# Chain rule independent variables

Does the following identity $$I(X_1,X_2 ; Y_1, Y_2) = I(X_1; Y_1) + I(X_2; Y_2)$$ hold for mutual information for $$(X_1, Y_1)$$ and $$(X_2, Y_2)$$ independent?

Attempt: $$p(x_1, x_2, y_1, y_2) = p(x_1, y_1) \cdot p(x_2, y_2)$$ implies $$p(x_1, x_2) = p(x_1) \cdot p(x_2)$$ and $$p(y_1, y_2) = p(y_1) \cdot p(y_2)$$, where I use lower case $$p$$ to denote the probability of specific outcomes, e.g. $$p(x)$$ for $$\Pr\{X=x\}$$. Thus $$\sum_{x_1, x_2, y_1, y_2} p(x_1, x_2, y_1, y_2) \cdot \log \frac{p(x_1, x_2, y_1, y_2)}{p(x_1, x_2) \cdot p(y_1, y_2)} = \\ \sum_{x_1, x_2, y_1, y_2} p(x_1, x_2, y_1, y_2) \cdot \log \frac{p(x_1, y_1)}{p(x_1) \cdot p(y_1)} \cdot \frac{p(x_2, y_2)}{p(x_2) \cdot p(y_2)}$$

• Could you show what you tried? It would make it easier for us to guide you towards the answer. Also I believe on the right-hand side you have a typo since you dropped indices on the $Y$. Jun 23 at 8:22
• @adrien_vdb Thanks! I corrected the typo Jun 23 at 9:17
• Great thanks for adding it, let me give you two hints. 1) You can use the following property of the logarithm $\log(x\cdot y) = \log(x) +\log(y)$. and 2) Use the fact that probability distributions sum up to 1 --> $\sum_{x, y} p(x, y) = 1$ Jun 23 at 13:47