I have a standard normal random variable $X\sim\mathcal{N}(0,1)$ and an event $E$ with $\mathbb{P}(E)=p$, and this event is about $X$ and some other variables. I am interested in upper-bounding the expression $\mathbb{E}[1_E\cdot X^2]$. The random variables $1_E$ and $X$ are $not$ independent.
One obvious solution is using the Cauchy-Schwarz inequality and it gives $\mathbb{E}[1_E\cdot X^2]\leq\sqrt{\mathbb{E}[1_E]\mathbb{E}[X^4]}=\sqrt{3p}$. This is, however, not strong enough for what I need. Ideally, I would want to have an upper-bound that involves $p$, not $\sqrt{p}$.
In order to get this, I was thinking about using Holder's inequality: for any $p,q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, $\mathbb{E}[|XY|]\leq(\mathbb{E}[X^p])^{1/p}(\mathbb{E}[X^q])^{1/q}$ with $p=1+\varepsilon$ any take the limit $\epsilon\rightarrow 0_+$. The main steps are as follows:
\begin{align*}
\mathbb{E}[1_E\cdot X^2]&\leq (\mathbb{E}[1_E])^{1/(1+\varepsilon)}(\mathbb{E}[X^{2(1+\varepsilon)/\varepsilon}])^{\varepsilon/(1+\varepsilon)}\\
&=p^{1/(1+\varepsilon)}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(x^2)^{(1+\varepsilon)/\varepsilon}e^{-x^2}dx\right)^{\varepsilon/(1+\varepsilon)}\\
&=p^{1/(1+\varepsilon)}\left(\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}x^{1/2+1/\varepsilon}e^{-x}dx\right)^{\varepsilon/(1+\varepsilon)}\\
&=p^{1/(1+\varepsilon)}\left(\frac{1}{\sqrt{2\pi}}\Gamma\left(\frac{3}{2}+\frac{1}{\varepsilon}\right)\right)^{\varepsilon/(1+\varepsilon)}
\end{align*}
Taking the limit $\varepsilon\rightarrow 0_+$, one gets $\mathbb{E}[1_E\cdot X^2]\leq p$.
I am not 100% sure about this very last step, is it correct? And if not, are there any other approaches that come to mind that I should try? Thank you!
Best Answer
This approach will fail in general. With regard to your specific computation, I believe the last limit $\varepsilon\to 0^+$ will converge to $+\infty$, based on the asymptotic approximation $$ \Gamma(x) \sim \sqrt{2\pi} x^{x-\frac{1}{2}}e^{-x} ~\text{as}~x\to\infty. $$ And in general, you should not expect this limiting approach to work with Holder's inequality - if it did it would obviate the need for the $(\infty,1)$ endpoint case for Holder. And clearly we can't use the $(\infty,1)$ endpoint.
More generally, you cannot obtain the inequality you're looking for without specifying more information about $E$. As an example, let $p(x)$ denote the PDF of $X$, and take $F(t)$ to be the complementary CDF of $X$, $$ F(t) = 1 - P(X\leq t) = P(X>t). $$ Define $E_p = (T_p,\infty)$ where $F(T_p) = p$; this is manifestly well-defined. Then by definition, $$ \mathbb{E}[1_{E_p}] = \int_{T_p}^\infty p(x)~dx = F(T_p) = p. $$ Now we compute: $$ \mathbb{E}[1_{E_p}X^2] = \int_{T_p}^\infty x^2p(x)~dx \geq T_p^2\int_{T_p}^\infty p(x)~dx = T_p^2F(T_p) = T_p^2p. $$ We can write $T_p = F^{-1}(p)$. Now send $p\to 1^-$. Since $F(t) = 1$ iff $t = -\infty$, we find that $T_p^2\to\infty$. We conclude that there is no constant $C$ independent of $p$ with $$ T_p^2p \leq\mathbb{E}[1_{E_p}X^2]\leq Cp. $$ Normality of $X$ was not invoked here other than $F(t)=1$ iff $t=-\infty$, but hopefully it is apparent that specifying $p(x) = (2\pi)^{-1/2}e^{-x^2/2}$ would not change anything.
The deeper issue you are facing is that the bound you are seeking is too strong if you are not willing to invoke something about how $E$ and $X^2$ are related. If $E$ and $X^2$ were independent, then you could trivially get your result by $$ \mathbb{E}[1_EX^2] = \mathbb{E}[1_E]\mathbb{E}[X^2] = p $$ where we have used $X\sim \mathcal{N}(0,1)$. This is significantly stronger than Cauchy-Schwarz or Holder, since $\mathbb{E}[Y] \leq \mathbb{E}[Y^q]^{1/q}$ (where $q\in[1,\infty]$ and $Y\geq 0$). So by seeking an upper bound proportional to $p$, you're trying to achieve a bound (up to constants) that normally only works because of independence - in a sense, what you are trying to show is that $1_E$ and $X^2$ are "close to independent." You shouldn't expect to be able to do this without using some specific information about the joint distribution of $1_E$ and $X^2$ that supports this "close to independence" notion.