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I would like to compute the PDF of the difference of the logarithms of two shifted Rayleigh laws ($Z$):

\begin{equation} Z = \log{X_{1}} - \log{X_{2}} \end{equation}

where $X_1 \sim R(\alpha_1, \beta_1)$ and $X_2 \sim R(\alpha_2, \beta_2)$ with \begin{equation} R(x, \alpha, \beta) = \begin{cases} \frac{x-\alpha}{\beta^{2}} e^{\frac{-(x-\alpha)^{2}}{2\beta^{2}}} \text{if } x \geq \alpha \\ 0 \text{ otherwise}. \end{cases} \end{equation}

$\textbf{How I've tried:}$

Thus, let's find out the PDF of $Y_{i} = f(X_{i}) = \log{X_{i}}$. According to the change-of-variable technique, we can write:

\begin{equation} f_{Y_{i}}(y) = f_{X_{i}}\left(f^{-1}(y)\right)*\left\lvert{\frac{dg^{-1}(y)}{dy}}\right\rvert. \end{equation}

It follows:

\begin{equation} f_{Y_{i}}(y) = \frac{e^{y}-\alpha_i}{\beta_{i}^2} e^{\frac{-(e^{y}-\alpha_{i})^{2}}{2\beta_{i}^{2}}}e^{y} \label{eq:pdf_log_x1}. \end{equation}

However, I do not know if I should have centered my law to make the following steps easier. And if I do so, how could I take this centering into account at the end of the computation to retrieve the true PDF (considering $\alpha_1$ and $\alpha_2$)?

$Y_{1}$ and $Y_{2}$ being two independent variables, I would have then computed the PDF of $Z = Y_{1}-Y_{2}$ like that:

\begin{equation} f_{Z}(z) = \int_{-\infty}^{+\infty}f_{Y_{1}}(y)f_{Y_{2}}(y-z)dy. \end{equation}

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  • $\begingroup$ It might be easier to work this out as $\log(X_1/X_2)$ but honestly it seems you're nearly done $\endgroup$ Jun 22, 2016 at 10:39

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