# Doubt in density function of scaled of random variables

The $$h$$ is normally distributed random variable, having PDF $$f_h(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/2\sigma^2}$$. Another random variable $$\rho$$ is related with $$h$$ by the relation $$\rho=h^2\rho_t$$. Then its PDF is given as $$f_\rho(x)=\frac{f_h(\sqrt{x/\rho_t})}{\sqrt{\rho_tx}}$$--------(1). I am having difficulty in understanding how eq (1) is derived. Because using the concept of scaling of random variable, the PDF of $$\rho$$ should be $$\frac{1}{\sqrt{\rho_t}}f_h(\sqrt{x/\rho_t})$$. Any help please as why extra $$\sqrt{x}$$ is coming in denominator of equation (1).

Since you have the non-linear form $$h^{2} \rho_t$$ the fomula you used is not correct. It is applicable only for finding the density of $$c h$$ where $$c$$ is a constant.
$$P(\rho \leq x)=P(h^{2} \leq \frac x {\rho_t} )=\int_{-\sqrt {\frac x {\rho_t}}}^{\sqrt {\frac x {\rho_t} }} f_h(y)dy=2\int_{0}^{\sqrt {\frac x {\rho_t} }} f_h(y)dy$$. To find the density function of $$\rho$$ you have to differentiate w.r.t. $$x$$. When you do that you have to apply Chain Rule for differentiation and the derivative of $$\sqrt {\frac x {\rho_t} }$$ is $$\frac 1 2 (\rho_t)^{-1/2} x^{-1/2}$$. I will let you finish the computation of $$f_\rho$$.