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5

Let $(X_t)_{t\geq 0}$ denote a Geometric Brownian Motion $$ \frac{dX_t}{X_t} = \mu_X dt + \sigma_X dW^X_t,\ \ \ X(0) = X_0$$ such that $X_t$ is lognormally distributed $\forall t > 0$ $$ X_t = X_0 e^{(\mu_X - \frac{1}{2}\sigma_X ^2)t + \sigma_X W_t^X}$$ Let $(Y_t)_{t\geq 0}$ denote an Arithmetic Brownian Motion $$ dY_t = \mu_Y dt + \sigma_Y dW_t^Y,\ \ \ ...


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One assumption is that both (or more) instruments are liquid enough to offer a market (both sides). You can use the bid/ask/mid (your choice) or "conflate" (implemented by the big boys on their data feeds). i.e. 1 second conflation: if no trade, send out last trade price (or assume so in your application).


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Humm....you must miss some hypothesis. Assuming $\alpha>0,\beta>0$, you can without loss of generality set $\alpha=1$. (since $\frac{r}{\alpha}$ and $\frac{V}{\alpha^2}$ being positively correlated is the same as $r$ and $V$ positively correlated) now, positive correlation is equivalent to positive covariance. Working with covariance and above ...


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[Edit] My "answer" below is not a really an answer for I have completely misinterpreted your original question. I thought you asked about the covariance of 2 processes over a given time horizon (i.e. for a fixed $\omega$) and not the covariance of two random variables (fixed $t$). Also note that $\text{cov}(x,y)=0$ does not mean that $x$ and $y$ are ...


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Hint: By Integration, we have $$x_t=x_{0}+\int_{0}^{t} \alpha_1(x_s,s)ds+\int_{0}^{t} \beta_1(x_s,s)dW_1(s)$$ $$y_t=y_{0}+\int_{0}^{t} \alpha_2(y_s,s)ds+\int_{0}^{t} \beta_2(y_s,s)dW_2(s)$$ then $$E[x_t]=x_0+E\left[\int_{0}^{t} \alpha_1(x_s,s)ds\right]$$ $$E[y_t]=y_0+E\left[\int_{0}^{t} \alpha_2(y_s,s)ds\right]$$ Now we apply Ito's lemma $$d(x_ty_t)=...



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