Deriving $X|\mu = 3$ where $X \sim N(\mu, 1)$ and $\mu \sim N(0,1)$

Question

Now I know it seems trivial, but I want to derive it using the standard conditional formula - but since I am conditioning on a continuous random variable assuming a specific value $\mu = 3$, how would I derive that using $f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$. $X$ and $\mu$ are not independent, so how would I obtain the joint distribution?

Answer 1

Your approach is fundamentally flawed because you are trying to derive something, whereas there is nothing to actually derive. To elaborate a little ...

The way to create independent random variables is product measure. One way to create dependent random variables is by using what is known as a markov kernel. Let $(E,\mathcal E)$ be a measurable space. We call the map $\nu:E\times \mathcal E$ a markov kernel if 1) $\nu(x,\cdot)$ is a probability measure on $(E,\mathcal E)$ for all $x\in E$ and 2) $\nu(\cdot,A): (E,\mathcal E)\to([0,1],\mathcal B([0,1]))$ is measurable for all $A\in \mathcal E$. Then, given a markov kernel $\nu$ and an initial probability measure $\lambda$ on $(E,\mathcal E)$, one constructs the dependent random elements $X,Y$ on $(E,\mathcal E)$ by defining $$P(Y\in B_2,X\in B_1) = \int_{B_1}\nu(x,B_2)\lambda(dx)$$ Moreover, by definition, $\nu$ is the regular conditional distribution of $Y$ given $X$

In particular, the notation $X\sim \mathcal N(\mu,1)$ with $\mu \sim \mathcal N(0,1)$ means that $$P(X\in B_2,\mu \in B_1) \overset{\text{def}}{=} \frac{1}{2\pi}\int_{B_1}\left\{\int_{B_2}e^{-(x-u)^2/2}dx\right\}e^{-u^2/2}du$$ Essentially, it means that by definition, $X$ given $\mu$ has regular conditional distribution $e^{-(x-\mu)^2/2}/\sqrt{2\pi}$dx.