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Is this the correct use of Girsanov's transformation where $B_{n}$ is a discrete Brownian motion?

For example computing:

$E[(B_{n}+2n)^{2}]$

Set: $\widetilde{B_{n}}=B_{n}+2n$

And use:$\frac{dQ}{dP}= Z = \exp(-2B_{n}-\frac{1}{2}(4)n)$

$E_{P}[(Z\widetilde{B_{n}})^2] = \widetilde{E}_{Q}[\widetilde{B_{n}}^2] = n$

since $\widetilde{B_{n}}$ has become a discrete Brownian motion under measure $Q$ and $\widetilde{E}_{Q}[\widetilde{B_{n}}^2] = \mathrm{Var}(\widetilde{B_{n}})$ under measure $Q$?

Thanks, trying to get a handle on this...

share|improve this question
    
What is a "discrete" Brownian motion? –  Nate Eldredge Nov 25 '12 at 14:21
1  
If you replace every $B_t$ by $B_n$ and every $\widetilde B_t$ by $\widetilde B_n$, then this is correct. But why would anyone would compute $E((B_n+2n)^2)$ using Girsanov escapes me. –  Did Nov 25 '12 at 14:23
    
Just exploring the example to see if I understand the method correctly... t's changed to n's thanks. –  Dirk Calloway Nov 25 '12 at 14:28

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