I am trying to prove the statement in the title. However, I want some help to make my derivation mathematically rigorous.
I have $X_{1}$ and $X_{2}$ which are two independent rvs.
In addition, I have $Y=X_{1}+X_{2}$.
I find it hard to prove: $f_{Y|X_{1}}(y|x_{1})$ = $f_{X_{2}}(y-x_{1})$. (1)
Intuitively, I can maybe see why this is true, since $Y|x_{1}$ is the same as getting $X_{2}$ given $x_{1}$ and since $X_{2} = Y-x_{1}$ (for a specific $x_{1}$, the above seems true.
However, trying to proof it doesn to seem right, since using: $$\frac{d(F_{Y}(y))}{dy} = \frac{d(F_{X_2}(y-x_{1}))}{dx_{2}}\cdot\frac{dx_{2}}{dy} \to f_{Y}(y) = f_{X_{2}}(y-x_{1})$$ (2)
You want $$\small\begin{align}f_{Y\mid X_1}(y\mid s) &=\dfrac{\partial F_{Y\mid X_1}(y\mid s)}{\partial y}&&\text{by definition of a pdf}\\&=\dfrac{\partial\mathsf P(Y\leq y\mid X_1=s)}{\partial y}&&\text{by definition of a CDF}\\&=\dfrac{\partial\mathsf P(X_1+X_2\leq y\mid X_1=s)}{\partial y}&&\text{by definition of }Y\\&=\dfrac{\partial\mathsf P(X_2\leq y-s\mid X_1=s)}{\partial y}&&\text{the events are identical, under that condition}\\&=\dfrac{\partial \mathsf P(X_2\leq y-s)}{\partial y}&&\text{by independence of }X_1,X_2\\&=\dfrac{\partial F_{X_2}(y-s)}{\partial y}&&\text{by definition of a }CDF\\&=\left.\dfrac{\mathrm dF_{X_2}(t)}{\mathrm d t}\right\vert_{t=y-s}\cdot\dfrac{\partial (y-s)}{\partial y}&&\text{by the chain rule}\\&=f_{X_2}(y-s)\cdot 1&&\text{by definition of a pdf}\end{align}$$