Find the distribution with the following Laplace transform. Is anybody aware of the distribution whose Laplace transform is the following.
\begin{equation}
\mathbb{E}[e^{-tX}] = \frac{e^{-t}}{(1+2t)}
\end{equation}
Note: The Laplace transform of the indicator random variable is $e^{-t}$ and of exponential distribution with mean $1/2$ is $\frac{1}{1+2t}$.

You just need to compute a convolution. If $X$ is a random variable having pdf: $$ f_X(x) = \frac{1}{2}e^{\frac{1-x}{2}}\cdot \mathbb{1}_{[1,+\infty)}(x) $$
then, for every $t>-\frac{1}{2}$,
$$ \mathbb{E}[e^{-tX}]=\frac{e^{-t}}{1+2t} $$
as wanted.
