Distribution function of the random variable $R_2=e^{-R_1}$

Question

An absolutely random variable $R_1$ is uniformly distributed betweem $-1$ and $+1$, find the density and the distribution function of the random variable $R_2$, where $R_2=e^{-R_1}$.

$R_1$ is uniformly distributed, which means it is constant, and $\displaystyle\int_{-1}^1c=1\Rightarrow c=\frac{1}{2}$

The distribution function of $R_2$ is $\displaystyle P(R_2\le y)=P(e^{-R_1}\le y)=P(R_1\ge-\ln(y))=1-P(R_1\le-\ln(y))=1-\int_{?-1}^{-\ln(y)}\frac{1}{2}dx(\star)=1-(1-\ln(y))\frac{1}{2}=\frac{\ln(y)}{2}-\frac{1}{2}$

Is the result and lower limit of the integral $(\star)$ correct ?

Thanks.

Answer 1

The lower limit is correct.

You should consider over which domain the pdf will be non-zero.

Note that the pdf will be the derivative of the cdf you have obtained.