Let $x$ be a random variable and $\mathbf{Y}$ be a random vector. Is there a general formula to derive the probability density function of $$ \mathbf{Z} = \frac{1}{\sqrt{x}} \mathbf{Y} \; ? $$
I could only find the rule for the case where $\mathbf{Y}$ is also a random variable (Wikipedia).
FYI: In the end I would like to use it for the case where $x \sim \Gamma(k = \nu/2, \theta = 2 / \nu)$ and $\mathbf{Y} \sim N(\boldsymbol{\mu}, \mathbf{\Sigma})$ to arrive at the multivariate t-distribution. And I know about the easier derivation in this case by utilizing $\int_0^{\infty} f(\mathbf{Z} | x) f(x) dx$, but I just want to do it another way just for fun.
The formula should be $$ \begin{align} f(\mathbf{Z}) &= \int_{-\infty}^{\infty} f_x(x) f_\mathbf{Y}(\mathbf{Z} \sqrt{x}) x^{p/2} dx, \end{align} $$ where the $x^{p/2}$ part is the Jacobian of the transformation from $\mathbf{Y}$ to $\mathbf{Z}$, $J(\mathbf{Y} \rightarrow \mathbf{Z})$.
Using this formula to derive the multivariate-$t$ distribution from its stochastic representation given in the question above we get $$ \begin{align} f(\mathbf{Z}) &= \int_{-\infty}^{\infty} f_x(x) f_\mathbf{Y}(\mathbf{Z} \sqrt{x}) x^{p/2} dx, \\ &= \int_{0}^{\infty} \frac{1}{\Gamma(\nu/2) (2/\nu)^{\nu/2}} x^{\nu/2-1} \exp(- x \nu/2) \frac{\exp\left(- \frac{\mathbf{Z}' \sqrt{x} \boldsymbol{\Sigma}^{-1} \mathbf{Z} \sqrt{x}}{2}\right)}{(2 \pi)^{p/2} |\boldsymbol{\Sigma}|^{1/2}} x^{p/2} dx \\ &= \frac{1}{\Gamma(\nu/2) (2/\nu)^{\nu/2} (2 \pi)^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \int_{0}^{\infty} x^{\frac{\nu + p}{2} - 1} \exp\left[- x\frac{\nu}{2} \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)\right] dx \\ &= \frac{1}{\Gamma(\nu/2) (2/\nu)^{\nu/2} (2 \pi)^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \frac{\Gamma\left((\nu + p)/2\right)}{\left(\frac{\nu}{2}\right)^{(\nu + p)/2} \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)^{(\nu + p)/2}} \\ &= \frac{\Gamma\left((\nu + p)/2\right)}{\Gamma(\nu/2) \nu^{p/2} \pi^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \frac{\nu^{\nu/2} 2^{(\nu + p)/2}}{2^{\nu/2} 2^{p/2} \nu^{\nu/2}} \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)^{-(\nu + p)/2} \\ &= \frac{\Gamma\left((\nu + p)/2\right)}{\Gamma(\nu/2) \nu^{p/2} \pi^{p/2}} |\boldsymbol{\Sigma}|^{-1/2} \left(1 + \frac{\mathbf{Z}' \boldsymbol{\Sigma}^{-1} \mathbf{Z}}{\nu}\right)^{-(\nu + p)/2}, \end{align} $$ where in the third step can be checked here.