Krdytkyu

I have this question and not too sure where to proceed.

A point starts from the origin and on any move is equally likely to go one unit up, down, left or right, independently of previous moves. Let $X_1, X_2, X_3$ and $X_4$ be random variables giving the number of moves up, down, left and right in a sequence of $n$ moves.

If $D$ is the distance from the origin after $n$ moves, show that $mathsf{E}(D^2)=n$.

I know that$ D^2=(X1 - X2)^2 + (X3- X4)^2$ and that each of $X1, X2$ will have a probability of $0.25$ but am not sure how to find the expected value of this.

It is easy to know that
$$
X_1 + X_2 + X_3 + X_4 = n
$$
and thus
$$
mathsf{E}((X_1 + X_2 + X_3 + X_4)^2) = sum_{i=1}^4 mathsf{E}(X_i^2) + sum_{i neq j} mathsf{E}(X_iX_j) = n^2 tag{$spadesuit$}
$$
Since each $X_i sim mathsf{Binomial}(n, 1/4)$, we have $mathsf{E}(X_i^2)=frac{3}{16}n + frac{1}{16}n^2$. Moreover, by symmetry, we have
$$
mathsf{E}(X_1X_2) = mathsf{E}(X_1X_3) =cdots = mathsf{E}(X_3X_4)
$$
Therefore, by $(spadesuit)$, we obtain
$$
mathsf{E}(X_1X_2) = mathsf{E}(X_1X_3) =cdots = mathsf{E}(X_3X_4) = frac{n^2 - frac{3}{4}n - frac{1}{4}n^2}{12} = frac{1}{16}(n^2 - n)
$$
Finally, we have
$$
mathsf{E}(D^2) = sum_{i=1}^4mathsf{E}(X_i^2) - 2mathsf{E}(X_1X_2) - 2mathsf{E}(X_3X_4) = frac{3}{4}n + frac{1}{4}n^2 - frac{1}{4}(n^2 - n) = n
$$

Alternative way is to express each of $X_i$ as the sum of indicator r.v.: let $U_i, D_i, L_i, R_i$ equal to $1$ if at the $i$th step particle moves up, down, left and right correspondingly. For all $i$
$$U_i+ D_i+ L_i+ R_i = 1, U_iD_i=0, L_iR_i=0$$
$$X_1=sum_{i=1}^n U_i, X_2=sum_{i=1}^n D_i, X_3=sum_{i=1}^n L_i, X_4=sum_{i=1}^n R_i.$$
Calculate the expected value of $D^2$:
$$mathbb ED^2=mathbb Eleft(sum_{i=1}^n (U_i-D_i)right)^2+mathbb Eleft(sum_{i=1}^n (L_i-R_i)right)^2=2mathbb Eleft(sum_{i=1}^n (U_i-D_i)right)^2.$$
Use $mathbb EX^2=text{Var} X + (mathbb EX)^2$:
$$mathbb ED^2=2 text{Var}left(sum_{i=1}^n (U_i-D_i)right)+2left(mathbb Esum_{i=1}^n (L_i-R_i)right)^2=$$
$$mathbb ED^2=2left(sum_{i=1}^n text{Var}(U_i-D_i)right)+2biggl(sum_{i=1}^n underbrace{mathbb E(L_i-R_i)}_{0}biggr)^2=2n text{Var}(U_1-D_1)=2n mathbb E(U_1^2+D_1^2-2underbrace{U_1D_1}_0)=4n mathbb EU_1^2=4n mathbb EU_1=4ncdot 0.25=n.$$