 Solution

The exact condition is again ${\sum }_k{a}_k^2<\infty$, with the terminal almost-sure limit used to define $M_T$ when $T=\infty$. In that case ${\sup }_n\mathbb{E}{M}_n^2<\infty$, andproves uniform integrability. The preceding unbounded optional stopping argument yields $\boxed{\mathbb{E}{M}_T=0}$.

For necessity, suppose $v_n\to \infty$. For every fixed $R>0$, the normal distribution of $M_n$ gives $\mathbb{P}(M_n\ge R)\to 1/2$. Hence $\mathbb{P}({\sup }_n{M}_n\ge R)\ge 1/2$, and intersecting over positive integer $R$ shows $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. Unboundedness above is unchanged by altering finitely many of the independent $Z_k$, because that changes all sufficiently late sums by one finite constant. It is therefore a tail event; the Kolmogorov zero-one law makes its probability one. Consequently $T<\infty$ almost surely and $M_T\ge 1$. Its expectation is at least one, possibly infinite, and cannot equal zero. This proves the necessity assertion in threshold stopping of an independent Gaussian series.