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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?

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1 Answer 1

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 $\sqrt{\frac{n}{\pi}}$.

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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

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