Suppose $X_t$ is simple random walk on $\mathbb{Z}$ in continuous time. Time between jumps is distributed as ${\rm Exp}(\lambda)$, $X(0) = 0$.
I am interested in calculation of $\mathbb{E} e^{X_t}$ and in general $\mathbb{E} f\left(X_t\right)$.
I do not know where to start, since $\mathbb{E} e^{X_t}$ is the sum of a random number of random variables.
Please advise me where I can read about this problem.
Many thanks to @MatthewH.!
$$\mathbb E (e^{X_t}) = \mathbb E \left(\mathbb E (e^{X_t}|N_t)\right), $$ where $N_t$ is number of jumps during time $t$.
$$\mathbb E (e^{X_t}|N_t) = \sum\limits_{x = -\infty}^{\infty}e^x\cdot P(X_n = x),$$ where $X_n$ is sum of $n$ jumps.
$$P(X_n = x) = \frac 12 P(X_{n-1} = x-1) + \frac 12 P(X_{n-1} = x+1).$$
Therefore
$$\mathbb E (e^{X_t}|N_t) = \sum\limits_{x = -\infty}^{\infty}e^x\cdot P(X_n = x) = \\= \frac12 \left( \sum\limits_{x = -\infty}^{\infty}e^x\cdot P(X_{n-1} = x-1) + \sum\limits_{x = -\infty}^{\infty}e^x\cdot P(X_{n-1} = x+1) \right) = \\ = \frac12 \left( \sum\limits_{x = -\infty}^{\infty}e^{x+1}\cdot P(X_{n-1} = x) + \sum\limits_{x = -\infty}^{\infty}e^{x-1}\cdot P(X_{n-1} = x) \right) = \\ = \frac12 \left( e \sum\limits_{x = -\infty}^{\infty}e^{x}\cdot P(X_{n-1} = x) + e^{-1}\sum\limits_{x = -\infty}^{\infty}e^{x}\cdot P(X_{n-1} = x) \right) = \\ = \frac{e^{-1} + e}2 \cdot \mathbb E (e^{X_t}|N_t-1) = \dots = \left( \frac{e^{-1} + e}2 \right)^{N_t} = (\cosh 1 )^{N_t}. $$
Going back to $\mathbb E (e^{X_t})$:
$$\mathbb E (e^{X_t}) = \mathbb E \left(\mathbb E (e^{X_t}|N_t)\right) = \mathbb E \left( (\cosh 1 )^{N_t} \right) = \sum\limits_{k=0}^{\infty} \left( \cosh 1 \right)^k \cdot e^{-\lambda t}\frac{(\lambda t)^k}{k!} = \\ = e^{-\lambda t }\cdot \sum\limits_{k=0}^{\infty} \frac{(\lambda t \cosh 1)^k}{k!} = e^{-\lambda t} \cdot e^{\lambda t\cosh 1} = e^{ \lambda t (\cosh 1 - 1)}. $$