Let $U$ and $V$ be two independent random variables where $E[U]=$ and $E[U^2]=1$ and where $V$ is standard Gaussian.
We also let $W=U+V$. How to compute the following expectation or find an an upper bound \begin{align} E\left[ e^{{(W-U)^2-(W-aU)^2}}\right] \end{align} where $a>0$. Observer, that the above can be re-written like \begin{align} E\left[ e^{{(W-U)^2-(W-aU)^2}}\right]&=E\left[ e^{{V^2-(1-a)^2U^2-2(1-a)UV-V^2}}\right]\\ &=E\left[ e^{{-(1-a)^2U^2-2(1-a)UV}}\right] \end{align}
The question is what to do next? For example, since we want to upper bound Jensen's inequality does not apply.
I was thinking Cauchy Schwarz
\begin{align} E\left[ e^{{-(1-a)^2U^2-2(1-a)UV}}\right] \le \sqrt{E\left[ e^{{-2(1-a)^2U^2}}\right] E \left[ e^{e^{{-4(1-a)UV}}} \right]} \end{align}
But what to do next? Also, is there a way of not using Cauchy Schwarz?
Thank you
From the MGF for the standard normal random variable, $$ \eqalign{E\left[ \exp(-(1-a)^2 U^2 - 2(1-a)UV)\right] &= E \left[ \exp(-(1-a)^2 U^2)\; E\left[\exp(- 2(1-a)UV)|U\right]\right]\cr &= E\left[ \exp((1-a)^2 U^2)\right]} $$
Unfortunately, you didn't give us enough information about $U$ to say much about this. It could be infinite.