Let $X,Y\sim\mathcal N(0,1)$ be independent standard normal random variables. Compute the expectation $E[e^{X+Y}1_{X+Y>a}]$ where $1$ is an indicator function, and $a$ is some constant.
Without the indicator function it is straightforward to compute the expectations due to independence. That is I can write $E[e^{X+Y}] = E[e^{X}]E[e^Y]$ which gives me $e$ as the answer. However, the presence of indicator function makes this slightly complicated. Am I allowed to use law of iterated expectations and start with $E[E[e^{X+Y}1_{X+Y>a}|Y]]$? Any hint is appreciated. Thanks.
Well, yes, you may do that. $~\mathsf E\big(\mathrm e^{X+Y}\mathbf 1_{X+Y>a}\big) = \mathsf E\big(\mathrm e^Y\,\mathsf E(\mathrm e^{X}\mathbf 1_{X>a-Y}\mid Y)\big)$
However, if you can evaluate $\mathsf E(\mathrm e^{X}\mathbf 1_{X>a-Y}\mid Y)$, you could instead evaluate $\mathsf E(\mathrm e^{Z}\mathbf 1_{Z>a})$ when $Z\sim\mathcal{N}(0,2)$
Which is useful since $X+Y\sim\mathcal N(0,2)$