A mouse walks in a maze that is an orthogonal grid made of corridors that intersect at crossings onefoot apart, stopping at every intersection. Using coordinates (with units in feet), it starts at the origin,then moves equally likely up to $(0,1)$ or down to $(0,−1)$ or right to $(1,0)$ or left to $(−1,0)$ by one unit until the next crossing. Then it stops, picks another random direction (up, down, right or left) equally likely and moves by another unit till the next crossing. Every time it stops, its position is a vector $(a,b)$ where both $a$ and $b$ are integers. Scientists let the mouse walk $n$ feet, after which its position is $(X,Y)$and it is at distance $D$ from the origin $(D=\sqrt{X^2+Y^2}$).
a)What is $Cov(X,Y)$?
b) Are $X$ and $Y$ independent?
c) What is $E(D^2)$?
My approach involves defining 4 random variables, $a,b,c,d$ to denote the number of left, right, up and down turns respectively. $a,b,c,d$ ~ $Binomial(n,\frac{1}{4})$, and if the mouse ends at <$X$,$Y$>, then $X=B-A$ and $Y=C-D$. It was not too difficult to show that $Cov(X,Y) = 0$, but how do I approach b? The covariance being zero does not imply independence, and besides the intuitive explanation that since each turn is equally likely the number of right, left, up and downs must be independent, is there a way to mathematically show independence?
Best Answer
Hints:
For (b) calculate the probabilities of $X=n$ and of $Y=n$ and of $X=n,Y=n$. What does this tell you about independence?
For (c) you have $D^2=X^2+Y^2$ so $E[D^2]=E[X^2]+E[Y^2]$. So all you need to find is $E[X^2]$ and $E[Y^2]$. If it helps, you also know that $E[X]=E[Y]$ and that every step is independent of all the other steps.