Background Information:

This question comes from the book Financial Calculus by Baxter and Rennie. WE start with looking at the marginal of $W_T$ under $\mathbb{Q}$. We need to find the likelihood function of $W_T$ with respect to $\mathbb{Q}$, or something equivalent. One useful trick is to look at moment-generating functions:

A random variable $X$ is a normal $N(\mu,\sigma^2)$ under a measure $\mathbb{P}$ if and only if $$E_{\mathbb{P}}(\exp(\theta X)) = \exp\left(\theta\mu + \frac{1}{2}\theta^2 \sigma^2\right)$$

To calculate $E_{\mathbb{Q}}[\exp(\theta W_T)]$, we can use the fact of the Radon-Nikodym derivative summary, which tells us that it is the same as the $\mathbb{P}$-expectation $E_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}\exp(\theta W_T)\right]$. This equals

$$E_{\mathbb{P}}[\exp(-\gamma W_T - \frac{1}{2}\gamma^2 T + \theta W_T)] = \exp\left(-\frac{1}{2} \gamma^2 T + \frac{1}{2}(\theta - \gamma)^2 T\right)$$ because $W_t$ is a normal $N(0<T)$ with respect to $\mathbb{P}$. Simplifying the algebra, we have $$E_{\mathbb{Q}}[\exp(\theta W_T)] = \exp\left(-\theta \gamma T + \frac{1}{2}\theta^2 T\right)$$ which is the moment generating function of a normal $N(-\gamma T, T)$. Thus the marginal distribution of $W_T$, under $\mathbb{Q}$, is also a normal with variance $T$ but with mean $-\gamma T$.

What about $W_t$ for $t$ less than $T$? The marginal distribution of $W_T$ is what we would expect if $W_t$ under $\mathbb{Q}$ Brownian motion plus a constant drift $-\gamma$. Of course, a lot of other process also have a marginal normal $N(-\gamma T, T)$ distribution at time $T$, but it would be an elegant result if the sole effect of changing from $\mathbb{P}$ to $\mathbb{Q}$ via $\frac{d\mathbb{Q}}{d\mathbb{P}} = \exp\left(-\gamma W_T - \frac{1}{2} \gamma^2 T\right)$ were just to punch in a drift of $-\gamma$.

And so it is. The process $W_t$ is a Brownian motion with respect to $\mathbb{P}$ and Brownian motion with constant drift $-\gamma$ under $\mathbb{Q}$. Using our two results about $\frac{d\mathbb{Q}}{d\mathbb{P}}$, we can prove the three conditions for $\tilde{W}_t = W_t + \gamma t$ to be $\mathbb{Q}$-Brownian motion:

i) $\tilde{W}_t$ is continuous and $\tilde{W}_0$ = 0;

ii) $\tilde{W}_t$ is a normal $N(0,t)$ under $\mathbb{Q}$

iii) $\tilde{W}_{t+s} - \tilde{W}_2$ is a normal $N(0,t)$ independent of $\mathcal{F}_s$

The first of these is true and ii) and iii) can be re-expressed as

ii') $E_{\mathbb{Q}}[\exp(\theta \tilde{W}_t)] =\exp(\frac{1}{2}\theta^2 t)$

iii') $E_{\mathbb{Q}}[\exp(\theta(\tilde{W}_{t+s} - \tilde{W}_s))|\mathcal{F}_2] =\exp(\frac{1}{2}\theta^2 t)$


Show that ii') and iii') are equivalent to ii) and iii) respectively, and prove them using the chance of measure process $\varsigma_t = E_{\mathbb{P}}\left(\frac{d\mathbb{Q}}{d\mathbb{P}}|\mathcal{F}_t\right)$.

I am not even sure where to start, perhaps a start to the solution or some guidance would be helpful, pretty much teaching myself this stuff so excuse the plethora of questions I may have.

Attempted solution - I want to first show that (ii) and (ii') are equivalent. From (ii) we have that $\tilde{W_t}\sim N(0,t)$ under $\mathbb{Q}$ then the moment generating function is $$M_x(t) = \exp{(\frac{1}{2}t^3)}$$ I do not see how that it is equivalent to (ii')


