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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 ...


3

I would not say there is no link to what you say but here would be my view. Intuitive explanation If you wait for a delay $h$ before exercising, you lose your exercise right between $t$ and $t+h$, this leads to a loss in value. Supermartingale property proof (to apply it in your case : $\phi_t=e^{-rt}(L-S_t)^+$) If we denote $\phi$ the obstacle, and ...


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 - ...


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) ...


1

Equations (1) to (3) are correct. Your investment strategy is then, $\forall t > 0$ $$ X_t = \theta _ t S_t $$ Provided you use this strategy as part of self-financing portfolio you can write the P&L over an infinitesimal time interval as $$ dV_t = \theta_ t dS_t $$ assuming zero safe rate, i.e. that any cash required to finance your long stock ...


1

Just to add an intuitive argument to @MJ73550's already very nice answer: When holding an American option - or any option callable by the holder for that matter -, the question you ask yourself before exercising it is whether the proceeds from early exercise (i.e. exercise now to get the option's intrinsic value) are greater than what you could expect to ...


1

Applying Itô's lemma to $$ Y_t := e^{\int_0^t b(v) dv} r_t $$ You get \begin{align} dY_t &= b(t) e^{\int_0^t b(v) dv} r_t dt + e^{\int_0^t b(v) dv} dr_t + 0\\ &= e^{\int_0^t b(v) dv} (b(t) r_t dt + dr_t) \\ &= e^{\int_0^t b(v) dv} (a(t) dt + \sigma(t) dW_t) \end{align} where the last line is obtained by using the fact that $$ dr_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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