*Bounty: 50*

*Bounty: 50*

Suppose I have a random variable $X sim f_{X}(x mid lambda)$ with support over $(0, infty)$ and I find the Fisher information in $X$ about $lambda$, i.e.,

$$I_{X}(lambda)=mathbb{E}left[left(dfrac{partialell_X}{partiallambda}right)^2midlambda right]$$

where $ell_X$ is the log-likelihood of $X$, which is just merely $ell_X(lambda) = log f_{X}(x mid lambda)$.

Now let $Y = text{floor}(X)$, i.e., the rounded-down-to-the-nearest-integer version of $X$. Can I make any claims about $I_Y(lambda)$?

This arose in a qualifying exam solution as follows: suppose $X sim text{Exp}(lambda)$, i.e.,

$$f_{X}(x) = lambda e^{lambda x}cdot mathbf{1}*{(0, infty)}(x)$$
and let $Y = text{floor}(X)$. Then $I*{X}(lambda) = 1/lambda^2$ and $I_{Y}(lambda) = e^{-lambda}/(1-e^{lambda})^2$.

**Furthermore, since $Y$ is a function of $X$, $I_{Y}(lambda) leq I_{X}(lambda)$. Why is this? Is there a theorem that I don’t know about?**

I’ve tried asking about how to compute this inequality directly, but showing this isn’t easy given timing on a qualifying exam, and it would be more useful if I understood why $I_{Y}(lambda) leq I_{X}(lambda)$ follows from $Y$ being a function of $X$.

**EDIT**: I have managed to find one mention of this inequality at http://cs.stanford.edu/~ppasupat/a9online/1237.html:

For other statistics $T(X)$, $I_{T}(theta) leq I_{X}(theta)$.

Alas, no proof.