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

Question:

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')

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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.

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  • $\begingroup$ 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 '16 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 '16 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 '16 at 16:16
  • $\begingroup$ @Wolfy: my answer above for (ii) and (ii') is based on the moment generating function. $\endgroup$ – Gordon Dec 13 '16 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 '16 at 16:35

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