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Let $X$ and $Y$ be two random variables, where $Y=f(X)$ is a deterministic function of $X$. Furthermore suppose $X,Y$ are continuous and that $f$ is smooth.

Is the mutual information between $X$ and $Y$ well defined in this case? I have the intuition that it can only be zero (if $f$ is a constant function) or infinity.

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    $\begingroup$ The usual definitions of MI apply only to discrete variables or jointly continuous variables or, more generally, measures for which one is absolutely continuous with respect to the other. This case is none of the above. What definition do you have in mind, then? $\endgroup$
    – whuber
    Commented May 7, 2020 at 12:42
  • $\begingroup$ It shouldn't be infinite because for example $I(X,X) = H(X)$. $\endgroup$
    – Simone
    Commented May 7, 2020 at 12:54
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    $\begingroup$ @whuber Say we take a limit, $Y = f(X) + \epsilon z$ where $z$ is a standard normal and $\epsilon \rightarrow 0$. In that case the mutual information tends to infinity, correct? $\endgroup$
    – a06e
    Commented May 7, 2020 at 13:23
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    $\begingroup$ @Simone That is not correct for continuous distributions. $\endgroup$
    – a06e
    Commented May 7, 2020 at 13:26
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    $\begingroup$ I think your intuition is correct because under independence and non-constancy of a smooth $f,$ the joint distribution has a continuous component (in its Lebesgue decomposition as a measure) and that will be infinitely distant from the singular distribution of $(X,f(X)).$ $\endgroup$
    – whuber
    Commented May 7, 2020 at 21:49

2 Answers 2

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Let's suppose that $Y=f(X)+\epsilon Z$ where $Z$ is a standard normal, and we take the limit $\epsilon\rightarrow 0$

The mutual information equals $$I = H(Y) - H(Y|X),$$ where $H(Y)$ is the entropy of $Y$ and $H(Y|X)$ is the conditional entropy of $Y$ on $X$. Since $Z$ is normal, $H(Y|X)=\frac{1}{2}\ln(2\pi e\epsilon^2)$.

If $f$ is not constant and smooth, then $H(Y)$ stays positive and finite. But as $\epsilon\rightarrow 0$ we find $H(Y|X)\rightarrow-\infty$ and therefore $I=\infty$ in this limit.

If $f(X)=c$ is a constant, then $H(Y)=H(Y|X)$ for any positive $\epsilon$, and therefore in this case the limit is $I=0$.

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Although $f$ is stated as neural network, I think theorem 1 of paper https://arxiv.org/pdf/1802.09766.pdf will answer part of your question, i.e. , if $f$ is bi-Lipschitz then $I(Y, X) = \infty$.

Furthermore, $f(X) = c$ is not a bi-Lipschitz function. A example of piecewise linear function, a general form of constant function, is present in section 4.3; I think at least for that example the mutual information is a finite constant for piecewise linear function.

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