Ok, suppose there's a random walk U which starts at 0 and has a variance of 1 over a time of 1. In other words, U(0)=0, and the probability density of the value of the function at U(1)=x is $\frac{e^{\frac{-x^2}{2}}}{(2*\pi)^{\frac{1}{2}}}$. Naturally the variance is directly proportional to time, being a random walk. What is the probability density on the highest value attained by the random walk in this interval of 1? Is it a gaussian distribution itself, or something else? Ideally, I would like to know the multivariate distribution on the high, low and end value (the supremum value in the interval [0,1], the infimum, and the value at 1). Naturally if it IS indeed a gaussian distribution, I might expect the answer would be defined by a 3x3 covariance matrix with a mean>0 on the high and a mean<0 on the low, but that might not be possible even if they are all gaussians if the relationships between them are not linear.

  • $\begingroup$ Probability distribution on the highest value? I think this problem is ill-posed: is there a highest value? $\endgroup$ – parsiad Aug 8 '13 at 19:56
  • $\begingroup$ Yes, there will be a highest value. And a lowest value. If you think otherwise, that only demonstrates you don't understand what a random walk is. I'm just talking about the maximum INSIDE THE TIME INTERVAL FROM 0 TO 1 mind you. Not the maximum over all time. Over all time, yes, it will attain any arbitrarily large positive and arbitrarily negative number, but I just am talking about in [0,1]. I can say that the mean of the lowest value will be the negative of the mean of the highest but that's about all I can say. Well, I'm confident the mean of the max is between .2 and 2. $\endgroup$ – zortharg Aug 8 '13 at 21:07
  • $\begingroup$ I think you can show that for all $M>0$, $\mathbb{P}\left(\sup_{t\in\left[0,1\right]}\left|U\left(t\right)\right|< M \right)>0$ (this is pretty intuitive). I don't think, however, that you can show that there exists some $M \geq0$ for which $\mathbb{P}\left(\sup_{t\in\left[0,1\right]}\left|U\left(t\right)\right|< M \right)=1$ (this is pretty intuitive too). $\endgroup$ – parsiad Aug 8 '13 at 21:24
  • $\begingroup$ math.stackexchange.com/questions/38642/… $\endgroup$ – parsiad Aug 8 '13 at 21:26
  • $\begingroup$ No no no. I'm not asking for a number which the random walk is guaranteed to never surpass in [0,1]. Indeed there is no such number, there will always be a nonzero chance of it passing anything. But that is irrelevant! For any given instance of the random walk, there WILL BE a maximum value it attains. For crying out loud, it starts at 0 and has a standard deviation of 1 in the first 1 time unit. Suppose it starts at 0 and ends at 1. Do you really think there IS no number it doesn't attain in the middle? Really? Do you think that it went all the way up to a billion and then back to 1? $\endgroup$ – zortharg Aug 8 '13 at 22:32

It is not a multivariate Gaussian: the maximum cannot be below $0$ and the minimum cannot be above zero, so neither have Gaussian distributions.

Wikipedia gives expressions for the distribution of the running maximum of a standard Wiener process, so letting $t=1$, and $M$ the maximum and $W$ the end value, you have

$$f_{M,W}(m,w) = \frac{2(2m - w)}{\sqrt{2 \pi}} e^{-\frac{(2m-w)^2}{2}}, \qquad m \ge 0, w \leq m$$

with an unconditional distribution of $M$ of the form $$f_{M}(m) = \sqrt{\frac{2}{\pi }}e^{-\frac{m^2}{2}} \qquad m \ge 0$$ which has a mean of $\frac{2}{\pi}\approx 0.79788$.

  • $\begingroup$ Ahhh, yes, of course, you're awesome, of course it can't be Gaussian. It could maybe be a Rayleigh distribution or something, but it has to be 0 at 0. How did I miss this answer? It didn't appear until just now. $\endgroup$ – zortharg Aug 8 '13 at 22:40
  • $\begingroup$ Huh. I just needed to search for Wiener process apparently. I had never heard a random walk called that. Shame on me, being ignorant on some basic terminology like that. Well, thanks. Though it does still leave as a mystery the JOINT distribution of the maximum, the minimum, and the final value (the value at t=1), since these are without a doubt all dependent parameters. Certainly if it raises to a very high value, the running maximum must be bigger than that. $\endgroup$ – zortharg Aug 8 '13 at 22:45
  • $\begingroup$ And if the minimum is very far from 0, that probably pushes down the expected value of the maximum, because it means the random walk wasted no time in going down and probably didn't go up very much. $\endgroup$ – zortharg Aug 8 '13 at 22:52
  • $\begingroup$ Ok, I have looked around and I absolutely cannot find where this joint distribution on the maximum and end value is derived. I'd REALLY like the joint distribution on all 3 variables - maximum, minimum, AND end value. It's easy enough to adapt it to be the joint distribution of the minimum and end value, and I'm not really getting anywhere taking that in conjunction with the joint distribution of the maximum and end value and trying to extrapolate the answer to get the distribution of all 3. By any chance do you have an answer to that? $\endgroup$ – zortharg Aug 9 '13 at 1:33

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