Radon-Nikodym derivative and risk natural measure

I need help with my understanding of changing probability measure. Im not a mathematician so I hope for answers that are not too technical.

As shown in this Wikipedia article http://en.wikipedia.org/wiki/Risk-neutral_measure you can change the drift of a GBM with the following procedure:

$$dS_t = \mu S_t dt + \sigma S_t dW_t$$

Introducing a new process:

$$d\tilde{W_t} = dW_t - \frac{\mu - r}{\sigma}dt$$

I understand that now the discounted value of the following process:

$$dS_t = r S_t dt + \sigma S_t d \tilde{W_t}$$

is a martingale if $\tilde{W_t}$ is a standard Brownian motion. OK, so we change to a new probability measure Q and now $\tilde{W_t}$ is a standard Brownian motion.

My first question is, $W_t$ can no longer be a standard Brownian motion under Q, because it now has non-zero expectation, is this true?

If the probability of an event, $dW_t=x$ under the physical measure, P, is $dP(x)$, then the probability for that same event under Q is $dQ(x)=dP(x) \Phi(x)$, where $\Phi(x)$ is what i think is called the Radon-Nikodym derivative. For $d\tilde{W_t}$ to have zero expectation, then under Q $E_Q[dW_t]=\frac{\mu-r}{\sigma}t$, am I right?

If this is true, can we then find $\Phi(x)=\frac{dQ(x)}{dP(x)}$ by dividing the density function of a Brownian motion with expectation $\frac{\mu-r}{\sigma}t$ with the density function for a for a standard Brownian motion?

$$\frac{e^{\frac{-(x-\frac{\mu-r}{\sigma}t)^2}{2t}}}{e^{\frac{-x^2}{2t}}}$$ $$e^{\frac{x^2}{2t}\frac{-(x-\frac{\mu-r}{\sigma}t)^2}{2t}}$$ $$e^{\frac{x^2-(x-\frac{\mu-r}{\sigma}t)^2}{2t}}$$ $$e^{\frac{x^2-x^2+2x\frac{\mu-r}{\sigma}t-\frac{\mu^2-2\mu r+r^2}{\sigma^2}t^2}{2t}}$$ $$e^{x\frac{\mu-r}{\sigma}-\frac{\mu^2-2\mu r+r^2}{2 \sigma^2}t}$$

Denoting $\frac{\mu-r}{\sigma}$, the market price of risk, as $\lambda$ and substituting we get

$$\Phi(x)=e^{x\lambda-\frac{1}{2}\lambda^2 t}$$

The problem is that most references I have looked at states that the Radon-Nikodym derivative as something like:

$$\Phi(t)=e^{-\int^t_0 \lambda dW(u)-\frac{1}{2}\int^t_0 \lambda^2 du}$$

I cannot seem to see the link between this expressions. Is the last expression even possible to solve?

"Introducing a new process: $d\tilde{W}_t = dW_t +\frac{\mu-r}{\sigma} dt$"

This is incorrect. Rather, $d\tilde{W}_t = dW_t -\frac{\mu-r}{\sigma} dt$

Otherwise, your derivation is correct. After correcting for the sign error, your final equation becomes $\Phi(x)=e^{-\lambda x-\frac{1}{2}\lambda^2 t}$. Notice that when $\lambda$ is a constant, $\int_0 ^t \lambda dW_t =\lambda x$ where $x$ is a normally distributed random variable. Hence your derivation agrees with your references. However, the references are more general since they do not require $\lambda$ to be a constant.

"Is the last expression even possible to solve?"

In general, there is no reason to "solve" the Radon-Nikodym derivative. Simply knowing it exists allows us to price contingent claims as expected values of functions of the underlying, which can be often be efficiently computed. Indeed, if the goal is to price assets $\lambda$ can be safely ignored.

