Let $\mathcal{B}$ be a $\sigma$-algebra and $U$ and $V$ be random variables such that :
- $V$ is nonnegative and $B$-measurable
- $U \sim \mathcal{N}(0,1)$ and $U$ is independent of $\mathcal{B}$.
From the second condition, we know that for every nonnegative $\lambda$, $\mathbb{E}[e^{\lambda U} | \mathcal{B}]=\mathbb{E}[e^{\lambda U}]=e^{\lambda^2/2}$. But can we also say that $$\mathbb{E}[e^{\lambda VU} | \mathcal{B}]=e^{\lambda^2V^2/2},$$ by treating $V$ as a nonnegative constant in the conditional expectation ?
Just so this question may leave the Unanswered queue:
The answer by Nate Eldredge in the post https://math.stackexchange.com/a/3221336/1104384 outlines a proof (and backs it up with a reference from Resnick's A Probability Path) for the fact that in our context, and given a (Borel) measurable function $f(\cdot, \cdot)$, then, if we take $g : v \mapsto \mathbb{E}[f(U,v)]$, we have: $$g(V) = \mathbb{E}[f(U,V) \mid \mathcal{B}]$$ This applies in particular to the function $f : (u,v) \mapsto e^{\lambda u v}$, and thus OP's desired result is true.