Expected distance from origin of random walk on Z^d using multivariate CLT

Question

I am trying to prove that $E[D(X_n, X_0)]$ is roughly equals $ \sqrt{n}~$ using Multivariate Central limit theorem. $$$$ So I represent a d-dimensional simple symmetric random walk $~S~$ such that $~S_n = X_1 + ... + X_n$ where $X_k~'s$ are i.i.d random vectors for all $k~ {\geq}~ 0$ which are uniformly distributed on $~~~~~~~~~~~~~~~~~~~~~$ {$\pm{e_j,~ j= 1,...,d}$} where $~e_1,...,e_d$ are the canonical basis of $Z^d$. Obviously the expected value is zero component-wise, I know $$\frac{S_n}{\sqrt{n}} \rightarrow N(\overrightarrow{0}, \Sigma)$$ here comes my struggle to apply multivariate CLT, and then to arrive to $E(D) \sim \sqrt{n} \\$ . $~~~~~~~~~~~~~~~~~~$Can you help me with that?, Thank you in advance

Answer 1

If I understand your question correctly, the missing ingredient is the continuous mapping theorem. The assertion $$ \frac{S_n}{\sqrt n}\stackrel d\to Z,\tag1 $$ where $Z$ has multivariate normal $N(\vec0,\Sigma)$ distribution, is already the multivariate CLT, so apply the continuous mapping theorem to (1) using the function $g(t):=\|t\|:=(t_1^2+t_2^2+\cdots + t_d^2)^{1/2}$ (i.e., $g$ is the $d$-dimensional Euclidean norm) to conclude $$ \left\|\frac{S_n}{\sqrt n}\right\|\stackrel d\to \| Z\|.\tag2$$

The LHS of (2) equals $D/\sqrt n$. This may be enough for you to assert (heuristically) that $E(D)$ is roughly $\sqrt n$. In fact, you can argue further that $$E\left(\left\|\frac{S_n}{\sqrt n}\right\|\right)\to E(\| Z\|),\tag3 $$ an assertion that does not follow immediately from (2). A rigorous proof of (3) would note that the LHS of (2) has constant variance independent of $n$, and therefore is uniformly bounded in $L^2$, which implies uniform integrability and justifies (3).