Thread: Compare the expectation of two functions of same r.v.

Hello everybody,

does anyone have an idea how to prove the following inequality

$\displaystyle E(\log(1+c_1(e^{Z+b}-1)))\geq E(\log(1+c_2(e^{Z+b}-1)))$?

where $\displaystyle Z$ is normally distributed with a non-negative mean and $\displaystyle c_1,c_2$ are constants with $\displaystyle c_1,c_2 \in [0,1],c_1\geq c_2$
$\displaystyle b>0$ is also a constant.

Thanks in advance

Hey Juju.

Try showing that the function is monotonic increasing and using that, show that if CDF_X(p) > CDF_Y(p) for all p in [0,1] then E[X] > E[Y].

Thanks!

Unfortuanately, I do not completely understand your hint. The function $\displaystyle log(1+c(e^{Y-b}+1)))$ is increasing and convex in $\displaystyle Y.$

The expectation is given by
$\displaystyle \int log(1+c(e^{y-b}+1)))\frac{1}{\sqrt{2 \pi \sigma^2}}e^{-0.5\frac{(x-\mu)^2}{\sigma^2}} dx$

Or do you mean that I should consider the cdf of the function of r.v. $\displaystyle log(1+c(e^{Y-b}+1)))$?
$\displaystyle P(log(1+c_1(e^{Y-b}+1))) \leq y)=P(e^Y\leq \frac{e^y-1-c_1}{c_1})$
But now it does not hold
$\displaystyle \frac{e^y-1-c_1}{c_1}\leq \frac{e^y-1-c_2}{c_2}\; \forall y$

Yes, consider the CDF of the log function of the random variable. That was my initial intent and idea for you,