# Tag Info

## Hot answers tagged stochastic-processes

6

$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process) $$df(t,X_t) = ... 5 What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments: The expected values - it is zero ... easy to see. Next what you did not specify is that the correlation between ... 2 What is written in attached slides is correct. However, what you have written is not correct. Setting M_t=\frac{X_t}{Y_t}, and applying Ito formula will lead to :$$dM_t=\frac{dX_t}{X_t} M_t -\frac{dY_t}{Y_t} M_t + M_t \frac{d<Y>_t}{Y^2_t}-\frac{d<X,Y>_t}{Y^2_t}$$which gives you in your case :$$dM_t = (\mu_x dt+\sigma_x dZ^1_t)M_t - ...

1

Let $Y_t := 2 S_t^1 S_t^2$. Applying (multivariate) Itô to the function $f(t,S_t^1,S_t^2)=2 S_t^1 S_t^2$ yields a stochastic differential equation for $Y_t$ $$\frac{dY_t}{Y_t} = \frac{dS_t^1}{S_t^1} + \frac{dS_t^2}{S_t^2} + \rho \sigma_1 \sigma_2 dt$$ Re-applying Itô's lemma to the function $f(t,Y_t) = \ln(Y_t)$ then yields  d\ln Y_t = (\mu_1 + \mu_2 ...

1

I think you got it. Wrapping up: Usually denoted by $(\mathcal {F}_t)_{t \geq 0}$, a filtration is a series of adaptive subsets of the $\sigma$-algebra $\mathcal{F}$ that keeps track of what really happened as time went by (i.e. fixed $\omega$). Over the probability space $(\Omega, \mathcal{F}, \mathbb{P})$, a random variable $X_t$ is measurable iff ...

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