Find the distribution with the following Laplace transform.

Question

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 $\frac{1}{1+2t}$.

Answer 1

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.