# Tag Info

Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with $$Y_t:=W_t,$$$$X_t:=-\int_0^t\Theta_udW_u$$ We can show: $$[Y,X]=-\int_0^t\Theta_udu$$ by using general Stochastic Calculus rules (e.g. p.37, 6.6 here): $$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\... 5 It depends on the purpose of your simulation. If you want to model the asset price path for pricing some derivative then you need the risk-neutral measure (thus you take the risk-less rate as drift). Why? Because the risk-neutral measure makes your pricing compatible with the pricing of other contracts in the market. It makes the prices consistent. If ... 5 IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure \mathbb{Q} equivalent to \mathbb{P} and under which X_t is a martingale. Just take$$\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$for any a \in \... 4 It doesn't imply$$ \ln S_T=\ln S_0+rT+σW^Q_T,$$it implies$$ \ln S_T=\ln S_0+(r-0.5\sigma^2)T+σW^Q_T.$$Look up Ito's lemma. This is covered in just about any book on financial maths including my own Concepts etc. 4 Bond Price Dynamics I do not know the source of the bond dynamics you show above but seeing how we are dealing with an affine model there is a very elegant way to derive those. Due to the model being affine the bond price is given by$$P(t,T)=A(t,T)e^{-r(t)B(t,T)}you can find the exact formulas for A(t,T) and B(t,T) in this document (or just read ... 3 Your notations are really hard to follow as you define \mathbb{P} twice at the beginning. The notation \mathbb{P} = \mathbb{\hat{P}} and \mathbb{P} =\mathbb{\tilde{P}} is not meaningful as the probability measure \mathbb{P} is already fixed and used for the real world probability measure. I think that this is the reason why you are getting confused. ... 3 Your mistake is actually made at the beginning: "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}. ... 1 Note that, under measure Q, the dynamics is of the form \begin{align*} dS_t = S_t \big[(r+ \sigma \theta_t) dt + \sigma dW_t^Q \big]. \end{align*} Then, for \Delta>0 sufficiently small, \begin{align*} S_{t+\Delta} &= S_te^{\left(r-\frac{1}{2}\sigma^2\right)\Delta + \sigma \int_t^{t+\Delta} \theta_s ds + \sigma \left(W_{t+\Delta}^Q-W_t^Q\right)}\\ &... 1 in the new measure, the stock has drift r + \sigma \theta so yes you just proceed with that drift as you say. If \theta is time dependent, it gets more complicated. 1 A martingale must have constant expectation, such that adding a deterministic finite variation process (b-r)dt would break the martingale property (except for when its a constant, which it is not by multiplication with dt). Hence the finite variation process must be eliminated under Q for LRS to be an (equivalent) martingale measure, and as shown the ... 1 I saw a quote from Brigo & Mercurio "IR models" (page 26, 2.1 No-Arbitrage in Continuous Time) . May be it will help you to find answer: Harrison and Pliska (1983) proved the following fundamental result. A financial market is (arbitrage free and) complete if and only if there exists a unique equivalent martingale measure. 1 The error is in the application of Girsanov theorem. We have multivariate Black-Sholes market, however I apply one-dimensional Girsanov theorem. I should apply multi-dimensional Girsanov theorem. Then there would be now such equations, except the case for \rho=1. The alike task is formulated here http://wwwf.imperial.ac.uk/~mdavis/course_material/... 1 If dS_t = r S_t \, dt + \sigma S_t \, dW_t^Q,S_T = S_0 \, e^{\sigma W_T^Q + \left( r - \frac{1}{2} \sigma^2\right) T}\, .$$Hence \mathbb{E}\left[ S_T \right] = S_0 \, e^{rT} \,. 1 Note that$$\frac{dQ_{T_p}}{dQ}|_{T_0} = \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}$$. Then$$E^{Q_{T_p}}\big(S(T_0, T_n)\big) = E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{P(0, T_p)}\frac{A(0, T_0, T_n)}{A(T_0, T_0, T_n)}\bigg) \\ = \frac{A(0, T_0, T_n)}{P(0, T_p)} E^Q\bigg(S(T_0, T_n) \frac{P(T_0, T_p)}{A(T_0, T_0, T_n)}\bigg). That ...
No, that's the point of Girsanov's theorem. If $Q$ is equal to $P_1$, then nothing has changed. In order to make $B_1(t)$ a standard BM we need to transition to a new Law. Namely, $Q$.