# Conditional entropy - solve example

Given a random variable $$X$$ that is uniformly distributed on $$[-b,b]$$ and $$Y=g(X)$$ with $$g(x) = \begin{cases} 0, ~~~ x\in [-c,c] \\ x, ~~~ \text{else}\end{cases}$$

Now I want to compute the information dimension $$d(X), d(Y)$$ and the conditional information dimension $$d(X|Y)$$ and show that $$d(X) = d(X|Y) + d(Y)$$ in this case.

The information dimension is defined as $$d(X) = \lim_{m\rightarrow \infty} \frac{H(\hat{X}^{(m)})}{m}$$ with $$\hat{X}^{(m)} := \frac{\lfloor2^m X \rfloor}{2^m}$$ the quantization of $$X$$.

For a discrete distribution, $$d(X) = 0$$, and for a continuous one-dimensional distribution, $$d(X) = 1$$. For a mixed distribution with discrete and continuous components of the form $$P_X = d P_X^{(ac)} + (1-d) P_X^{(d)}$$, the information dimension is $$d(X)=d$$.

Now I know, that the random variable X has a continuous component $$\Rightarrow d(X) = 1$$. The distribution $$P_Y$$ is a discrete-continuous mixture: $$P_Y = \begin{cases} \frac{c}{b}, ~~~Y=0\\ \frac{1}{2b},~~~Y \in [-b,-c] \cap [c,b]\\ 0,~~~\text{else} \end{cases}$$ Therefore, $$d(Y)=\frac{b-c}{b}$$.

Now my question is the following: how do I compute the conditional information dimension? $$d(X|Y) = \lim_{m \rightarrow \infty} \frac{H(\hat{X}^{(m)}|Y)}{m} = \int_\mathcal{Y} d(X|Y=y)dP_Y(y) = \mathbb{E}_{Y\sim P_Y}(d(X|Y=Y))$$

From the context, it appears that $$b\in(0,\infty)$$ and $$c\in[0,\infty)$$ (and, likely, $$c\le b$$). Anyway, let $$c_1:=\min(b,c)$$. Note that
• With probability $$1$$, either $$Y=0$$ or $$c_1<|Y|\le b$$;
• The conditional distribution of $$X$$ given $$Y=0$$ is the uniform distribution on the interval $$[-c_1,c_1]$$ and hence $$d(X|Y=0)=1$$.
• The conditional distribution of $$X$$ given $$Y=y$$ with $$c_1<|y|\le b$$ is the Dirac distribution supported at point $$y$$ and hence $$d(X|Y=y)=0$$.
So, $$d(X|Y)=\int_{\mathbb R}d(X|y)P(Y\in dy)=P(Y=0)=\frac{c_1}b=\frac{\min(b,c)}b.$$