Given $X=U_1$, $Y=U_1+U_2$ and $U_1,U_2$ are independent. How to find find $f_{Y\mid X}(y\mid x)$?
This is the problem from Probability. Statistics. and Random Processes for EE Problem 6.12.
The answer is $f_{U_2}(y-x)$. But I have no idea how to find it.
Based on the basic definition: $f_{Y\mid X}(y\mid x)=\frac{f_{XY}(x,y)}{f_X(x)}$, then I have no idea of what the next step is.
Based on the answer, that could be $f_{Y-X}(y-x)$. However, I cannot find the relation between the two.
You can use the "replacement rule" and that $U_1$ and $U_2$ are independent:
$$F_{Y|X}(y|x)=P\{U_1+U_2\le y\mid U_1=x\}=P\{U_2\le y-x\}=F_{U_2}(y-x)$$
Edit: Let $\mathcal F$ be the $\sigma$-field, $U_1\in \mathcal{F}$ and $U_2$ is independent of $\mathcal{F}$. For a function $\varphi$ with $\mathbb{E}|\varphi(U_1,U_2)|<\infty$ and $g(u_1)=\mathbb{E}\varphi(u_1, U_2)$ we have
$$\mathbb{E}[\varphi(U_1,U_2)\mid \mathcal F]=g(U_1).$$
You can find the reference in Durrett's book (section 6.2). In your case $\mathcal{F}=\sigma(U_1)$ and
$$\varphi(U_1,U_2)=1\{U_1+U_2\le y\}.$$
$\begin{align}f_{U_1+U_2\mid U_1}(y\mid x) & = \dfrac{f_{U_1+U_2,U_1}(y, x)}{f_{U_1}(x)} \\[1ex] & = \dfrac{f_{U_2,U_1}(y-x, x)}{f_{U_1}(x)} \\[1ex] & =\dfrac{f_{U_2}(y-x)f_{U_1}( x)}{f_{U_1}(x)} \\[1ex] & =f_{U_2}(y-x)\end{align}$
By analogy to discrete random variables:
$\begin{align}\mathsf P(U_1+U_2=y\mid U_1= x) & = \dfrac{\mathsf P(U_1+U_2=y,U_1= x)}{\mathsf P(U_1=x)} \\[1ex] & = \dfrac{\mathsf P(U_2=y-x,U_1=x)}{\mathsf P(U_1=x)} \\[1ex] & =\dfrac{\mathsf P(U_2=y-x)\mathsf P(U_1=x)}{\mathsf P(U_1=x)} \\[1ex] & =\mathsf P(U_2=y-x)\end{align}$
