Here
$$
S^\theta=(1+t)^\theta e^{-t\theta X}.
$$
The <Laplace transform of an exponential distribution> is $\mathbb E[e^{-sX}]=(1+s)^{-1}$ for $\operatorname{Re}s>-1$. Since $\operatorname{Re}(t\theta)\geq0$,
$$
\boxed{M(\theta)=\frac{(1+t)^\theta}{1+t\theta}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 211 / 2 / b / Solution

{c}

If $X\sim\operatorname{Exp}(\lambda)$, then direct integration gives
$$
\mathbb E[e^{-sX}]=\frac{\lambda}{\lambda+s},
\qquad \operatorname{Re}s>-\lambda.
$$


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