Problem 16.18 Easy Free 1 variation

Exponential Martingale of a Walk

Let $X_1,X_2,\dots$ be i.i.d. random variables, each equal to $+1$ or $-1$ with probability $\tfrac12$ each, and let $\mathcal{F}_n=\sigma(X_1,\dots,X_n)$ . Fix a constant $c$ , set $M_0=1$ , and define $M_n=M_{n-1}e^{X_n+c}$ for $n\ge 1$ . Find the unique value of $c$ for which $(M_n)$ is a martingale with respect to $(\mathcal{F}_n)$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Each $M_n=\exp\bigl(\sum_{k\le n}X_k+nc\bigr)$ is a function of $X_1,\dots,X_n$ , so $(M_n)$ is adapted. It is also bounded by $e^{n(1+|c|)}$ , so it is integrable. The only condition left to check is the martingale property.

Since $M_{n-1}$ is $\mathcal{F}_{n-1}$ -measurable and $X_n$ is independent of $\mathcal{F}_{n-1}$ ,

\mathbb{E}[M_n\mid\mathcal{F}_{n-1}] = M_{n-1}\,e^{c}\,\mathbb{E}\bigl[e^{X_n}\bigr] = M_{n-1}\,e^{c}\cdot\frac{e+e^{-1}}{2} = M_{n-1}\,e^{c}\cosh 1 .

We need this to equal $M_{n-1}$ . Because $M_{n-1}>0$ , that holds if and only if $e^{c}\cosh 1=1$ . The value of $c$ is therefore unique: dividing by the moment generating function of a step makes the multiplicative factor $e^{X_n+c}$ have conditional mean $1$ .

\boxed{c=-\log\cosh 1=-\log\!\Bigl(\frac{e+e^{-1}}{2}\Bigr)\approx -0.4338}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 16.18a

Let $N(t)$ be a Poisson process with rate $\lambda>0$ , and fix $c>0$ . Define $S(t)=e^{N(t)\log c-\lambda t}$ . Find the value of $c$ for which $(S(t))_{t\ge 0}$ is a martingale.

Work it out, then compare with the answer.

Solution

Note that $S(t)=c^{N(t)}e^{-\lambda t}$ , and let $\mathcal{F}_s$ be the filtration generated by the process.

Fix $0\le s\le t$ . The increment $N(t)-N(s)\sim\mathrm{Poisson}(\lambda(t-s))$ is independent of $\mathcal{F}_s$ , and $\mathbb{E}[c^{N}]=e^{\mu(c-1)}$ for $N\sim\mathrm{Poisson}(\mu)$ . Hence

\mathbb{E}[S(t)\mid\mathcal{F}_s] = S(s)\,e^{-\lambda(t-s)}\,\mathbb{E}\bigl[c^{N(t)-N(s)}\bigr] = S(s)\,e^{\lambda(t-s)(c-2)} .

This equals $S(s)$ for all $s\le t$ if and only if $c=2$ . Integrability holds because $\mathbb{E}[S(t)]=e^{\lambda t(c-2)}<\infty$ .

\boxed{c=2}

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