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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-211/2/b/solution