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Suppose we estimate the regression model $$\triangle y_{t}=\alpha + \beta y_{t-1}+\varepsilon_{t}$$ This is actually quite similar to the Dickey-Fuller test. If $\beta=0$, then the process has a unit root. Let's proceed assuming that $\beta<0$, i.e. that the process is stationary. The first equation is also similar to the continuous time ...


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What you are saying might be correct for discrete time processes. In continuous time the process $$ dX_t = X_t^2 dW_t,\quad X_0 > 0 $$ is stationary but not mean reverting.


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try to calculate the moment generating function and show that it correspond to the one of a beta distributed random variable


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As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: \begin{equation} d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du}) \end{equation} For the first two terms on R.H.S, we have: \begin{equation} d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + ...


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$F=0$ seems like a good choice.


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I am not absolutely sure what you mean by diffusion-jump but if you mean jump-diffusion. Here are some references: Chapter 15, Concepts and Practice of Mathematical Finance, Joshi Cont and Tankov, Financial Modelling with Jump Processes Using Monte Carlo Simulation and Importance Sampling to Rapidly Obtain Jump-Diffusion Prices of Continuous Barrier ...


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The average of the exponentials is not the exponential of the average. It is always higher due to convexity (Jensen inequality). So there is no contradiction between the average of $X_T$ being negative and the average of $S_T$ being $S_0$. So the question is: are your results really significantly different from what you would expect? Have you tried ...


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To shorten the notation, let's write $T_t = T(D_t,y_t)$ and $\delta_t = \delta(D_t,y_t)$. There are two ways to show that, in fact, the dynamics of $$ \xi_t = \xi(D_t, y_t,t) = e^{-\int_0^t \delta_s ds}\, T_t $$ is given by $$ \frac{d\xi_t}{\xi_t} = \left( -\delta_t + \frac{\mathscr{L} T_t}{T_t} \right)dt \quad+\quad \text{diffusion terms}. $$ First way ...


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I think you are having trouble differentiating the integral of $\delta$. You should remember the differential notation is just notation for an integral: $A_td B_t = A'_t dB'_t$ just means $\int_0^T A_td B_t = \int_0^T A'_t dB'_t$. In particular, $d\int_0^t A'_s dB'_s = A'_t dB'_t$ is a tautology. So $$ d ( e^{\int_0^t \delta(s,X_s) ds} )= e^{\int_0^t ...



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