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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).

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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 $f_\rho$.

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  • $\begingroup$ Ok....will do.... $\endgroup$
    – Pranu
    Dec 3 '20 at 8:47
  • $\begingroup$ Derived and done completely...... $\endgroup$
    – Pranu
    Dec 4 '20 at 5:34

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