# Conditional entropy for independent random variables

Is it true that for $$Y$$ and $$Z$$ independent, the conditional entropy $$H(X\mid Y,Z)$$ satisfies $$H(X\mid Y,Z) = H(X\mid Y) + H(X\mid Z)$$ where $$H(X\mid Y) = \sum_{y\in\mathcal{Y}}p(y)H(X\mid Y=y)$$ (similarly for $$H(X\mid Z)$$)?

Is this obvious? Can someone point me in the direction of a proof?

• No, it is not true. Take the example of $Y$ and $Z$ being independent Bernoulli-1/2 bits, and $X = Y \oplus Z$, where $\oplus$ is the XOR function. It is easy to show that $H(X|Y) = H(X|Z) = 1$ but $H(X|Y,Z)$ is obviously $0$. – stochasticboy321 Nov 8 at 16:11
• Thanks @stochasticboy321. Do you know of any identities that allow me to rewrite $H(X\mid Y,Z)$ in a different form? – jonem Nov 8 at 16:14
• Sure, but their usefulness will depend on what you're trying to do with them, so I'm not sure if what I think of will be helpful. For example, using the chain rule of entropies, $H(X|Y,Z) = H(X,Y,Z) - H(Y,Z) = H(X,Y,Z) - H(Y) - H(Z)$ (the last equality uses independence of $Y,Z$). – stochasticboy321 Nov 8 at 16:21
• @stochasticboy321 That's useful, thanks! – jonem Nov 8 at 16:29