Conditional expectation of a function of X and Y given just X Let $X$ and $Y$ be independent standard normal random variables.
Is $E(e^{XYt} | X) = e^{\frac{1}{2}Y^{2}t^{2}}$
or $E(e^{XYt} | X) = e^{\frac{1}{2}X^{2}t^{2}}?$
I strongly believe that it's the second one because $E(g(X,Y)|X)$ is usually a function of X, it's just that my textbook has $E(e^{XYt} | X) = e^{\frac{1}{2}Y^{2}t^{2}}$.
 A: We have a standard result: Let $\mathcal F, \mathcal G$ be two independent $\sigma$ algebras on a space $(\Omega,\mathbb P)$. Suppose that $X$ is a $\mathcal F$ measurable r.v. and $Y$ a $\mathcal G$ measurable r.v. Now let $f:\mathbb R^2\to \mathbb R$ be be a $\mathcal B(\mathbb R^2)$ measurable function such that $\mathbb E(|f(x,y)|)<\infty$. Then there exists a measurable $g$ such that $\mathbb E(f(X,Y)|\mathcal F)=g(X)$, and
$$g(x)=\int_\mathbb R f(x,y)d\mathbb P_Y(y).$$
This actually holds for arbitrary state spaces, but I formulated it in terms of $\mathbb R$ valued random variables for simplicity. This confirms that you are right, and it is indeed the second one (I'll leave the actual messy integration to you if you would like to verify that plugging in the normal distribution actually does lead to the second option).
A: Well, $\mathsf E(e^{tXY}\mid X)$ will be a function of $X$, rather than $Y$, because $\mathsf E(\mathcal{Whatever}\mid X)$ always shall be.

The moment generating function of a normal distributed random variable $Y$ is $\mathsf M_Y(t)=\mathsf E(e^{tY})=e^{t\mathsf E(Y)+t^2\mathsf{Var}(Y)/2}$.
Therefore, when $X\perp Y$, we can say: $$\mathsf E(e^{(tX)Y}\mid X)~{=\mathsf M_{Y\mid X}(tX)\\=e^{tX\mathsf E(Y\mid X)+t^2X^2\mathsf {Var}(Y\mid X)/2}\\=e^{tX\mathsf E(Y)+t^2X^2\mathsf {Var}(Y)/2}}$$ 
And when $Y$ is standard normal, this becomes: $$\mathsf E(e^{tXY}\mid X)~=e^{t^2X^2/2}$$ 
