Consider a Gaussian whose mean is the realization of another random variable: $p(x|y) = N(x; y, 1)$. I believe this is equivalent to saying $X \sim Y + N(0, 1)$.
Intuitively, this makes sense. If we first sample from Y and then choose a point near that by centering a Gaussian on it, it seems sensible that this is equivalent to choosing Y and then adding some noise.
In python, I tried simulating this by sampling many points from a triangular distribution (y) and then either adding standard normal noise to it ($x_2$) or using it as the mean of another normal distribution, and then sampling form that ($x$). As you can see, the histograms indicate that they are indeed equivalent.
However, I am having trouble seeing how to prove this statement. Does anybody have guidance or a resource that proves this?
Thank you!
Best Answer
\begin{align} p_X(x)&=\int_{-\infty}^\infty p_Y(y)p_{X|Y}(x|y)dy=\int_{-\infty}^\infty p_Y(y)\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-y)^2}dy \\&=(p_Y*\mathcal{N})(x) \end{align}
where $*$ denotes the convolution, and $\mathcal{N}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$. Therefore, since in general $Z+W=U\iff p_U(u)=(p_Z*p_W)(u)$, we have that $X=N(0,1)+Y$.