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

Is there a relation between the max of a Gaussian random walk of 10 steps vs the max of 10 Gaussian random walks? Specifics (in Mathematica notation):

(* a Gaussian random walk with standard deviation 1 *) 
a[0] := 0 
a[n_] := a[n-1] + RandomReal[NormalDistribution[0, 1]] 

(* the max of the walk over 10 steps *) 
b := Max[Table[a[i],{i,1,10}]] 

(* calculate max many times to get good sample set *) 
(* Mathematica "magic" insures we're not using the same random #s each time *) 
c = Table[b,{i,1,10000}] 

(* distribution isn't necessarily normal, but we can still compute mu + SD *) 
Mean[c] (* 3.66464 *) 
StandardDeviation[c] (* 1.61321 *) 

Now, consider 10 people doing a Gaussian random walk of 1 step and we take the max of these 10 values.

(* max of 10 standard-normally distributed numbers *) 
d := Max[Table[RandomReal[NormalDistribution[0, 1]],{i,1,10}]] 

(* get a good sample set *) 
f = Table[d,{i,10000}] 

(* and now the mean and SD *) 
Mean[f] (* 1.54843 *) 
StandardDeviation[f] (* 0.580593 *)

The two means/SDs are obviously different, but I sense they're related somehow, perhaps by Sqrt[10], since the sum (not max) of 10 random walks is normal with SD of Sqrt[10], and I sense that somehow the cumulative sum of the first 9 somehow cancel out.

Are these known distributions?

share|cite|improve this question

I don't have an answer but only some trivial observations about your question that I post below. It got too big to be left as a comment. If I understand your question correctly, you want to compare the following two problems:

1) Let $X_i$ be i.i.d unit normals and you are interested in $X^{*} = \max(X_i, 1 \leq i \leq 10)$.

2) Let $Y_i$ be i.i.d unit normals. Define $S_0 = 0$ and $S_i = S_{i-1} + Y_i$ for $i > 0$ and $S^{*} = \max(S_i, 1 \leq i \leq 10)$.

It is easy to see that $P(X^{*} \leq x) = P(X_i \leq x)$ for all $1 \leq i \leq 10$ and therefore the distribution of $X^{*}$ is given by

$f_{X^{*}}(x) = 10 \Phi(x)^9 \phi(x)$ where $\phi$ and $\Phi$ are the pdf and cdf of the standard normal. From this, you can technically compute the mean of $X^{*}$ though I don't know if the messy integration yields a nice solution.

The case of determining the distribution of $S^{*}$ seems much trickier. I found some references online that study the asymptotic behavior for long random walks and even that seems very hard. You can argue that $P(S^{*} \leq s) = P(S_i \leq s)$ for all $1 \leq i \leq n$ but the $S_i$ are dependent random variables. Their joint distribution is easy to derive but once again, I don't know if the integration is manageable.

Of course, if you are only interested in comparing the first moment of $X^{*}$ and $S^{*}$, there might be a clever way to do it that avoids all the integrations and such. If there exists such a proof, I would love to learn it.

Update: I found some links that give the asymptotics for these random variables as the number of random variables $n \rightarrow \infty$. See this and this. As best as I can determine, $E(X^{*})$ grows as $\sqrt{2\ln(n)}$ and $E(S^{*})$ grows as $\sqrt{\frac{n}{\pi}}$.

share|cite|improve this answer
Dinesh, the backticks are only for code here; if something doesn't render in the $\TeX$, try prepending a backslash to the offending character. – J. M. Oct 30 '10 at 0:32
@J.M. - Thanks. Fixed the backticks. – Dinesh Oct 30 '10 at 0:50
clearly you meant $P(X^*<x)=P(\cap_{i=1}^n\{X_i<x\})$ – mpiktas Dec 29 '10 at 4:30

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.