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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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