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I am trying to show that the discounted Geometric Brownian Motion SDE is a martingale, but I must be doing something wrong.

GBM SDE is given by $$X_t=X_0+\int_{h=0}^{h=t}X_hr dh+\int_{h=0}^{h=t}X_h\sigma dW_h$$ and the martingale condition should be that $$\mathbb{E}[e^{-rt}X_t|e^{-rs}X_s]=e^{-rs}X_s$$

Applying this condition (taking $t=s+u)$:

$$\mathbb{E}[e^{-rt}X_t|e^{-rs}X_s]=e^{-r(s+u)}\mathbb{E}\left[X_0+\int_{h=0}^{h=s+u}X_hr dh+\int_{h=0}^{h=s+u}X_h\sigma dW_h|e^{-rs}X_s \right]=\\=e^{-r(s+u)}X_s+e^{-r(s+u)}\mathbb{E}\left[\int_{h=s}^{h=s+u}X_hr dh+\int_{h=s}^{h=s+u}X_h\sigma dW_h \right]=\\=e^{-r(s+u)}X_s+e^{-r(s+u)}\int_{h=s}^{h=s+u}r\mathbb{E}[X_h] dh =\\=e^{-r(s+u)}X_s+e^{-r(s+u)}\int_{h=s}^{h=s+u}rX_0e^{rh} dh =\\=\left(e^{-rs}X_s\right)e^{-ru}+X_0e^{-r(s+u)}\left[e^{rh}\right]_{h=s}^{h=s+u}=\\=\left(e^{-rs}X_s\right)e^{-ru}+X_0e^{-r(s+u)}\left(e^{r(s+u)}-e^{rs}\right)=\\=\left(e^{-rs}X_s\right)e^{-ru}+X_0-X_0e^{-ru}\color{red}{\neq}e^{-rs}X_s$$

Conductor
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  • Somehwere your $\mu$ disappeared. The $X_t$ you are starting with will be a martingale after discounting with $e^{-rt}$ if and only if $\mu=r,.$ – Kurt G. Nov 17 '22 at 14:13
  • Thanks @KurtG., I corrected it, it was a typo. Much closer but still not the desired result. – Conductor Nov 17 '22 at 15:48
  • There are still too many $\mu$s in those equations. The problem however is that in this case the conditional expectation does not become an expectation because your $dW_h$ integral does not have independent increments. This would be the case if the integrand were deterministic but it has an $X_s$. Snoop's approach is much more elegant. – Kurt G. Nov 17 '22 at 16:50
  • Thanks @KurtG., those remaining $\mu$s were not meant to be there, removed now. I think the following is true: $$\mathbb{E}\left[\int_{h=0}^{h=s+u}rX_hdh+\int_{h=0}^{h=s+u}\sigma X_hdW_h|X_s\right]=X_s+\mathbb{E}\left[\int_{h=\color{red}s}^{h=s+u}rX_{\color{red}h}dh+\int_{h=\color{red}s}^{h=s+u}\sigma X_{\color{red}h}dW_h\right]=\=X_s+\int_{h=\color{red}s}^{h=s+u}r\mathbb{E}[X_{\color{red}h}]dh=\=X_s+\int_{h=\color{red}s}^{h=s+u}rX_0e^{rh}dh$$ – Conductor Nov 17 '22 at 17:09
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    I was saying that this were true when $X_s$ was deterministic. It is not true and leads ultimately to your confusion. – Kurt G. Nov 17 '22 at 18:17

1 Answers1

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It is better to use the solution instead of the SDE. Under the risk neutral measure: $$X_t=X_0e^{(r-(1/2)\sigma^2)t+\sigma W_t^Q},\,X_0=x_0>0$$ Of course $E^Q[X_t]=X_0e^{r t}<\infty,\,\forall t$. So $$\begin{aligned}E^Q[e^{-r(t-s)}X_tX_s^{-1}|\mathscr{F}_s]&=e^{-(1/2)\sigma^2(t-s)}E^Q[e^{\sigma(W_t^Q-W_s^Q)}|\mathscr{F}_s]=\\ &\stackrel{W_t^Q-W_s^Q\,\stackrel{Q}{\sim}\, \sqrt{t-s}Z\sim \mathcal{N}(0,t-s)}=e^{-(1/2)\sigma^2(t-s)}E^Q[e^{\sigma\sqrt{t-s}Z}]=\\ &=e^{-(1/2)\sigma^2(t-s)}e^{(1/2)\sigma^2(t-s)}=1\end{aligned}$$ So $X_te^{-rt}$ is a martingale.

Snoop
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  • Thank you (+1). I know it's easier to use the solution, but I deliberately wanted to try the SDE as a challenge. – Conductor Nov 17 '22 at 15:49
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    If you want to use the SDE, first note $d(X_te^{-rt})=\sigma e^{-rt} X_t dW_t$ so we have that $X_te^{-rt}$ is at least a local martingale. Then, you may have to use this machinery to formally conclude. @Conductor – Snoop Nov 17 '22 at 16:23