= Solution
The <moment-generating function> satisfies $0<\phi(\lambda)<\infty$. The proposed process is nonnegative and adapted to the <natural filtration> of the <random walk>, and <independence> of the increments gives
$$
\mathbb EM_n^\lambda=\frac{\prod_{j=1}^n\mathbb Ee^{\lambda X_j}}{\phi(\lambda)^n}=1.
$$
Moreover $X_{n+1}$ is independent of $\mathcal F_n$, so
$$
\mathbb E[M_{n+1}^\lambda\mid\mathcal F_n]
=\frac{e^{\lambda S_n}}{\phi(\lambda)^{n+1}}\mathbb Ee^{\lambda X_{n+1}}
=M_n^\lambda.
$$
Therefore \b[$M^\lambda$ is a mean-one nonnegative martingale], the <exponential martingale of a random walk>. For $\lambda=0$ it is the constant process one.
Back to article page
