# Conditional expectation of exponential function

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!

$$E(e^{X_1Y_2}|\sigma (X_1))=(e^{X_1}+e^{-X_1})) /2$$ 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 $$Ee^{X_2Y_1}$$.