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I am interested in justify the well known result about the process $M^\lambda _t =\exp\left(\lambda B_t - \frac{\lambda^2}{2} t\right)$ being $\mathcal F_t$-martingale in the filtered probability space $(\Omega,\mathcal F, (\mathcal F_t), \mathbb P)$ where $(B_t)$ is a standard Brownian motion and $(\mathcal F_t)$ is it's canonic filtration.

I starte by noting that since $M^\lambda > 0$

\begin{align} \mathbb E \left\{ {M^\lambda _t}\ / \ {M^\lambda _s}\ |\mathcal F_s \right\}&= \mathbb E \left\{ \exp(\lambda(B_t -B_s))\exp(-\frac{\lambda^2}{2}(t-s))\ |\mathcal F_s \right\} \\&= \exp\left(-\frac{\lambda^2}{2}(t-s)\right) \mathbb E \left\{ \exp(\lambda(B_t -B_s)) \right\}=1 \end{align}

The first line can be written because $x \rightarrow\exp(\lambda x) \in \mathbb L ^1(d\Phi(x))$, where $\Phi$ is the CDF of the gaussian law of $(B_t-B_s)$ under $\mathbb P$. The last line comes from independence of brownian increments and normal distribution.

I need it to be a very complete justification. Is there any mistake or omission on my proof?

Thanks in advance

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The assertion that "x→exp(λx)∈L1(P)" is absurd since "x→exp(λx)" is defined on the real line while P is a measure on Ω. –  Did Oct 4 '13 at 12:26
    
@Did: My mistake, I mean this function is in $\mathbb L (d\Phi(x))$, where $\Phi$ is a gaussian CDF –  Paul Oct 4 '13 at 17:12
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1 Answer 1

When you want to check some process is a martingale.

you need to check

it is adapted ?

it is integrable ? (since the conditional expectation is defined for integrable random variables[although you can define by approximation for nonnegative random variables of infinite expectation, if you want to.])

it has the martingale property ?

In your solution, you have not checked the first two requirements (as you said, you want a very complete justification.)

By the way, the reason you gave to the first equality can be changed to 1. that random variable is integrable. 2. explicit calculation

Hope this would be helpful.

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