Proof of $\mathbb E[\exp( \lambda XY)]=\mathbb E[\exp( \lambda^2 X^2)/2]$, where $X,Y$ are independent standard normal random variables I have a heuristic proof of the fact $\mathbb E[\exp( \lambda XY)]=\mathbb E[\exp( \lambda^2 X^2)/2]$,where $X,Y$ are independent standard normal random variables as follows:
Using the fact that $\mathbb E[\exp( \lambda Z)]=\exp(\lambda^2/2)$, where $Z$ is standard normal.
By conditioning, we have 
$\mathbb E[\exp( \lambda XY)]=\mathbb E (\mathbb E[\exp( \lambda XY)\mid X])$, at this point, one may treat $X$ as a constant (heuristically I take $\mathbb E[\exp( \lambda XY)\mid X]=\exp(\lambda^2 X^2/2) $, but is it true and why is it true? ) and use the formula above to get $E[\exp( \lambda^2 X^2)/2]$. 
But how to make this argument formal and rigorous? Any other proof will also be appreciated!
 A: Write 
$$\begin{align}
E(\exp(\lambda XY)) &= \int \exp(\lambda xy)dP_{(X,Y)}(x,y) \tag 1 \\
&= \int \int \exp(\lambda xy)dP_X(x) dP_Y(y) \tag 2\\
&= \int \left(\int \exp(\lambda xy)dP_Y(y)\right) dP_X(x) \tag 3\\
&= \int E(\exp(\lambda xY)) dP_X(x)\\
&= \int \exp( \lambda^2 x^2)/2\; dP_X(x)\\
&= E(\exp( \lambda^2 X^2)/2)
\end{align}$$
$(1)$: law of the unconscious statistician
$(2)$: independence of $X$ and $Y$
$(3)$: Tonelli's theorem
A: There is not even a need for integrals. One can perfectly rigorously manipulate expectations as long as one specifies the 'space' over which the expectation is taken. And this algebraic proof applies equally to all random variables, whether discrete or continuous or mixed. In particular,$\def\ee{\mathbb{E}}$ let $\ee_{Y|X}(f(X,Y))$ denote the expectation of $f(X,Y)$ over the conditional space of $Y$ given $X$ (i.e. treating $X$ as constant in evaluating the expectation). Similarly for more variables.
Firstly, note that if $Y$ is independent of $X$ then $\ee_{Y|X}(f(X,Y)) = \ee_Y(f(X,Y))$. If $X,Y$ are not independent this may not hold, for example if $X = Y$ then $\ee_{Y|X}(X-Y) = 0$ whereas $\ee_Y(X-Y) = X-\ee_Y(Y)$.
Secondly, the law of total expectation can be used as an axiom, namely that for any random variables $X,Y,...$ we have $\ee_{X,Y,...}(f(X,Y,...)) = \ee_X(\ee_{Y,...|X}(f(X,Y,...)))$.
With this, it is trivial to prove the desired result purely algebraically.
For any independent $X,Y \sim N(0,1)$, we have $\ee_{X,Y}(\exp(λXY))$ $= \ee_X(\ee_{Y|X}(\exp(λXY)))$ $= \ee_X(\ee_Y(\exp(λXY)))$ $= \ee_X(\exp((λX)^2/2))$.
Your result is wrong because you put the brackets in the wrong place.
