Conditional expectation of exponential function

Question

I want to calculate the conditional expectation: $$\mathbb{E}\left(e^{X_1 Y_2 + X_2 Y_1}\mid \sigma(X_1)\right)$$ Here the $X_i$ are independent identical standard normal distributions and the $Y_i$ are independent symmetric Bernoulli random variables.

I started by slightly rewriting this conditional expectation to: $$\mathbb{E}(e^{X_1 Y_2}e^{X_2 Y_1}\mid \sigma(X_1))$$ Then I used independence to get: $$\mathbb{E}(e^{X_1 Y_2}\mid \sigma(X_1))\mathbb{E}(e^{X_2 Y_1}\mid \sigma(X_1))$$

For the second part I again used independence such that the conditional expectation turns into a regular expectation: $\mathbb{E}(e^{X_2 Y_1})$. If this is still correct, my difficulty lies in analysing the first expression $\mathbb{E}(e^{X_1 Y_2}\mid \sigma(X_1))$. The take out what is known (TOK) property is defined by: $$\mathbb{E}(XY\mid \mathcal{G}=Y\mathbb{E}(X\mid \mathcal{G})$$ whenever $Y$ is $\mathcal{G}$ measurable. So I would like to take out the $X_1$, though the multiplication in the exponential prohibits the splitting as in the definition.
I hope it is clear where I am stuck. All help would be appreciated!

Thanks in advance

Answer 1

$E(e^{X_1Y_2}|\sigma (X_1))=(e^{X_1}+e^{-X_1})) /2$ since $Ee^{xY_1}=(e^{x}+e^{-x})) /2$ for all $x \in \mathbb R$. Hence the answer is $[(e^{X_1}+e^{-X_1})) /2] Ee^{X_2Y_1}$. I think you know how to compute $Ee^{X_2Y_1}$.