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Intuitively it seems likely that the expected whereabouts of Brownian motion on the unit circle would be the origin $\left(0,0\right)$, at least in the limit as $t\to\infty$. Is this right? Are there any techniques for determining the expected location for finite $t$ given some starting point $p$?

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  • $\begingroup$ Brownian motion lies on the circle $x^2+y^2=1$ for all time. $\endgroup$ – user339017 May 12 '16 at 9:15
  • $\begingroup$ You can show that the wrapped normal distribution will converge in distribution to a uniform distribution as $t$ increases making the mean approach the origin along the radius from $p$, though the mode will stay at $p$. You might be able to calculate intermediate values by messy integration or simulation $\endgroup$ – Henry May 12 '16 at 9:28
  • $\begingroup$ I'll have a look into this. Thanks. $\endgroup$ – user339017 May 12 '16 at 9:49
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Suppose $p$ is $(1,0)$ - if it is not then simply rotate the answer - and that the underlying Brownian motion is a standard Wiener process wrapped around the unit circle

A standard Wiener process has $W_t$ normally distributed with mean $0$ and variance $t$, i.e. standard deviation $\sqrt{t}$. Wrap this round the unit circle and you get $X_t = \cos W_t$ and $Y_t= \sin W_t$. You then have an mean expected value given by $$\mathbb{E}[X_t] = \int_{-\infty}^{\infty} \cos (w) \frac{1}{\sqrt{2\pi t}}e^{-\frac{w^2}{2t}}\,dw $$ $$\mathbb{E}[Y_t] = \int_{-\infty}^{\infty} \sin (w) \frac{1}{\sqrt{2\pi t}}e^{-\frac{w^2}{2t}}\,dw $$

By symmetry this makes $\mathbb{E}[Y_t] =0$, which is not a surprise

Empirically, it seems $$\mathbb{E}[X_t]=e^{-t/2}$$ or at least very close to that expression. Attempting to do the integration numerically in R with

cases <- 100000
n <- ((1:cases)-1/2)/cases
g <- qnorm(n)
t <- (0:1000)/100
expectedx <- rep(NA,length(t))
for (j in 1:length(t)){
  expectedx[j] <- mean(cos(g*sqrt(t[j])))
  }
plot(expectedx ~ t, type="l")

gives a graph

E X versus t

and the maximum difference from $e^{-t/2}$ is less than $10^{-5}$, which is possibly due to the numerical approximation to the integral. This may also be related to the von Mises distribution, a close approximation to the wrapped normal distribution. Both converge in distribution to a uniform distribution on the circle as the underlying dispersion increases, and $\mathbb{E}[X_t] \to 0$ as $t \to \infty$

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