Deriving a Radon–Nikodym Density from a Dividend-Paying Asset
Summary
The document examines a change-of-measure density for an asset with constant dividend yield, whose price follows a geometric Brownian motion under the physical measure. It solves the price process and substitutes that expression into a proposed density, obtaining an exponential involving the volatility and Brownian motion. This is compared with the usual exponential-martingale form to ask how the market price of risk should be interpreted and why the proposed expression appears to use a value related to volatility.
The derivation is an algebraic observation, not a complete validation of the density as a risk-neutral measure. In particular, the stated density has a positive Brownian term, whereas the displayed general formula has a negative term; matching signs and conventions matters. The document raises the distinction between dividend yield, drift, and the risk-free rate, but does not resolve it or establish when the proposed density is valid.
Key ideas
- Solving the geometric Brownian motion gives an expression for the terminal asset price in terms of Brownian motion.
- Substituting the price solution into the proposed density yields an exponential with volatility in its Brownian term.
- The sign of the Brownian term must be reconciled with the chosen measure-change convention.
- The document raises questions about how dividend yield and the risk-free rate enter the market price of risk but does not answer them.
Tags
Full text
# Deriving an expression for the Radon-Nikodym Derivative with price of risk $-\sigma$
# Deriving an expression for the Radon-Nikodym Derivative with price of risk $-\sigma$
Consider an asset with a constant dividend yield $q$. Assume the price $S$ of the asset satisfies the SDE:
$$ \frac{dS}{S} = (\mu - q) dt + \sigma dB_t $$
where $B_t$ is a Brownian motion under the physical measure.
I am given the Radon-Nikodym derivative for this question but I wasn't told how to derive it.
$$e^{(q-\mu)T} \frac{S_T}{S_0} $$
However, we all learned that the Radon-Nikodym derivative by the following magical equation
$$\ln m_T = \int_0^T - \frac{1}{2}\big(\frac{\mu-r}{\sigma}\big)^2 dt- \int_0^T \frac{\mu-r}{\sigma}dB_t$$
However, after some algebra I was struggling to prove this answer and I came up with an interesting result suggesting that the price of risk is $\sigma$.
It is well known that we can solve the SDE I gave for $\frac{dS}{S}$ as the exponential function:
$$S_T = S_0 e^{(\mu - q - \frac{1}{2} \sigma^2)T + \sigma B_T}$$
Now rewrite
$$\ln m_T = \int_0^T - \frac{1}{2}\big(\frac{\mu-r}{\sigma}^2\big) dt- \int_0^T \frac{\mu-r}{\sigma}dB_t$$
As:
$$ m_T = \exp\bigg( - \frac{1}{2} \theta^2 T- \theta B_T \bigg) $$
Effectively all we need to do is find $B_T$ which is given by the equation for $S_T$. But a simpler way to approach is just to simply manipulate the equation for $S_T$ as follows.
\begin{equation} e^{(q-\mu)T} \frac{S_T}{S_0} = e^{(q-\mu)T} e^{(\mu - q - \frac{1}{2} \sigma^2)T + \sigma B_T}. \end{equation}
Simplifying we now see that the left hand side is precisely the formula I was given for $m_T$
$$e^{(q-\mu)T} \frac{S_T}{S_0} = e^{-\frac{1}{2} \sigma^2 T + \sigma B_T}. $$
This seems to be a special case of the Radon-Nikodym derivative with price of risk equal to $-\sigma$? What is the general interpretation of this and why was it chosen to be $-\sigma$? Generally, I thought that the price of risk should look something like one of the below two expressions:
$\frac{\mu +q -r}{\sigma}$ or $\frac{\mu -q -r}{\sigma}$.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.