Deriving the Risk-Neutral Martingale Property of the Normalized Gains Process
Summary
The document works through the normalized gains process for a stock paying continuous dividends, using a bank account as numeraire. It starts with geometric Brownian motion under the real-world measure, applies a change of measure, and identifies the risk-neutral stock drift as the interest rate less the dividend yield. Substituting those dynamics into the discounted stock and dividend gains process makes the drift terms cancel.
The derivation illustrates why discounted gains, rather than the discounted ex-dividend price alone, is the relevant martingale when dividends are present. It also asks whether simulation under the risk-neutral measure uses different Brownian increments; the update does not answer that question explicitly. The final calculation establishes zero drift, which directly gives a local martingale under suitable conditions; a true martingale conclusion additionally requires integrability assumptions. The setup also assumes a constant rate and yield and a diffusion model.
Key ideas
- Under the risk-neutral measure, the stock drift in the stated model equals the interest rate minus the continuous dividend yield.
- Discounted stock prices alone retain a dividend-related drift when the stock pays dividends.
- Adding discounted dividends to the discounted stock price cancels that drift in the gains process.
- Zero drift establishes a local martingale under suitable conditions, while a true martingale requires additional integrability assumptions.
- The document raises but does not resolve whether risk-neutral Brownian increments differ in practical simulation.
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# Normalized Gains Process is a Q-Martingale - Proof and Intuition
# 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-martingaleShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.