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I want to quantify how mutual information depends on the variance of one of the variables. Here's a simple test I came up with

$$X \sim U(0, 1)$$ $$Y \sim U(0, 1)$$ $$Z = (X + Y) / 2$$

where $U$ denotes the uniform distribution. I am interested in finding an analytical expression for the mutual information $I(\alpha X, Z)$ for some positive value $\alpha$. I need this test to check the performance of a library that performs numerical calculation of mutual information.

Edit: I don't actually care what $U$ is. If it is simpler to calculate the result for standard normal distributions, you may assume that instead.

Edit 2: Perhaps it is possible to produce a result for a general probability distribution. For example, according to wiki article,

$$H(\alpha X) = H(X) + \log(|\alpha|)$$

Perhaps anybody knows how to prove this? If one can prove this, and a similar result for $H(\alpha X, Z)$, then the mutual information would be a simple subtraction

Edit 3: The result for univariate entropy can be proven by considering a pdf transformation. If $y = \alpha x$, then $\rho_y(y) = \frac{1}{|\alpha|} \rho_x(y / \alpha)$. Then one can simply integrate the definition of the differential entropy to obtain the desired result. The extension to multivariate case appears to be somewhat more difficult

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As you weren't considerate about the underlying distributions, lets assume that $$X\sim \mathcal{N}(0,\sigma^2_X) \text{ , } Y\sim \mathcal{N}(0,\sigma^2_Y)$$

Just to ease out computation, for a non scaled additive Gaussian Channel, $Z=X+Y \sim \mathcal{N}(0,\sigma^2_X+\sigma^2_Y)$ whose differntial entropy is $$h(Z)=\frac{1}{2}\log\left[2\pi e (\sigma^2_X+\sigma^2_Y)\right]$$

By definition $I(X;Z)=h(Z)-h(Z|X)=\frac{1}{2}\log\left[2\pi e (\sigma^2_X+\sigma^2_Y)\right]-\frac{1}{2}\log\left[2\pi e (\sigma^2_Y)\right] = \frac{1}{2}\log\left(1+\frac{\sigma^2_X}{\sigma^2_Y}\right)$

Also note that, $\text{Var}(\alpha X)=\alpha^2\sigma^2_X$ and hence $$I(\alpha X;Z)=\frac{1}{2}\log\left(1+\frac{\alpha^2\sigma^2_X}{\sigma^2_Y}\right)$$

EDIT: Based on comments:

(TBA)

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  • $\begingroup$ I think you have a mistake. You put $\alpha^2$ in front of $\sigma_x^2$ in your final answer, but that $\sigma_x^2$ came from $H(Z)$, and $Z$ was not scaled. Note that we only scale the random variable X when we plug it into the mutual information equation, but when we use it to compute Z, we do not scale it. Would you be so kind to address this, and clean up the proof to make it more step-by-step. For example, it would be great to show an explicit derivation of $H(Z|X)$, it is the most important part of the proof $\endgroup$ Jun 25, 2020 at 15:27
  • $\begingroup$ I see what you are asking for. I have some mistakes in my recent calculations. I will add them later $\endgroup$
    – Dinesh
    Jun 26, 2020 at 7:23

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