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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): ...


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 ...


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


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 ...


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


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 $\mathbb{E}\left[ S_T \right] = S_0 \, e^{rT} \,.$


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.


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, $Q$.

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