A rabbit tries to escape its predator by zigzagging in the following way.

Question

This problem is about a rabbit that tries to escape its predator by zigzagging. It initially fixes a paramter $\alpha\in(0,\frac\pi2)$ (in radians), and in each second $n\geq1$ it picks $U_n$, a Uniform$(0,\alpha)$ angle. Between times $n-1$ and $n$ (seconds) it runs at unit speed in direction $U_n$ relative to East, then picks $U_{n+1}$ and runs in that direction relative to East, etc. For example, if $U_n=\frac\pi4$, then from time $n-1$ to time $n$ it exactly runs towards North-East. Hence, the position of the rabbit at the end of the $n$th second is $$ (S_n,T_n)=\left(\sum_{i=1}^n\cos U_i,\sum_{i=1}^n\sin U_i\right). $$ Assume that the $U_n$ are independent.

• Show that the rabbit has an asymptotic velocity in the sense that its distance $D_n=\sqrt{S_n^2+T_n^2}$ from its starting point at time $n$ divided by $n$ converges a.s. to a limit.
• Calculate this asymptotic velocity $lim_{n\to\infty}\frac{D_n}n$ as a function of the paramater $\alpha$. What happens if $\alpha$ is close to $0$?

So far I have made some progress. We have that $$ \begin{align*} \frac{D_n}n&=\sqrt{\left(\frac{S_n}n\right)^2+\left(\frac{T_n}n\right)^2}\\ &\overset{a.s.}\to\sqrt{(\mathbb{E}\cos U_i)^2+(\mathbb{E}\sin U_i)^2}.\tag{by the SLLN} \end{align*} $$ So now all I have to do is calculate this limit. So firstly $$ \begin{align*} \mathbb{E}\cos U_i&=\int_0^\alpha\frac1\alpha\cos xdx\\ &=\frac1\alpha[\sin x]_0^\alpha\\ &=\frac{\sin\alpha}\alpha \end{align*} $$ and $$ \begin{align*} \mathbb{E}\sin U_i&=\int_0^\alpha\frac1\alpha\sin xdx\\ &=\frac1\alpha[-\cos x]_0^\alpha\\ &=\frac1\alpha(1-\cos\alpha). \end{align*} $$ So now we have that $$ \begin{align*} \lim_{n\to\infty}\frac{D_n}n&=\sqrt{\frac{\sin^2\alpha+1-2\cos\alpha+\cos^2\alpha}{\alpha^2}}\\ &=\frac{\sqrt{2-2\cos\alpha}}\alpha \end{align*} $$ Then when it came time to analysing the behaviour of this limit at $\alpha=0$, I started to get suspicious that I had done something wrong in my workings, because I would have thought that this would be an obvious limit of $1$, since the rabbit would almost surely be going in the same direction at unit speed. However, when I tried to evaluate this limit, I couldn't work out how, which makes me suspicious I have done something wrong, since I was expecting this part of the question to be easy. So can anyone see where I have gone wrong, or help me with calculating this limit? Thanks!

Answer 1

Your argument and computations are correct up to your point of concern, as far as I've checked. Here is a simulation of $D_n/n$ as $n\to 9000$, for $\alpha = \pi/4$. It converges to the value $f(\pi/4) \approx 0.9744954$ given by your formula $f(\alpha)=\sqrt{2-2\cos(\alpha)}/\alpha$.

Here is the code to produce this in R,

n <- 9000
f <- function(a)  sqrt(2-2*cos(a))/a
alpha <- pi/4
u <- runif(n, 0, alpha)
x <- cumsum(cos(u))
y <- cumsum(sin(u))
d <- sqrt(x^2+y^2)
v <- d/(1:n)
plot(1:n, v, type = "l")
abline(h = f(alpha), col = "blue", lty = "dashed")
w <- c(exact = f(alpha), mc = sqrt(mean(cos(u))^2+mean(sin(u))^2))
print(w)

Finally, about the behavior as $\alpha \to 0$. We claim that $$\lim_{\alpha \to 0} \frac{\sqrt{2-2\cos(\alpha)}}{\alpha}=1.$$ For, if you know your Taylor series/little-o expansions, we have $\cos(\alpha)=1-\alpha^2/2+o(\alpha^4)$ as $\alpha \to 0$. Thus, we get $$\frac{\sqrt{2}}{\alpha} \sqrt{1-\left(1- \frac 12 \alpha^2+o(\alpha^4)\right)}$$ $$=\sqrt{2} \sqrt{\frac 12 + o(\alpha^2)}\to 1,$$ as $\alpha \to 0$. Numerically this checks out as well if we choose some small $\alpha$ near zero in the code above.

Please comment for clarifications or corrections.