i$$ which gives the correct answer, but is a nightmare of recursion to calculate. I'm hoping for something in a shorter form.

share|cite|improve this question
Is there are typo there? How did $32$ get into it? – joriki Mar 10 '11 at 8:58
Oh yes, whoops. This question arises from a concrete example, where n was 32. – Rawling Mar 10 '11 at 9:04
Is there a reason for the "RV" abbreviation? The main page has plenty of space for two lines... – Uticensis Mar 10 '11 at 9:15
Force of habit. I can now see all the other questions in "Related" with "random variable" in their titles :) – Rawling Mar 10 '11 at 9:17
up vote 15 down vote accepted

First principle:

To deal with maxima $M$ of independent random variables, use as much as possible events of the form $[M\leqslant x]$.

Second principle:

To compute the expectation of a nonnegative random variable $Z$, use as much as possible the complementary cumulative distribution function $\mathrm P(Z\geqslant z)$.

In the discrete case, $\mathrm E(M)=\displaystyle\sum_{k\ge0}\mathrm P(M>k)$, the event $[M>k]$ is the complement of $[M\leqslant k]$, and the event $[M\leqslant k]$ is the intersection of the independent events $[X_i\leqslant k]$, each of probability $F_X(k)$. Hence, $$ \mathrm E(M)=\sum_{k\geqslant0}(1-\mathrm P(M\leqslant k))=\sum_{k\geqslant0}(1-\mathrm P(X\leqslant k)^n)=\sum_{k\geqslant0}(1-F_X(k)^n). $$ The continuous case is even simpler. For i.i.d. nonnegative $X_1, X_2, \ldots, X_n$, $$ \mathrm E(M)=\int_0^{+\infty}(1-F_X(t)^n) \, \mathrm{d}t. $$

share|cite|improve this answer
These are very nice principles. I'd just add that what Didier has called "repartition function" is called "cumulative distribution function" in English. – Michael Lugo Mar 10 '11 at 19:27
@Michael Thanks, post modified. – Did Mar 10 '11 at 21:01
@Did sorry to bother you over such an old answer but I am unclear on what F_X(k) is. For a geometric distribution of parameter p, I think it is the c.d.f. or (1-(1-p)^{k+1}). Is this correct? – Dale M Apr 2 '13 at 1:52
@DaleM If $P(X=k)=p(1-p)^k$ for every $k\geqslant0$, yes. – Did Apr 2 '13 at 5:08

There is no nice, closed-form expression for the expected maximum of IID geometric random variables. However, the expected maximum of the corresponding IID exponential random variables turns out to be a very good approximation. More specifically, we have the hard bounds

$$\frac{1}{\lambda} H_n \leq E_n \leq 1 + \frac{1}{\lambda} H_n,$$ and the close approximation $$E_n \approx \frac{1}{2} + \frac{1}{\lambda} H_n,$$ where $H_n$ is the $n$th harmonic number $H_n = \sum_{k=1}^n \frac{1}{k}$, and $\lambda = -\log (1-p)$, the parameter for the corresponding exponential distribution.

Here's the derivation. Let $q = 1-p$. Use Did's expression with the fact that if $X$ is geometric with parameter $p$ then $P(X \leq k) = 1-q^k$ to get

$$E_n = \sum_{k=0}^{\infty} (1 - (1-q^k)^n).$$

By viewing this infinite sum as right- and left-hand Riemann sum approximations of the corresponding integral we obtain

$$\int_0^{\infty} (1 - (1 - q^x)^n) dx \leq E_n \leq 1 + \int_0^{\infty} (1 - (1 - q^x)^n) dx.$$

The analysis now comes down to understanding the behavior of the integral. With the variable switch $u = 1 - q^x$ we have

$$\int_0^{\infty} (1 - (1 - q^x)^n) dx = -\frac{1}{\log q} \int_0^1 \frac{1 - u^n}{1-u} du = -\frac{1}{\log q} \int_0^1 \left(1 + u + \cdots + u^{n-1}\right) du $$ $$= -\frac{1}{\log q} \left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right) = -\frac{1}{\log q} H_n,$$ which is exactly the expression the OP has above for the expected maximum of $n$ corresponding IID exponential random variables, with $\lambda = - \log q$.

This proves the hard bounds, but what about the more precise approximation? The easiest way to see that is probably to use the Euler-Maclaurin summation formula for approximating a sum by an integral. Up to a first-order error term, it says exactly that

$$E_n = \sum_{k=0}^{\infty} (1 - (1-q^k)^n) \approx \int_0^{\infty} (1 - (1 - q^x)^n) dx + \frac{1}{2},$$ yielding the approximation $$E_n \approx -\frac{1}{\log q} H_n + \frac{1}{2},$$ with error term given by $$\int_0^{\infty} n (\log q) q^x (1 - q^x)^{n-1} \left(x - \lfloor x \rfloor - \frac{1}{2}\right) dx.$$ One can verify that this is quite small unless $n$ is also small or $q$ is extreme.

All of these results, including a more rigorous justification of the approximation, the OP's recursive formula, and the additional expression $$E_n = \sum_{i=1}^n \binom{n}{i} (-1)^{i+1} \frac{1}{1-q^i},$$ are in Bennett Eisenberg's paper "On the expectation of the maximum of IID geometric random variables" (Statistics and Probability Letters 78 (2008) 135-143).

share|cite|improve this answer
Is the log here based 2 or mathematical constant? – Fan Zhang Apr 2 '12 at 4:35
@FanZhang: It's log base $e$. – Mike Spivey Apr 2 '12 at 13:59

$$\begin{align} P(\max Y_i=k)&=P(\max Y_i\leq k)-P(\max Y_i<k)\\\\&=F(k)^n-(F(k)-f(k))^n. \end{align}$$ Thus $$\begin{align} E(\max Y_i) &= \sum_{k=0}^{\infty} k\left[F(k)^n-(F(k)-f(k))^n\right] \\\\ &=\sum_{k=1}^{\infty}k\left[\left(1-(1-p)^k\right)^n-\left(1-(1-p)^{k-1}\right)^n\right]. \end{align}$$

Not a closed form though.

See also Order statistic for both continuous and discrete case. The formula for the continuous case appears in Shai Covo's post here.

share|cite|improve this answer
Very nice. No recursion required to calculate, works with any distribution, and gives me a whole new concept to look into. I'll have to try with the continuous case to see if it gives me the same answer as I have above. – Rawling Mar 10 '11 at 9:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.