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Let $Y_{i}$ be IID with the property $P(Y_{i}=1)=\frac{1}{2}$ and $P(Y_{i}=-1)=\frac{1}{2}$ then if $T=\min\{n:K_n=-c$ or $K_n =c\}$ with $K_n=\sum_{i=1}^{n}Y_{i}$, is a stopping time. Find the Moment Generating Function of $T$.

I am pretty sure that the Optional Stopping Theorem is applicable here but I'm not entirely sure how to find the MGF for this particular situation.

1 Answers 1

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Hints:

  1. Fix $\lambda \in \mathbb{R}$. Show that $$M_n := e^{\lambda K_n} r^n$$ is a martingale for $r := 1/\cosh(\lambda)$.
  2. Using the optional stopping theorem and the dominated convergence theorem show that $$\mathbb{E}(M_T) = \mathbb{E}(e^{\lambda K_T} r^T)=1.$$ Deduce that $$e^{\lambda c} \mathbb{E}(1_{\{K_T=c\}} r^T) + e^{-\lambda c} \mathbb{E}(1_{\{K_T=-c\}} r^T) = 1. \tag{1}$$
  3. Plug in $\pm \lambda$ in $(1)$ to obtain a system of linear equations. Solve this system to show that $$\mathbb{E}(1_{\{K_T=-c\}} r^T) = \frac{1}{e^{\lambda c}-e^{-\lambda c}}= \mathbb{E}(1_{\{K_T=c\}} r^T). \tag{2}$$
  4. It follows from $(2)$ that $$\mathbb{E}(r^T) = 2 \frac{1}{e^{\lambda c}-e^{-\lambda c}} = \frac{1}{\sinh(\lambda c)}.$$
  5. Recall that $r=1/\cosh(\lambda)$. For given $t \geq 0$ choose $\lambda$ in such a way that $r = e^{-t}$ to calculate $\mathbb{E}(e^{-t T}) = \mathbb{E}(r^T)$.