# Joint differential entropy of sum of random variables: $h(X,X+Y)=h(X,Y)$?

I see the following simplification used frequently in the literature, but I have not been able to verify it.

Let $X$ and $Y$ be absolutely continuous (i.e. they have pdfs) $\mathbb{R}^d$-valued random variables. Assume the joint variable $(X,X+Y)$ is absolutely continuous on $\mathbb{R}^{2d}$. Then $$h(X,X+Y)=h(X,Y).$$

Here $h$ signifies differential entropy, defined by $$h(W)=-\int_{\mathbb{R}^{d_W}}f_W(w)\log(f_W(w))\ dw$$ whenever $W$ is an $\mathbb{R}^{d_W}$-valued random variable with pdf $f_W$.

Note1: $X$ and $Y$ are not assumed to be independent.

Note2: Examples where the lhs is finite but the rhs is not defined would be accepted as a counterexample.

I am also wondering, if the statement can be proved, then is it more generally true that $$h(X,g(X,Y))=h(X,Y)$$ where $g$ is a deterministic function of its arguments?

This question is similar, but seems to concern Shannon entropy (i.e. discrete variables). Shannon Entropy and Differential Entropy have different sets of properties as discussed in these links answer1, answer2, question1,and question2.

Another proof to that of @HaarD is the following.

Using the chain rule,

\begin{align} h(X, X+Y) &= h(X) + h(X+Y\mid X)\\ &=h(X)+h(Y|X)\\ &=h(X,Y), \end{align} where the first equality is application of the chain rule, the second equality holds because adding a constant to a random variable does not change its entropy, and the third equality is again by application of the chain rule.

This result does not generalize for arbitrary $g(X,Y)$.

• I'm a little unsure about the reasoning behind the second equality (adding a constant to a random variable does not change its entropy). Would $h(X^2+Y|X)=h(Y|X)$ for the same reason? Commented Feb 5, 2018 at 22:33
• @cantorhead yes Commented Feb 5, 2018 at 22:38
• Then by doing the chain rule in the reverse order we get $$h(X,X^2+Y)=h(X,Y).$$ If this is true then it seems more general than the other answer based on linear change of variables. Commented Feb 5, 2018 at 22:42
• @cantorhead indeed Commented Feb 5, 2018 at 22:52
• Can you give more detail in your answer about why the second equality holds? My original motivation for asking the question was actually to understand why $h(X+Y|X)=h(Y|X)$. You've said that its related to translation invariance. Can you show step by step how translation invariance implies $h(X+Y|X)=h(Y|X)$? Commented Feb 6, 2018 at 2:31

Use the fact that if $W$ has a pdf and $A$ is a linear transformation then \begin{align*} h(AW)&=h(W)+\log|\mathrm{det}A|. \end{align*}

In this case, let $W=[X \ \ Y]^T$, a vector valued variable in $\mathbb{R}^{2d}$. Let \begin{align} A=\left[\begin{array}{c c} I_d & 0 \\ I_d & I_d \end{array}\right], \end{align} where $I_d$ is the $d\times d$ identity block. Then $AW=[X \ \ X+Y]^T$. Therefore, \begin{align} h(X,X+Y)&=h(AW) \\ &= h(W)+\log|\mathrm{det}A| \\ &= h(X,Y) + \log(1) \\ &= h(X,Y) \end{align}