I have to find the density function of the inverse Burr distribution with $Y = 1/X $ but by plugging it in the CDF of $F(y)$ and then deriving it to get $f(y)$, I don't get the supposed answer which is:
I'm confused since I usually do this method and it works. How can I find its inverse density function? Thanks in advance for your help!
Because $Y = g(X) = 1/X$ is a monotone transformation, we use the formula $$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right|,$$ where $$f_X(x) = \frac{\alpha \gamma (x/\theta)^\gamma}{x(1+(x/\theta)^\gamma)^{\alpha+1}}.$$ This yields $$\begin{align} f_Y(y) &= f_X(1/y) \left|\frac{d}{dy}\left[y^{-1}\right]\right| \\ &= \frac{\alpha \gamma (y\theta)^{-\gamma}}{(1/y)(1 + (y \theta)^{-\gamma})^{\alpha + 1}}\cdot \frac{1}{y^2} \\ &= \frac{\alpha \gamma (y \theta)^{-\gamma}}{y(1 + (y\theta)^\gamma)^{\alpha+1} (y\theta)^{-\gamma(\alpha+1)}} \\ &= \frac{\alpha \gamma (y \theta)^{\alpha \gamma}}{y(1+(y\theta)^\gamma)^{\alpha+1}}. \tag{1} \end{align}$$ Note that the chosen parametrization for the inverse Burr is a bit confusing because we have $y\theta$ instead of $y/\theta$. In other words, the density $$f_X(x) = \frac{\tau \gamma (x/\theta)^{\tau \gamma}}{x (1 + (x/\theta)^\gamma)^{\tau+1}} \tag{2}$$ is using a scale parametrization, whereas in the density we derived in $(1)$ is a rate parametrization. This is perhaps the source of your confusion: they are not the same $\theta$.