# Bounds on the expectation of a binary random variable function

Consider R independent binary random variables $Y^1, \ldots, Y^R$ over the space $\{-1,+1\}$ such that $\Pr(Y^j = 1) = p^j \geq 0.5$ and $\Pr(Y^J = -1) = 1 - p^j$, $\forall j = 1, \ldots, R$.

Consider also the following expectation:

\begin{align*} e = E[|\sum_{j=1}^R \text{logit}(p^j) Y^j|] \end{align*} where $\text{logit}(p^j) = \log(\frac{p^j}{1-p^j})$.

By applying the Jensen's inequality it is possible to obtain a lower bound $LB_e$ for $e$:

\begin{align*} E[|\sum_{j=1}^R \text{logit}(p^j) Y^j|] \geq |E[\sum_{j=1}^R \text{logit}(p^j) Y^j]| \end{align*}

How to identify an upper bound $UB_e$ for $e$ as tight as possible? Moreover, is it possible to express $UB_e$ as $k\cdot LB_e$? Thanks.

• With no further information on $(p^j)$, Cauchy-Schwarz inequality might yield the best one can get. – Did Jun 27 '13 at 6:31
• Thank you for your comment. The Cauchy-Scharwz inequality states that $E[XY]^2 \leq E[X^2]E[Y^2]$. In my case, I have a sum of (weighted) random variables, rather than a product. How can I use the inequality in this case? – burton0 Jun 27 '13 at 13:13
• Use $E[|X|]\leqslant E[X^2]^{1/2}$ and expand $X^2=\left(\sum\limits_j\ell_jY^j\right)^2$. – Did Jun 27 '13 at 13:22
• I assume your are using $E[|XY|] \leq \sqrt{E[X^2]E[Y^2]}$ where $Y = 1$ almost surely, right? If this is the case, I don't get how you can go from $|E[XY]| \leq \sqrt{E[X^2]E[Y^2]}$ (Cauchy-Schwarz inequality) to $E[|XY|] \leq \sqrt{E[X^2]E[Y^2]}$. The only way I see is to use the Jensen's inequality, but in that case you have $|E[XY]| \leq E[|XY|]$ and $|E[XY]| \leq \sqrt{E[X^2]E[Y^2]}$ and I don't see how to combine them. – burton0 Jun 27 '13 at 14:33
• Apply C-S to |X| and 1. – Did Jun 27 '13 at 14:41

Hint: if $X$ is a positive random variable, then $E[X]\geq 0$. In our case $X:=|\sum_{j=1}^R l_j Y^{j}|\leq \sum_{j=1}^R |l_j Y^{j}|:=Z$, with $l_j=logit(p_j)$ and so
$$E[Z-X]\geq 0$$
$$E[X]\leq E[Z].$$
The expectation value $E[Z]= \sum_{j=1}^R E[|l_j Y^{j}|]$ is an upper bound for $e$.
• Thank you for your answer. The bound you suggest reduces to $\sum_j^R l^j$, which is correct. However, I'm afraid it is not tight enough for my application scenario. Is there a way to tighten the bound? – burton0 Jun 26 '13 at 8:22