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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 $$d\ln Y_t = (\mu_1 + \mu_2 ... 6 X_t being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function f of t and X_t. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming X_t is here a continuous, square integrable stochastic process)$$ 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 ...

0

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 ...

1

I edit this answer to give more details. The process for $r$ above is the geometric Brownian motion (GBM) used to model stock prices in the Black-Scholes framework. Thus the question is about (the expectation of the) exponetial of the integral of GBM. The intergral of GBM is closely connected to Asian options. Thus one can study the literature about this ...

0

It seems that uniformly integrable martingales, as described by quasi, account for a specific class of strict local martingales. A martingale on $[0,\infty)$ that is not uniformly integrable (like geometric Brownian motion) is a uniformly integrable martingale on $[0,t]$ for every $t\in [0,\infty)$. Consequently, mapping $[0,\infty]$ onto $[0,1]$ as done ...

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If you model the spot price of the stock, then it is just the spot price (what else could be more accurate?). If you model the forward price of a stock, then you most probably want to apply cost-of-carry (in order to avoid arbitrge). If there are no dividends in your spot, then the forward price for time $T$ is $$F_T = S_0 rT$$ where $r$ is a rate that ...

2

Could it be that your problem is only due to the $t^-$ notation convention? Think of it that way, it is only worth distinguishing $S_{t^-}$ from $S_t$ at a jump time. Elsewhere, knowing that Brownian motion paths are continuous, you'll always have $S_t = S_{t^-}$. Thus you could also write the SDE: $$\frac {dS_t}{S_{t^-}} = \alpha dt+\sigma dW_t+ ... 2 I don't think you can have an explicit form. Let Y_t= e^{at}X_t then :$$ Y_t -Y_0 =\sum_{i=1}^{N_t}e^{aT_i} $$where (T_i)_{i=1...N_t} are the jump times of your poisson process. then$$P(Y_t\leq x)=\sum_{n\geq 0}\frac{(mt)^n}{n!}e^{-mt}P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n)P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n) ...

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