I am struggling with some basic argument in probability. Let $Z\sim N(0,1)$ and $W\sim \text{Unif}(1,2)$, and they are independent. Define $T = \frac{Z}{W}$. They we consider the conditional distribution of $T$ with respect to $W$.
Intuitively, it is obvious that $\mu(\cdot, w)$ is the same distribution as $\frac{Z}{w}$. But I am having trouble to see why. If we know the joint distribution of $T,W$ ($f(x,y)$), then we can compute the density of the conditional distribution by \begin{align*} g(x,y)=\frac{f(x,y)}{\int_{\mathbb{R}}f(x,t)dt} \end{align*}
However, we don't know the joint distribution of $T,W$ in this case. So I have no idea how can we calculate the conditional distribution formally.
I also tried to use \begin{align*} P(T\in H\,|\, W)=P\left(\frac{Z}{W}\in H\,\big|\, W\right) \end{align*} and evaluate the conditional probability at $W=w$. However, I do not know how to proceed to end up with the normal distribution.
I tried to use the uniqueness of conditional distribution, but it seems does not work. Does anyone have any idea on this?
Thanks in advance!
The distribution of $T$ conditioned on the value of $W$ is normal since we consider $W$ as a constant when evaluating the conditional distribution. Since $T=Z/W$, and $Z/a$ is normal for any nonzero value of $a$, then the distribution of $T$ given the value of $W$ must be normal.
Further, consider the CDF of $T$ conditional on the value of $W$:
$$\mathbb{P}(T \leq t | W=w) = \mathbb{P}(Z/w \leq t | W=w) = \mathbb{P}(Z \leq tw | W=w)$$
which is the CDF of a gaussian distribution given any fixed value of $w$. The existence of this distribution is proved by the Radon-Nikodym theorem.
Note that $\mathbb{P}(Z/w \leq t | W)$ is simply the random variable that takes on the value $\mathbb{P}(Z/w \leq t | W=w)$ whenever the event $\{W=w\}$ occurs.
Also note that your expression for the conditional density only holds if each of $Z$ and $W$ have a density (in this case they do).