Let $W_n$ be a chi square random variable with $n$ degrees of freedom and $Z\sim N(0,1)$ be a standard normal random variable with $Z$ and $W_n$ independent. Now take $$U_n=\frac{Z(n^{1/2})}{W_n^{1/2}}.$$
An exercise in my lecture notes asks me to find a formula for the conditional density $$h_n(x,z) = \frac{d}{dx}P(U_n ≤x |Z = z),\quad x,z \in \mathbb{R}$$
However I'm not too sure how I would go about this task, I know that given that $Z$ and $W_n$ are independent I can find the density of $U$ using $Z$ and $W_n$'s respective densities however I'm not sure how to simplify said density now that I have found it nor do I know how to use this density to find the answer to this exercise, I would probably assume a certain inequality for $x$ and $z$ each and deduce the function from there however I can't do any of this without the first part.
I can see in the conditional expectancy $Z$ is fixed at $z$, would that mean the joint fixed density also follows chi-squared density with $f(n)$ degrees of freedom (with $f(n)$ to be found)?
Any pointers would be greatly appreciated I would rather understand where I'm going wrong instead of being given the answer immediately, thank you.
By definition,
$$\Pr[U_n \le x \mid Z = z] = \Pr\left[\frac{Z \sqrt{n}}{\sqrt{W_n}} \le x \mid Z = z\right] = \Pr\left[ W_n \ge \frac{nz^2}{x^2} \right] = 1 - F_{W_n}(n z^2/x^2);$$ therefore, the conditional density of $U_n$ is simply $$\frac{d}{dx}\left[1 - F_{W_n}(nz^2/x^2)\right] = \frac{2nz^2}{x^3} f_{W_n}(nz^2/x^2),$$ where $f_{W_n}$ is the chi-squared density.