Solution
= Solution
Using part ii and symmetry of the <simple random walk>,
$$
\mathbb EX_n^2-\mathbb EM_n^2
=\sum_{k\geq1}(2k-1)\mathbb P(X_n=k)>0
$$
for $n\geq1$. Thus \b[$\mathbb EM_n^2<\mathbb EX_n^2=n$].
= Solution
Using part ii and symmetry of the <simple random walk>,
$$
\mathbb EX_n^2-\mathbb EM_n^2
=\sum_{k\geq1}(2k-1)\mathbb P(X_n=k)>0
$$
for $n\geq1$. Thus \b[$\mathbb EM_n^2<\mathbb EX_n^2=n$].