# Help me understand this line of proof which concerns a marginal expectations in the presence of independent variables

The following comes From Martin Wainwright's book on High-Dimensional Statistics, page 41 on Lipschitz functions of Gaussian variables.

It first begins by the following Lemma:

Suppose $$f:\mathbb{R}^n\to\mathbb{R}$$ is differentiable. Then for any convex function $$\phi:\mathbb{R}\to\mathbb{R}$$, we have $$$$\mathbb{E}[\phi(f(X)-\mathbb{E}[f(X)])]\leq\mathbb{E}\left[\phi\left(\frac{\pi}{2}\langle\nabla f(X),Y\rangle\right)\right]$$$$ where $$Y,X\sim N(0,I_{n})$$ are standard multivariate Gaussian, and independent.

It then goes on to show this:

For any $$\lambda\in\mathbb{R}$$, applying the inequality above to the convex function $$t\to\exp(\lambda t)$$ yields \begin{align} \mathbb{E}_X\left[\exp\left(\lambda\{f(X)-\mathbb{E}[f(X)]\}\right)\right]&\leq \mathbb{E}_{X,Y}\left[\exp\left(\frac{\lambda\pi}{2}\langle \nabla f(X),Y \rangle\right)\right]\\ &=\mathbb{E}_X\left[\exp\left(\frac{\lambda^2\pi^2}{8}\lVert\nabla f(X)\rVert_2^2\right)\right] \end{align} where we have used the independence of $$X$$ and $$Y$$ to first take the expectation over $$Y$$, and the fact that $$\langle Y,\nabla f(X)\rangle$$ is a zero-mean Gaussian variable with variance $$\lVert \nabla f(x)\rVert_2^2$$.

I do not follow the equality in the second line of the most recent lines of the equations. Some clarification here would be very much appreciated.

• I think you must have omitted a "$\lambda$" from the inequality at the end. You might find this problem to be more accessible by first considering the case $n=1.$
– whuber
Mar 16, 2021 at 17:50
• Good point. I have added the lambda. I shall give it a try by considering n=1. Thanks.
– Carl
Mar 16, 2021 at 17:54