# What is the probability distribution on the high, low and end values of a random walk?

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.

• Probability distribution on the highest value? I think this problem is ill-posed: is there a highest value? – parsiad Aug 8 '13 at 19:56
• 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. – zortharg Aug 8 '13 at 21:07
• 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). – parsiad Aug 8 '13 at 21:24
• math.stackexchange.com/questions/38642/… – parsiad Aug 8 '13 at 21:26
• 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? – 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 $\frac{2}{\pi}\approx 0.79788$.