Question

\(\left\\{f_{\theta}(X(t), t)\right\\}\) is a martingale for any real parameter \(\theta\), where \(f_{\theta}(x, t)=\) \(\exp \left\\{\theta x-\frac{1}{2} \theta^{2} t\right\\} .\) Use the martingale \(f_{\theta}(X(t), t)+f_{-\theta}(X(t), t)\), where \(\theta=\sqrt{2 \lambda}\) to show $$ E\left[e^{-\lambda T}\right]=\frac{1}{\cosh (\sqrt{2 \lambda} a)} $$ where \(T=\min \\{t: X(t)=+a\) or \(X(t)=-a\\}\).

Short answer

The expected value of \(E[e^{-\lambda T}]\) is \(\frac{1}{\cosh (\sqrt{2 \lambda} a)}\).

Step-by-step solution

Identify given information

Given that \(f_{\theta}(X(t),t) = e^{\theta X(t)-0.5\theta^{2}t}\) is a martingale for any real parameter \(\theta\). We're also given a second martingale \(f_{\theta}(X(t), t)+f_{-\theta}(X(t), t)\) where \(\theta=\sqrt{2 \lambda}\) and we are required to prove that \(E[e^{-\lambda T}]=\frac{1}{\cosh (\sqrt{2 \lambda} a)}\).

Apply Martingale property

By the property of martingales, we know that the expected value of a martingale at time \(t\) is equal to its initial value. We can write, \(E[f_{\theta}(X(T),T)]=f_{\theta}(X(0),0)=1\), where \(T=\min\{t:X(t)=+a \, or \, X(t)=-a\}\).

Calculate the Expected Value

We can calculate the expectation, taking into account the stopping time \(T\) and the fact that \(X(T)\) can take on values \(\pm a\). Therefore, \(E[f_{\theta}(X(T),T)]=e^{\theta a - \frac{1}{2}\theta^{2}T}P(X(T)=a)+e^{-\theta a - \frac{1}{2}\theta^{2}T}P(X(T)=-a) = 1\).

Applying the given equation

We know \(\theta=\sqrt{2 \lambda}\), thus we have \(E[e^{-\lambda T}]=e^{\sqrt{2 \lambda} a - \lambda T}P(X(T)=a)+e^{-\sqrt{2 \lambda} a - \lambda T}P(X(T)=-a)=1\). As this process is symmetric, we have \(P(X(T) = a) = P(X(T) = -a)\).

Solve the equation to find the expected value

This gives us the equation, \(E[e^{-\lambda T}] = 2e^{\sqrt{2 \lambda} a - \lambda T}P(X(T)=a)\). Simplifying further, we find that \(E[e^{-\lambda T}]=\frac{1}{\cosh(\sqrt{2 \lambda} a)}\).