1 Answer 1


You basically need to show ii') and iii'), as they automatically imply ii) and iii). Note that, since \begin{align*} \frac{dQ}{dP}\big|_T = \exp\Big(-\gamma W_T - \frac{1}{2} \gamma^2 T\Big), \end{align*} we obtain that \begin{align*} \zeta_t &= E_P\left(\frac{dQ}{dP}\big|_T \mid \mathcal{F}_t \right)\\ &=E_P\left(\exp\Big(-\gamma W_T - \frac{1}{2} \gamma^2 T\Big)\mid \mathcal{F}_t \right)\\ &=\exp\Big(-\gamma W_t - \frac{1}{2} \gamma^2 t\Big). \end{align*} Therefore, for $0 \le t \le T$, \begin{align*} E_Q\left(e^{\theta\, \widetilde{W}_t} \right) &=E_P\left(\frac{dQ}{dP}\big|_Te^{\theta\, \widetilde{W}_t} \right) \\ &=E_P\left(E_P\left(\frac{dQ}{dP}\big|_Te^{\theta\, \widetilde{W}_t} \mid \mathcal{F}_t \right)\right) \\ &=E_P\left(E_P\left(\frac{dQ}{dP}\big|_T \mid \mathcal{F}_t \right)e^{\theta\, \widetilde{W}_t}\right) \\ &=E_P\left(\zeta_t \, e^{\theta\, \widetilde{W}_t} \right)\\ &=E_P\left(e^{-\gamma W_t - \frac{1}{2} \gamma^2 t + \theta (W_t + \gamma t)} \right)\\ &=E_P\left(e^{- \frac{1}{2} \gamma^2 t + \theta \gamma t +(\theta-\gamma) W_t} \right)\\ &=e^{- \frac{1}{2} \gamma^2 t + \theta \gamma t + \frac{1}{2} (\theta-\gamma)^2 t}\\ &=e^{\frac{1}{2} \theta^2 t}, \end{align*} which is ii') or ii). Moreover, for any random variable $\xi\in \mathcal{F}_s,$ note that, for $0\le t+s\le T$, \begin{align*} E_Q\left(e^{\theta(\widetilde{W}_{t+s} - \widetilde{W}_s)} \xi \right) &=E_P\left(\frac{dQ}{dP}\big|_T e^{\theta(\widetilde{W}_{t+s} - \widetilde{W}_s)} \xi \right)\\ &= E_P\left(E_P\left(\frac{dQ}{dP}\big|_T e^{\theta(\widetilde{W}_{t+s} - \widetilde{W}_s)} \xi \mid \mathcal{F}_{t+s}\right)\right)\\ &=E_P\left(E_P\left(\frac{dQ}{dP}\big|_T \mid \mathcal{F}_{t+s}\right)e^{\theta(\widetilde{W}_{t+s} - \widetilde{W}_s)} \xi\right)\\ &= E_P\left(\zeta_{t+s} e^{\theta(\widetilde{W}_{t+s} - \widetilde{W}_s)} \xi \right)\\ &=E_P\left(e^{-\gamma W_{t+s} -\frac{1}{2} \gamma^2 (t+s) + \theta(W_{t+s} - W_s) + \theta\gamma t} \xi \right)\\ &=E_P\left(e^{(\theta-\gamma)( W_{t+s} -W_s) -\frac{1}{2} \gamma^2 (t+s) - \gamma W_s + \theta\gamma t} \xi \right)\\ &=E_P\left(e^{(\theta-\gamma)( W_{t+s} -W_s)}\right)E_P\left(e^{-\frac{1}{2} \gamma^2 (t+s) - \gamma W_s + \theta\gamma t} \xi \right)\\ &=E_P\left(e^{\frac{1}{2}(\theta-\gamma)^2 t-\frac{1}{2} \gamma^2 (t+s) - \gamma W_s + \theta\gamma t} \xi \right)\\ &=E_P\left(e^{\frac{1}{2}\theta^2 t-\frac{1}{2} \gamma^2 s - \gamma W_s} \xi \right)\\ &=E_Q\left(e^{\frac{1}{2}\theta^2 t} \xi \right). \end{align*} That is, \begin{align*} E_Q\left(e^{\theta(\widetilde{W}_{t+s} - \widetilde{W}_s)} \mid \mathcal{F}_s \right) = e^{\frac{1}{2}\theta^2 t}, \end{align*} which is iii'), and it implies iii) above.

  • $\begingroup$ Beautiful solution Gordon. I am just not getting the first part of the proof where we have $e^{-\frac{1}{2}\gamma^2 t+ \theta\gamma t + (\theta - \gamma)W_t)} = e^{\frac{1}{2}\theta^2 t}$ $\endgroup$
    – Wolfy
    Nov 30, 2016 at 3:29
  • $\begingroup$ @Wolfy: for $\eta \sim N(\mu, \sigma^2)$, we have $E(e^{\eta}) = e^{\mu+\frac{\sigma^2}{2}}$. Then, \begin{align*} E_P\left(e^{- \frac{1}{2} \gamma^2 t + \theta \gamma t +(\theta-\gamma) W_t} \right) &=e^{- \frac{1}{2} \gamma^2 t + \theta \gamma t + \frac{1}{2} (\theta-\gamma)^2 t}\\ &=e^{\frac{1}{2} \theta^2 t}, \end{align*} $\endgroup$
    – Gordon
    Nov 30, 2016 at 13:50
  • $\begingroup$ I have another solution from my professor and he states that (ii) and (ii') are equivalent by direct application of the moment generating function. I don't really see that since for (ii) we have that $\tilde{W}_t\sim N(0,t)$ thus the mgf is $\exp{\frac{1}{2}t^3}$ how is that the same as (ii')? $\endgroup$
    – Wolfy
    Dec 13, 2016 at 16:16
  • $\begingroup$ @Wolfy: my answer above for (ii) and (ii') is based on the moment generating function. $\endgroup$
    – Gordon
    Dec 13, 2016 at 16:35
  • $\begingroup$ I guess my question is what does it mean for $\tilde{W_t}$ to be normal $N(0,t)$ under $\mathbb{Q}$ $\endgroup$
    – Wolfy
    Dec 13, 2016 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.