• If we substitute your proposed expression for $dW_t = d\tilde{W}_t + \frac{\mu-r}{\sigma}dt$ into the SDE for $dS_t = \mu S_t dt + \sigma S_t dW_t$ then the $\mu S_t dt$ term will not cancel. Jan 15 '19 at 12:06
• hi @user9403, doesn't the expectation of exp{integral[sigma]dW} equate to exp{0.5 * sigma^2 * W}? How do we get phi(x) = exp{-lambda*x - 0.5 * phi^2*t} though? Apr 2 '19 at 9:07

Five years late to the party, but let me put my two cents in for intuition.

Let $$W^P_t \equiv W_t$$, and $$W^Q_t \equiv \tilde{W_t}$$ to easier remember under what measure either one is a standard Brownian motion.

Also, let $$\lambda := \frac{\mu-r}{\sigma}$$, which will be used as abbreviation below.

Then,

$$\frac{dS_t}{S_t} = \mu dt + \sigma dW^P_t$$

$$\phantom{\frac{dS_t}{S_t}}= rdt + \sigma \cdot\left(dW^P_t + \frac{\mu-r}{\sigma}dt \right)$$

$$\phantom{\frac{dS_t}{S_t}}= rdt + \sigma \cdot d \left(W^P_t + \lambda t\right)$$

$$\phantom{\frac{dS_t}{S_t}}= rdt + \sigma dW^Q_t$$,

where $$W^Q_t:=W^P_t + \lambda t \iff W^P_t=W^Q_t - \lambda t.$$

We already know that $$W^P_t$$ is a standard Brownian motion under $$\mathbb{P}$$, that is $$W^P_t|\mathcal{F_0} \sim \mathcal{N}^P(0, t).$$

Now, what is the distribution of $$W^P_t$$ under $$\mathbb{Q}$$? This is given by the above relation between the two Brownian motions, e.g. our intuition, taking expectation on both sides yields $$E^Q[W^P_t|\mathcal{F_0}] = E^Q[W^Q_t|\mathcal{F_0}] - \lambda t = - \lambda t$$. Here we can see that $$W^Q_t$$ is not a standard Brownian motion under $$\mathbb{Q}$$ since the drift is not zero.

$$W^P_t|\mathcal{F_0} \sim \mathcal{N}^P(0, t)$$ $$W^P_t|\mathcal{F_0} \sim \mathcal{N}^Q(-\lambda t, t)$$

The likelihood ratio between these two densities is $$\Phi_1(t) := \frac{f^Q(W^P_t)}{f^P(W^P_t)} = \frac{\exp(-\frac{(W^P_t+\lambda t)^2}{2t})}{\exp(-\frac{(W^P_t)^2}{2t})} = e^{-\lambda W^P_t - \frac{1}{2}\lambda^2 t} = e^{-\lambda \int_0^t dW^P_u - \frac{1}{2}\lambda^2 \int_0^t du}.$$

You can also do the same thing but for $$W^Q_t|\mathcal{F_0}$$ which has distributions $$\mathcal{N}^Q(0,t)$$ and $$\mathcal{N}^P(\lambda t,t)$$, but the result will be the same after a substitution. $$\Phi_2(t) = \frac{f^Q(W^Q_t)}{f^P(W^Q_t)} = e^{-\lambda W^Q_t + \frac{1}{2}\lambda^2 t} = e^{-\lambda W^P_t - \frac{1}{2}\lambda^2 t} = \Phi_1(W^P_t).$$

The above is what you were after. Note that the above only works if $$\lambda$$ is constant.

If is not, i.e. $$\lambda(t)$$, you can discretize the Brownian motions relationship, get a similar expression as before but for the interval $$(t,t+\Delta t)$$ as $$\Phi(t,t+\Delta t) = e^{-\lambda_t \Delta W^P_t - \frac{1}{2}\lambda^2 \Delta t}$$, aggregate up from $$0$$ to $$T$$ as $$\Phi(t_0,t_0+\Delta t)\cdot .. \cdot \Phi(t_T,t_T+\Delta t)$$, and let $$\Delta t \to 0$$ which will yield $$\Phi(0,t) = e^{-\int_0^t \lambda_u dW^P_u - \frac{1}{2} \int_0^t \lambda_u^2 du} .$$