Pdf of sum of independent rvs is the convolution of pdfs proof

Solution 1:

You want $$\small\begin{align}f_{Y\mid X_1}(y\mid s) &=\dfrac{\partial F_{Y\mid X_1}(y\mid s)}{\partial y}&&\text{by definition of a pdf}\\&=\dfrac{\partial\mathsf P(Y\leq y\mid X_1=s)}{\partial y}&&\text{by definition of a CDF}\\&=\dfrac{\partial\mathsf P(X_1+X_2\leq y\mid X_1=s)}{\partial y}&&\text{by definition of }Y\\&=\dfrac{\partial\mathsf P(X_2\leq y-s\mid X_1=s)}{\partial y}&&\text{the events are identical, under that condition}\\&=\dfrac{\partial \mathsf P(X_2\leq y-s)}{\partial y}&&\text{by independence of }X_1,X_2\\&=\dfrac{\partial F_{X_2}(y-s)}{\partial y}&&\text{by definition of a }CDF\\&=\left.\dfrac{\mathrm dF_{X_2}(t)}{\mathrm d t}\right\vert_{t=y-s}\cdot\dfrac{\partial (y-s)}{\partial y}&&\text{by the chain rule}\\&=f_{X_2}(y-s)\cdot 1&&\text{by definition of a pdf}\end{align}$$