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9

First, note $$\mathbb{E^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=0$$ and $$\mathbb{Var^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=\mathbb{E^Q}\left[\int_{0}^{t} e^{-2a(t-s)}ds\right]=\frac{1}{2a}(1-e^{-2at})$$ therefore $$\mathbb{E^Q}[r_t]=r_0 e^{-at} + \frac{b}{a}(1 - e^{-at})$$ $$\mathbb{Var^Q}(r_t)=\frac{\sigma^2}{2a}(1-e^{-2at})$$ second The ItÃ´ integral ...

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 $a \in \... 2 Assuming$\theta>0$(take$\tilde{X}=\mu-X$if it is not the case) Let us denote$\text{erfi}(x)$the imaginary error function Let us denote$\tau_L$,resp.$\tau_U$the hitting time of$L$resp.$U$where$L<U1) Using Ito's lemma, prove that : $$Y_t = \text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(X_t-\mu\right)\right) \text{ is a martingale}$$ ... 2 :D Is it a joke? $$\underset{\kappa \to 0 }{\mathop{\lim }}\,\frac{1-e^{-\kappa(T-t)}}{\kappa}=\underset{\kappa \to 0 }{\mathop{\lim }}\,\frac{\frac{d}{d\kappa}\left(1-e^{-\kappa(T-t)}\right)}{\frac{d}{d\kappa}\kappa}=\underset{\kappa \to 0 }{\mathop{\lim }}\,(T-t) e^{-\kappa(T-t)}=T-t$$ 2 By the usual integrating factor method, \begin{align*} r_t = r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds +\sigma \int_0^t e^{-\beta(t-s)}dW_s. \end{align*} Let \begin{align*} x_t &=\sigma \int_0^t e^{-\beta(t-s)}dW_s, \textrm { and}\\ y_t &=r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds. \end{align*} Thenr_t = x_t + y_t\$, ...

2

\begin{align*} \int_0^T W(t)\, dt &{}= \int_0^T\!\!\int_0^t dW(u)\,dt \\ &{}= \int_0^T\!\!\int_u^T dt\, dW(u) \\&{}= \int_0^T (T - u)\,dW(u) \\&{}= TW(T) - \int_0^T u\, dW(u) \end{align*} however i am not sure if it is what you are asking for

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