Is $g(u)= \frac{E [ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} ] }{E [ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} ]}$ decreasing in $u$

Question

Let $X$ be a positive random variable, let us define a function \begin{align} g(u,a)= \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]}. \end{align}

Question: Can we show that the above integral is monotonically decreasing in $u$ ( for $u>0$ ) for all $a > 1$.

Note that $X$ here represents the variance of standard normal. That is we consider the variance to be a random variable.

I can show that $g(u,a)$ is bounded by $1$ and continuous but can not establish that it is decreasing. Also, note that the function $g(u,a)$ is symmetric around $u=0$.

What I tried:

I was able to show that for $p,q\ge 1$ and $\frac{1}{q}+\frac{1}{p}=1$ and $a^2 \ge \frac{1}{p}$ we have \begin{align} g(u,a) \le \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}}, \end{align}

where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$.

Proof: By using Holder's inequality \begin{align} E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] &=E \left[ \frac{1}{\sqrt{X}} e^{-\frac{ (a^2-\frac{1}{p})u^2}{2X}} e^{-\frac{ \frac{1}{p}u^2}{2X}} \right] \\ &\le E^\frac{1}{q} \left[ \frac{1}{\sqrt{X}} e^{-\frac{ q(a^2-\frac{1}{p})u^2}{2X}} \right] E^\frac{1}{p} \left[ \frac{1}{\sqrt{X}} e^{-\frac{ u^2}{2X}} \right]. \end{align} Therefore, \begin{align} g(u,a) \le \left( \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{q(a^2-\frac{1}{p})u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]} \right) ^\frac{1}{q} &= \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}}, \end{align} and

\begin{align} g(u,a) \le \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}}, \end{align}

where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$. Thank you. Looking forward to seeing your approaches.

Answer 1

This is false in general.

Change the variables as in my comment: $1/(2X) \to X$, $a^2 \to a$, $u^2 \to u$. Then the problem is to show that $$ f(u) = \frac{\mathsf{E}[\sqrt{X}e^{-auX}]}{\mathsf{E}[\sqrt{X}e^{-uX}]}. $$ decreases.

Set $a=1.1$ and let $X=1$ or $100$ with probability $1/2$. Then $$ f(u) = \frac{e^{-1.1u} +10e^{-110u}}{e^{-u} +10e^{-100u}}. $$ However, it is not decreasing for small values of $u$: Wolfram Alpha

Answer 2

If by "monotonically decreasing" you mean decreasing for $u \in (-\infty, \infty)$, then the answer is trivially no: $g$ is an even function in $u$ on the reals, thus it is either constant or it has a local extremum. And we can already tell intuitively that $g$ is not constant.