Title is more of a generalization of my question.
Consider $Y=XW$, where $X$ is a positive constant, $W$ is from the standard normal distribution $N(w;0,1)$. I wonder what the pdf $f(y|x,w)$ is.
My answer is $f(y|x,w) = \delta(y-xw)$; however, this answer implies: $$f(y,x|w) = f(y|x,w)f(w) = \delta(y-xw) \cdot N(w;0,1)$$
When I integrate this joint distribution over $w$ to obtain the pdf $f(y|x)$, the result becomes $f(y|x) = xN(y;0,x^2)$, which is obviously wrong, since we know that $f(y|x)=N(y;0,x^2)$, clearly.
I feel very confused and hopeless right now. If you can help me, that'd be great. Thanks!
The dirac-delta impulse function has the property, for constant $c$ and function $g$, that $$\int_\Bbb R \delta_{w-c}\, g(w)\,\mathrm d w= g(c)$$
Also for non-zero $x$, we have $\delta_{y-xw}=\delta_{w-y/x}$ ... when integrating over $w$ the impulse occurs when $y-xw=0$, which is exactly when $w-y/x=0$ .
Thus for a positive constant $X$...
$$\begin{align}f_{Y\mid X}(y\mid x)&=\mathbf 1_{\small x=X}\,\int_{\Bbb R} \delta_{y-xw}\,\mathcal N(w;0,1^2)\,\mathrm d w\\[1ex]&=\mathbf 1_{\small x=X}\,\int_{\Bbb R}\delta_{w-y/x}\,\mathcal N(w;0,1^2)\,\mathrm d w\\[1ex]&=\mathcal N(y/x;0,1^2)\,\mathbf 1_{\small x=X}\\[2ex]&=\mathcal N(y;0,x^2)\,\mathbf 1_{x=X}\end{align}$$