Changing from Physical to Risk-Neutral Stock Price Dynamics
Summary
The note derives the log-price of a stock with constant drift and volatility under the physical probability measure, then explains how to express it under the risk-neutral measure. Ito's formula gives the log-price drift as the asset drift minus half the variance rate. The interval from the current time to maturity appears by integrating the differential equation over that period.
To change measures, the answer uses the Radon-Nikodym density and Girsanov's theorem. The Brownian motion is shifted by the market price of risk, changing the stock's drift from its physical expected return to the risk-free rate under the risk-neutral measure. The two log-price expressions therefore use different Brownian motions; this is a change of probability measure, not a direct substitution in the same stochastic process. The explanation assumes constant coefficients and a setting where the stated measure change is valid. It does not discuss extensions such as stochastic rates, dividends, or time-varying parameters.
Key ideas
- Ito's formula gives the log-price drift as the stock's drift minus half its variance rate.
- Integrating the log-price process from the current time to maturity produces the elapsed-time term.
- Girsanov's theorem shifts Brownian motion when moving from the physical measure to the risk-neutral measure.
- Under the risk-neutral measure, the stock's drift becomes the risk-free rate.
- The Brownian motions in the physical and risk-neutral expressions are different.
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# clarification to log-stock price formula
# clarification to log-stock price formula
Having financial market with safe rate r and risky asset S with dynamics under physical measure P $$\frac{dS_t}{S_t}=\mu dt +\sigma dW_t$$ what is the log-stock price?
Using Ito formula it is straightforward to derive the below equation $$log(S_T)=log(S_t) + (\mu - \frac{\sigma^2}{2})(T-t) + \sigma \ (W_T - W_t) \tag{1}$$
what should be equivalent to $$log(S_T)=log(S_t) + (r - \frac{\sigma^2}{2})(T-t) + \sigma \ (W_T^* - W_t^*) \tag{2}$$
Q what allows to formally transition (1) into (2)? I mean the change of dt into T-t and $\mu$ into r
## Answer by Gordon (score 5, accepted)
https://quant.stackexchange.com/a/26106
The dynamics \begin{align*} \frac{dS_t}{S_t} =\mu dt + \sigma dW_t. \end{align*} is under the real-world measure $\mathbb{P}$. Then, \begin{align*} d\ln S_t =\Big(\mu-\frac{1}{2}\sigma^2 \Big) dt + \sigma dW_t. \end{align*} Therefore, \begin{align*} \ln S_T = \ln S_t + \Big(\mu-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \big(W_T-W_t\big).\tag{1} \end{align*} To obtain the dynamics under the risk-neutral probability measure $\mathbb{Q}$, we employ the Radon-Nikodym derivative \begin{align*} \frac{d\mathbb{Q}}{d\mathbb{P}}\big|_{\mathcal{F}_t} = \exp\left(-\frac{1}{2}\lambda^2 t + \lambda W_t \right), \end{align*} where $\lambda = (r-\mu)/\sigma$ is the market-risk premium. Then, from Girsanov theorem, the process $\{\widehat{W}_t, t \ge 0\}$, where $$\widehat{W}_t = W_t -\lambda t,$$ is a standard Brownian motion under measure $\mathbb{Q}$. Moreover, under measure $\mathbb{Q}$, \begin{align*} \frac{dS_t}{S_t} &=\mu dt + \sigma dW_t\\ &=rdt + \sigma d\widehat{W}_t. \end{align*} Consequently, similar to $(1)$ above, \begin{align*} \ln S_T = \ln S_t + \Big(r-\frac{1}{2}\sigma^2 \Big)(T-t) + \sigma \left(\widehat{W}_T-\widehat{W}_t\right).\tag{2} \end{align*} Note that, in $(1)$ and $(2)$, the Brownian motions $W$ and $\widehat{W}$ are different.Shown 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.