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I have $X_t = e^{5B_t}$ and Where $B_t$ is brownian motion at time $t$. $M_t = X_t \cdot e^{-bt}$

I need to find a value for $b$ such that $M_t$ is a martingale.

I am encountering difficulty, however.

$$\mathbb{E}[ e^{5B_t}e^{-bt} | \mathcal{F}_s] \text{for} s\leq t$$ $$= e^{-bt}\mathbb{E}[ e^{5B_t} | \mathcal{F}_s]$$ $$= e^{-bt}\mathbb{E}[ e^{5(B_t-B_s)+5B_s} | \mathcal{F}_s]$$ $$=\exp\left\{\frac{25(t-s)}{2}+5B_s-bt\right\}$$

Now since $M_t$ is a martingale if $E[M_t | \mathcal{F}_s] = M_s$

We require that $$\exp\left\{\frac{25(t-s)}{2}+5B_s-bt\right\} = \exp\{5B_s-bs\}$$

Isn't this impossible to solve? Or have I made a mistake?

edit: $b= 12.5$

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2 Answers 2

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You should apply Ito's lemma: Let $Y_t = 5 B_t - bt$ then $$ M_t = \exp(5 B_t -bt) = \exp(Y_t) $$ thus $$ d\exp(Y_t) = dM_t = M_t dY_t + M_t \frac{1}{2} 25 dt = M_t (5 dB_t - b dt + 12.5 dt). $$ It will be a martingale if the drift terms cancel. Thus if $b = 12.5$.

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    $\begingroup$ This was a far quicker solution than what I originally did. Thanks for the suggestion :) $\endgroup$
    – elbarto
    Apr 24, 2015 at 3:18
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You are almost there. Write $bt=bs+b(t-s)$ and think how the second summand relates to $25/2(t-s)$.

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  • $\begingroup$ Thanks, I actually had the answer, and then ruled it out for some silly reason! $\endgroup$
    – elbarto
    Apr 23, 2015 at 6:36
  • $\begingroup$ where did you get $b(t-s)$? I think OP should just solve $$\frac{25(t-s)}{2}-bt = -bs$$ then come up with $b = 12.5$ from $t-s \ne 0$? $\endgroup$
    – BCLC
    May 1, 2016 at 19:24
  • $\begingroup$ If you think about it, this is exactly what I was proposing. $\endgroup$
    – mpiktas
    May 2, 2016 at 6:00

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