Show $\mathbb{E}\left[\left|\sum_{i=1}^n X_iY_i\right|\right] \leq 2\mathbb{E}\left[\left|\sum_{i=1}^n Y_i\right|\right]$

In Exercise 5.9 of this book, it is stated that if $Y_1,\ldots,Y_n$ are independent random variables with finite variance and let $X_1,\ldots, X_n$ be independent Rademacher random variables (i.e., $\mathbb{P}[X_i = 1] = \mathbb{P}[X_i = -1] = 1/2$ for all $i$). Suppose that $(X_i)$ and $(Y_i)$ are independent as well, then $$\mathbb{E}\left[\left|\sum_{i=1}^n X_iY_i\right|\right] \leq 2\mathbb{E}\left[\left|\sum_{i=1}^n Y_i\right|\right].$$ I think the hint is to use a symmetrization argument but I still cannot figure out a way to do it. Thanks for any help!

Solution 1:

The result is false as stated (the book is wrong). Namely, let $n=2$, $Y_1 = -5$ and $Y_2 = 5$. However, if the random variables are $Y_i$ are zero-mean, then you can indeed use symmetrization, and it is done in the book in chapter 11. Let $Y_i'$ be an i.i.d. copy of $Y_i$. Then, as done in the book in the proof of lemma 11.4:

\begin{align*} E|\sum_{i=1}^n X_i Y_i| &= E|\sum_{i=1}^n X_i Y_i - X_i E Y_i'|\\ &\leq E|\sum_{i=1}^n X_i (Y_i - Y_i')|\\ &=E|\sum_{i=1}^n (Y_i - Y_i')|\\ &\leq 2 E|\sum_{i=1}^n Y_i| \end{align*}