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I have two independent random variables $X_1$ and $X_2$. I want to find the differential entropy defined as $$H(X_1+X_2\mid X_1)=\int_{X_1} \int_{X_2} p_{X_1,X_2}(x_1,x_2)\log\left(\frac{1}{p_{X_1+X_2\mid X_1}(x_1+x_2\mid x_1)}\right) \, dx_1 \, dx_2.$$ In order to find this I need the formula for $$p_{X_1+X_2\mid X_1}(x_1+x_2\mid x_1).$$ I know that the conditional cumulative distribution function of $X_1+X_2$ given $X_1$ is given as follows $$F_{X_1+X_2\mid X_1}(x_1+x_2\mid x_1) = F_{X_2}(x_2).$$ But I do not how to show that $p_{X_1+X_2\mid X_1}(x_1+x_2\mid x_1) = p_{X_2}(x_2).$ Any help in showing that $p_{X_1+X_2\mid X_1}(x_1+x_2\mid x_1) = p_{X_2}(x_2)$ will be much appreciated. Thanks in advance.

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  • $\begingroup$ @MichaelHardy thanks for correction. I have edited it in my question. $\endgroup$ – Frank Moses Aug 13 '18 at 2:56
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Differentiate both sides of $$F_{X_1 + X_2 \mid X_1} (x_1 + x_2 \mid x_1) = F_{X_2}(x_2)$$ with respect to $x_2$.

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  • $\begingroup$ Thank you for your answer. I know that the right hand side will result in $p_{X_2}(x_2)$ (by definition) but I do not know how to apply the differentiation on the left hand side. Please add some steps. Thank you. $\endgroup$ – Frank Moses Aug 13 '18 at 2:45
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    $\begingroup$ @FrankMoses : That's not by definition. It's a theorem, not a definition. $\endgroup$ – Michael Hardy Aug 13 '18 at 2:58

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