Expected value of point Consider a point starting at the origin.  It is equally likely to travel up, down, left or right.  Defining random variables $Y_1,Y_2,Y_3,Y_4$ to represent the number of moves up, down, left, right respectively in n moves.  Let $D$ be defined to be the Euclidean distance from the origin that the point ends up at.  Show that $E(D^2) = n$.
So far I have determined that $D^2 = (Y_2-Y_1)^2 + (Y_3-Y_4)^2$.  My initial hunch is to expand the right side of this and use linearity of expected value, however that does not get me very far.  Any suggestions as to a more streamlined approach to this?
Thanks!
 A: Your approach is a good one. Why it does not get you very far? You can model $Y_i$ as binomial random variables with $p = 1/4$
They are not independent though as  $Y_1 + Y_2 + Y_3 + Y_4 = n$
Expanding $E(D^2)$ one gets $E(Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2) - 2E(Y_1Y_2 + Y_3Y_4)$
Now the first term is easy because of linearity of expectation, the second is not that good.
But using a symmetry argument, one can see that $E(Y_1Y_2 + Y_3Y_4) = E(Y_1Y_3 + Y_2Y_4) = E(Y_1Y_4 + Y_2Y_3)$ 
(what I mean by this is that there is no reason why the pair $1-2$ and $3-4$ should be special. Those numbers are just label, those random variables are indistinguishable from one another, hence the symmetry)
Why does that help? Because those are the coefficient in the expansion of $(Y_1 + Y_2 + Y_3 + Y_4)^2$
Indeed, $(Y_1 + Y_2 + Y_3 + Y_4)^2 = Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2 + 2(Y_1Y_2 + Y_3Y_4 + Y_1Y_3 + Y_2Y_4 + Y_1Y_4 + Y_2Y_3)$
But $(Y_1 + Y_2 + Y_3 + Y_4)^2 = n^2$ . So take expectation to both sides and remember the symmetry to get 
$$n^2 = E(Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2) + 6E(Y_1Y_2 + Y_3Y_4) \implies $$$$E(Y_1Y_2 + Y_3Y_4) = \frac{n^2 -  E(Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2) }6$$
Substitute back into the original formula to find 
$$E(D^2) = E(Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2) - 2E(Y_1Y_2 + Y_3Y_4) =$$$$ E(Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2) - \frac{n^2 -  E(Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2) }3 = \frac 43 E(Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2) - \frac{n^2}3$$
Since $E(Y_i^2) = np(1-p+np)$, plug $p = \frac 14$ to get 
$$E(D^2) = n$$
