# Quadratic estimation error differential entropy

I am reading the book "Elements of information theory" by thomas M. Cover and Joy A. Thomas, second edition. In page 255 of the book there is a theorem:

For any random variables $X$ and estimator $\hat{X}$, $E(X - \hat{X})^2 \geq \frac{1}{2\pi e}e^{2h(X)}$. The proof uses at its core the fact that $var(X) \geq \frac{1}{2\pi e}e^{2h(X)}$.

Edit: I include here the proof:

$E(X-\hat{X})^2 \geq \min_{\hat{X}}E(X-\hat{X})^2 = E(X-E(X))^2 = var(X) \geq \frac{1}{2\pi e}e^{2h(X)}$

Right after the proof there is a corollary that says: Given side information $Y$ and estimator $\hat{X}(Y)$, it follows that: $E(X - \hat{X}(Y))^2 \geq \frac{1}{2\pi e}e^{2h(X|Y)}$, however, no further explanation is supplied for how to derive this corollary from the previous theorem.

I assume that if I fight the equations for a while I might be able to derive the corollary from something, however I can't manage to find a simple way to derive it from the previous theorem. I wonder if I miss something very simple here.

So far my best attempt was to split $(X - \hat{X}(Y))$ into $(X - Y) + (Y - \hat{X}(Y))$.

Apply the given result, which works for arbitrary $X$, to the conditional distribution $X(Y)$ whose differential entropy is $h(X|Y)$.

• Two questions: 1. Is $X(Y)$ in your notation some kind of a random variable? If so, how is it defined? 2. The given result assumes that $\hat{X}$, the estimator of $X$, is constant (The proof uses this fact). In the second result it seems to me that $\hat{X}(Y)$ doesn't have to be a constant. It depends on $Y$. Hence I can't see how to apply the first result to the second one using your suggested method.
– real
May 18, 2016 at 8:13

I propose here an answer using some calculations.

We want to prove that $E[(X-\hat{X}(Y))^2] \geq \frac{1}{2\pi e}e^{2h(X|Y)}$. Assuming that $X,Y$ are continuous random variables, we get:

$$E[(X-\hat{X}(Y))^2] = \int_{x,y} \left(x-\hat{X}(y)\right)^2 f(x,y)dxdy = \int_{x,y} \left(x-\hat{X}(y)\right)^2 f(y)f(x|y)dxdy$$

$$= \int_{y} f(y)\left(\int_{x}\left(x-\hat{X}(y)\right)^2 f(x|y)dx\right)dy$$

Using the given theorem, we know that $$\int_{x}\left(x-\hat{X}(y)\right)^2 f(x|y)dx \geq \frac{1}{2\pi e}e^{2h(X|Y=y)}$$

Hence $$E[(X-\hat{X}(Y))^2] \geq \int_{y} f(y)\frac{1}{2\pi e}e^{2h(X|Y=y)}dy = \frac{1}{2\pi e}\int_{y} f(y)e^{2h(X|Y=y)}dy$$

By Jensen inequality, $$\int_{y} f(y)e^{2h(X|Y=y)}dy \geq e^{\int_{y}2h(X|Y=y)f(y)dy}$$

And so

$$E[(X-\hat{X}(Y))^2] \geq \frac{1}{2\pi e}e^{\int_{y}2h(X|Y=y)f(y)dy} = \frac{1}{2\pi e}e^{2h(X|Y)}$$

As required.

$$h(X|Y) = h(X-E(X|Y)|Y),$$ since shift (as opposed to scaling) does not change differential entropy. Now apply previous result (and Jensen's inequality) in the following fashion: $$h(X|Y) = h(X-\hat X|Y) \leq E\left(\frac 12 \log (2\pi e E( (X - \hat{X})^2|Y)) \right) \leq \frac 12 \log (2\pi e E(X-\hat{X})^2)).$$