TITLE: A version of Wald identity
QUESTION [6 upvotes]: Let $W$ be a standard one-dimensional Brownian motion. Let $T$ be a stopping time with $\mathbb{E}\sqrt{T}<+\infty$. Then
$$\mathbb{E}W_T=0\quad \mathbb{E}W^2_T=\mathbb{E}T$$
I can prove these identies under the condition $\mathbb{E}T<+\infty$. But what about this one?

REPLY [3 votes]: Well, I provide a proof using Burkholder-Davis-Gundy(BDG) inequality.(http://en.wikipedia.org/wiki/Quadratic_variation#Martingales)
Let $M=\left\{M_t=W_{t\wedge T};0\le t<+\infty\right\}$, then $M$ is a martingale with $\langle M\rangle_t=t\wedge T$, so by BDG inequality, we have there exists a constant $K>0$ such that
$$\mathbb{E}M_T^*\le K\mathbb{E}\sqrt{T}<+\infty$$
where $M_t^*=\sup_{0\le s\le t}|M_s|$.
Since for all $t\ge0$
$$|M_t|\le\sup_{0\le s\le t}|M_s|=\sup_{0\le s\le t}|W_{s\wedge T}|\le\sup_{0\le s\le T}|W_{s\wedge T}|=\sup_{0\le s\le T}|M_s|=M_T^*$$
Hence $M$ is uniformly integrable, $\mathbb{E}W_T=\mathbb{E}M_T=0$.
Hence $\left\{M_t;0\le t\le+\infty\right\}$ is a martingale with last element $M_T=W_T$, $\left\{M_t^2;0\le t\le+\infty\right\}$ is a submartingale.
If $\mathbb{E}T<+\infty$, then we already know that $\mathbb{E}W_T^2=\mathbb{E}T$.
If $\mathbb{E}W_T^2<+\infty$, then for all $t>0$, since $M_t = \mathbb{E}[W_T \mid \mathcal{F}_t]$, the conditional Jensen inequality yields
$$\mathbb{E}M_t^2\le \mathbb{E}W_T^2<+\infty$$
hence $M$ is $L^2$ bounded, hence $L^2$ convergent.
By the monotone convergence theorem, we have
$$\mathbb{E}T=\mathbb{E}\lim_{t\to+\infty}(t\wedge T)=\lim_{t\to+\infty}\mathbb{E}(t\wedge T)=\lim_{t\to+\infty}\mathbb{E}W_{t\wedge T}^2=\lim_{t\to+\infty}\mathbb{E}M_{t}^2=\mathbb{E}M_T^2<+\infty,$$
so again $\mathbb{E}W_T^2=\mathbb{E}T$.