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?
$$d(X|Y) = \lim\_{m \rightarrow \infty} \frac{H(\hat{X}^{(m)}|Y)}{m} = \int\_\mathcal{Y} d(X|Y=y)dP\_Y(y) = \mathbb{E}\_{Y\sim P\_Y}(d(X|Y=Y))$$



Conditional entropy - solve example
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,
$$d(X|Y)=\int\_{\mathbb R}d(X|y)P(Y\in dy)=P(Y=0)=\frac{c\_1}b=\frac{\min(b,c)}b.$$


