Density of Y = log(X) for Gamma-distributed X

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Suppose I have a random variable $X \sim \text{Gamma}(k, \theta)$, and I define $Y = \log(X)$. I would like to find the probability density function of $Y$.

I had originally thought I would just define cumulative distribution function X, do a change of variable, and take the "inside" of the integral as my density, like so,

\begin{align} P(X \le c) & = \int_{0}^{c} \frac{1}{\theta^k} \frac{1}{\Gamma(k)} x^{k- 1} e^{-\frac{x}{\theta}} dx \\ P(Y \le \log c) & = \int_{\log(0)}^{\log(c)} \frac{1}{\theta^k} \frac{1}{\Gamma(k)} \exp(y)^{k- 1} e^{-\frac{\exp(y)}{\theta}} \exp(y) dy \\ \end{align}

Here I use $y = \log x$ and $dy = \frac{1}{x} dx$, then sub in definitions for $x$ and $dx$ in terms of $y$.

The output, unfortunately, does not integrate to 1. I'm not sure where my mistake is. Could some tell me where my error is?

Answer 1

Write the densities with the indicators to have a clear picture.

If $X\sim\mathrm{Gamma}(k,\theta)$, then $$ f_X(x) = \frac{1}{\theta^k \Gamma(k)} \;\; x^{k-1} e^{-x/\theta} \, I_{(0,\infty)}(x) \, . $$

If $Y=g(X)=\log X$, with inverse $X=h(Y)=e^Y$, then $$ f_Y(y) = f_X(h(y)) |h'(y)| = \frac{1}{\theta^k\Gamma(k)} \;\; \exp\left( ky-e^y/\theta\right)\,I_{(-\infty,\infty)}(y) \, , $$ and the CDF is obtained from the definition $$ P(Y\leq y) = \int_{-\infty}^y f_Y(t) \,dt \, . $$