Characteristic function of a normal variable and conditioning

Question

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 ?

Considering you are not considering different $n$s at any point, it could be a good thing to just drop the $n$s (and thus the filtration tag), that way we get to the core of the question. — Bruno B, Jul 17 '23 at 17:33
There is a property which states that if you let $g(v) := \mathbb{E}\left[f(U, v)\right]$ for some measurable function $f$, then we have $g(V) = \mathbb{E}[f(U,V) \mid V]$ (a proof is given here: Conditional expectation of a function with independent random variables), and since $\sigma(V) \subset \mathcal{B}$, we have: $$g(V) = \mathbb{E}[f(U,V) \mid V] = \mathbb{E}\left[\mathbb{E}[f(U, V) \mid \mathcal{B}] \mid V\right]$$ Hence $\mathbb{E}[f(U, V) \mid \mathcal{B}] = g(V)$ if and only if $\mathbb{E}[f(U, V) \mid \mathcal{B}] $ is $V$-measurable. — Bruno B, Jul 17 '23 at 18:14
However I do not see how to proceed further, which is why this is only a comment. — Bruno B, Jul 17 '23 at 18:15
This answer gives a reference and the outline of the proof that $\mathbb{E}[f(U, V) \mid \mathcal{B}] = g(V)$, which would answer your question: $E[f(X, Y)|\mathscr{F}]$ where $X$ is independent of $\mathscr{F}$ and $Y$ is $\mathscr{F}$ measurable. — Bruno B, Jul 17 '23 at 18:39

score 2 · Accepted Answer · answered Jul 17 '23 at 18:59

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.

For completeness, I’ll add that the result holds even if $U$ is not independent of $\mathcal B$, but when taking the expectation one needs to replace $U$ with $U$ conditional on $\mathcal B$ — Andrew, Jul 17 '23 at 19:02

Characteristic function of a normal variable and conditioning

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