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Deriving the Mean of a Time-Scaled Geometric Brownian Motion

Article Quant Q&A · Author: Siron
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Summary

The document considers a process with zero drift and proportional Brownian volatility, then asks for the expectation of that process after multiplying it by a deterministic time function and adding another deterministic function. Since the original process is a martingale, its conditional mean remains its initial value; linearity of expectation therefore gives the requested mean directly.

The follow-up applies Itô’s lemma to the scaled process. Its drift depends on the logarithmic derivative of the scaling function, so solving the stochastic differential equation requires integrating that time-varying term and using elapsed time in the volatility correction. Taking expectations cancels the Brownian contribution and recovers the same result. The example is limited to deterministic scaling and the stated zero-drift model; the derivation also implicitly requires the scaling function to be nonzero where its logarithm is used.

Key ideas

  • A geometric Brownian motion with zero drift is a martingale under the stated assumptions.
  • A deterministic multiplier and additive term can be handled directly using linearity of expectation.
  • The scaled process has a drift determined by the multiplier’s logarithmic derivative.
  • Solving its stochastic differential equation requires integrating the time-dependent drift.

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# SDE linear combination of stock price


# SDE linear combination of stock price












Assume that $X_t$ is a process with dynamics $dX_t = \sigma X_t dW_t$ is where $W_t$ is a standard Brownian motion. Given two deterministic functions $p(t)$ and $q(t)$, compute $\mathbb{E}[p(t)X(t)+q(t)]$.

Solution

Since $p$ and $q$ are deterministic we have $$\mathbb{E}[p(t)X(t)+q(t)] = p(t) \mathbb{E}[X(t)]+q(t)=p(t)X_0+q(t)$$ where the last equation from the fact that $X_t$ is a martingale. To be more precise, $\mathbb{E}[X(t)|X_0]=X_0$.

SDE dynamics

I want to derive the SDE for $p(t)X(t)+q(t)$ using Ito's lemma. First, I am considering the case $V(t)=p(t)X(t)$, then according to Ito's lemma: \begin{align} dV_t&=\frac{\partial p}{\partial t}X_tdt+p(t)\sigma X_tdW_t \\ &=\frac{1}{p(t)}\frac{\partial p}{\partial t}V_tdt+\sigma V_tdW_t \end{align} The SDE that remains is quite straightforward to solve then with solution: $$V(t) = V(0)\exp \left \{ \left(\frac{1}{p(t)}\frac{\partial p}{\partial t} - \frac{\sigma^2}{2}\right)T + \sigma W_t\right \} $$ This will give a different answer for $\mathbb{E}[p(t)X(t)]$, so I am wondering where my mistake is.

## Answer by Kurt G. (score 1)

https://quant.stackexchange.com/a/70973

Let's just write $p'(t)$ for $\frac{\partial p}{\partial t}$. Then your last expression for $V$ should rather be $$ V(t)=V(0)\exp\Bigg\{\int_0^t\frac{p'(s)}{p(s)}\,ds-\frac{\sigma^2}{2}\color{red}{t}+\sigma W_t\Bigg\}. $$ This gives \begin{align} E[V(t)]&=V(0)\exp\Bigg\{\int_0^t\frac{p'(s)}{p(s)}\,ds\Bigg\}\\[3mm] &=V(0)\exp\Bigg\{\int_0^t\frac{d}{ds}\log p(s)\,ds\Bigg\}\\[3mm] &=V(0)\exp\{\log p(t)-\log p(0)\}\\[3mm] &=\frac{V(0)p(t)}{p(0)}=X(0)p(t) \end{align} as it should.

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