# Normalized Gains Process is a Q-Martingale - Proof and Intuition

I'm trying to work the proof that the normalized gains process, $G^z_t = \frac{S_t}{B_t}+\int^t_0\frac{1}{B_s}dD_s$ is a Q-martingale under Q (the risk-neutral measure). I'll show what I've worked through, but it feels like I'm just invoking circular logic, so I'm not sure if I've followed the right path. Also, I have a question regarding the intuition of the whole thing, and also a practical question related to the Wiener process under Q that I'll place at the end. This will probably be long, so many thanks to anyone who takes the time to lend any insight.

My proof:

First, note that $\frac{1}{B_t}$ can be thought of as $e^{-rt}$, if we think of $B_0$ as \$1 in a bank account. Furthermore, we can also think of$dD_t=qS_tdt$, with q the continuously compounded dividend yield. Finally, also let the normalized price process$\frac{S_t}{B_t}=Z_t$. Now, by Ito,$\frac{S_t}{B_t}$, or equivalently$S_te^{-rt}$differentiates to$-r\frac{S_t}{B_t}dt+\frac{1}{B_t}dS_t$. Assuming a GBM for$dS_t=\alpha S_tdt+\sigma S_tdW_t$, we are led to the following for$Z_t$and$G^z_t$: $$dZ_t=Z_t(\alpha-r)dt+Z_t\sigma dW_t$$ $$dG^z_t=Z_t(\alpha-r+q)dt+Z_t\sigma dW_t$$ Now, by Girsanov, let$dQ=LdP$where$dL_t=L_t\phi dW_t$, we have$dW_t=\phi_tdt+dW^Q_t$. Plugging this into the dynamics of$dS_t$we have $$dS_t=S_t(\alpha+\sigma\phi_t)dt+S_t\sigma dW^Q_t$$ Making this the Q-dynamics for the stock price. Furthermore, plugging this into our result for$G^Z_t$we are led to $$dG^Z_t=Z_t(\alpha-r+q+\sigma\phi_t)dt+Z_t\sigma dW^Q_t$$ Now, to be a martingale, we need no drift, that is$a-r+q+\sigma\phi_t=0$, or$\alpha+\sigma\phi_t=r-q$. From this we see that we can go back to the Q-dynamics for the stock price and write $$dS_t=S_t(r-q)dt+S_t\sigma dW^Q_t$$ ...and this is where I sort of end. I don't know if I've proved what was meant to be proved or not. I know that I've ended up with the known Q-dynamics result, which is encouraging. Finally, for a practical question - is there any difference in simulating$W^Q_t$in practice versus$W_t$? For instance, suppose I decide to price a derivative via Monte Carlo simulation in excel. Would it be the exact same function (i.e. norm.inv(rand(0,1))? ***Update - ok, I think I took it a step further and have it now. So, now that I basically rigorously derived the Q-dynamics of the stock price price process in the last line, let's go back to the fact that the normalized gains process is$G^z_t = \frac{S_t}{B_t}+\int^t_0\frac{1}{B_s}dD_s$, or equivalently $$G^z_t=e^{-rt}S_t+\int^t_0e^{-rs}S_sqds$$ Let's focus on the undifferentiated part, that is,$e^{-rt}S_t$, call it$\tilde{S}_t$. By Ito, we have $$d\tilde{S}_t=-re^{-rt}S_tdt+e^{-rt}dS_t$$ Plugging in the before derived Q-dynamics for$dS_t$, we have $$d\tilde{S}_t=-r\tilde{S}_tdt+e^{-rt}[(r-q)S_tdt+\sigma S_tdW^Q_t]<=>$$ $$d\tilde{S}_t=(r-r-q)\tilde{S}_tdt+\sigma\tilde{S}_tdW^Q_t<=>$$ $$d\tilde{S}_t=-q\tilde{S}_t+\sigma\tilde{S}_tdW^Q_t$$ Now, going back to the full equation for the gains process, we have $$dG^z_t=-q\tilde{S}_tdt+q\tilde{S}_tdt+\sigma\tilde{S}_tdW^Q_t<=>$$ $$dG^z_t=\sigma\tilde{S}_tdW^Q_t$$ Proved Q-martingale • Yes, unfortunately you did not prove what you wanted. Instead, using the result that "normalised gain processes should be$\Bbb{Q}$-martingales", you've shown how the dynamics of a dividend paying stock (with continuous div yield) should write under$\Bbb{Q}$. For the second question, these are both standard Brownian motions under their respective measures meaning that simulating$W_t^\Bbb{Q}$under$\Bbb{Q}$or$W_t$under$\Bbb{P}$is exactly the same problem and you can use the exact same procedure (this is not the case when simulating$W_t$under$\Bbb{Q}\$ for instance though). – Quantuple Sep 18 '17 at 8:25
• Also the statement to be proven is true irrespective of the dynamics (no need to assume GBM). The ratio of a traded asset price to the bank account value is a Q-martingale by the fundamental theorem of asset pricing. – dm63 Sep 18 '17 at 11:17