7

This is impossible unless you are very intelligent with good memory-retention skills and already mathemathically proficient in the field of analysis and statistics (and no, a single course in basic probability theory does not suffice). And even if you are, two weeks is an extremely short amount of time. However, assuming these criteria are satisfied, I would ...


5

If $M$ and $N$ are independent (your references appear to make this assumption), then $M+N$ is also a Poisson process. So, using the polarization identity: $$ dMdN = 2^{-1}\left[(d(M+N))^2 - (dM)^2 - (dN)^2\right] $$ $$ = 2^{-1}\left[d(M+N) - dM - dN \right] = 0 $$ (A proof of $(dX)^2 = dX$ for a Poisson process $X$ is available here.)


3

$$ X_t = B_t 1_{t<0.5} + (x+ B_t) 1_{t\geq 0.5} = B_t + x1_{t\geq 0.5}$$ $$ [X, X]_t = [B, B]_t + x^2 1_{t\geq 0.5} = t+ x^2 1_{t\geq 0.5}$$ (the author probably intended to use $0.5$ as jump size too)


3

(Special case only.) One special way to create correlated Poisson processes is using a common 'shock' model idea. For $X$, $Y$, and $Z$ independent Poisson processes, let's define: $$ M = X+ Z, \; \; N = Y+Z.$$ We note that $M$ and $N$ are Poisson processes, but that $M+N$ is not ($2Z$ is not a Poisson process). We also note that the Pearson correlation ...


3

Summarizing the suggestions in comments: Nicolas Privault's chapter Stochastic Calculus of Jump Processes [available online] provides only a very brief overview. Chapter 11 of Shreve's II volume (Stochastic Calculus for Finance II: Continuous Time Models), called "Introduction to Jump Processes" is a good starting point. Then Cont and Tankov "Financial ...


3

http://www.math.tau.ac.il/~uriy/Papers/encyc57.pdf - page 5 in this does that do it?


2

First, we need to be careful about putting the condition at the right place: \begin{align} e^{kt}\mathbb{E}[\mu_t] -\mu_0 &= \mathbb{E}\bigg[\sum_{m=1}^{N_t} e^{k\tau_m}\eta_m\bigg]\\ &= \mathbb{E}\bigg[\sum_{m=1}^{N_t} \mathbb{E}\bigg[e^{k\tau_m}\eta_m|N_t\bigg]\bigg]\\ &=\mathbb{E}\Big[\eta_1\Big] \cdot \mathbb{E}\bigg[\sum_{m=1}^{N_t} \mathbb{...


1

The basic Ito formula for a Poisson process is $$ dY_t = \mu_t dt + g_t dN_t $$ $$ df(Y_t) = \mu_t f'(Y_t) dt + (f(Y_{t-}+g_t) - f(Y_{t-}))dN_t $$ (dropped $f$'s direct dependence on the time variable to avoid the partial derivative clutter). Case $\mu_t = -\lambda g_t$ (this is your original case): $$ df(Y_t) = -\lambda g_t f'(Y_t) dt+ (f(Y_{t-}+g_t) - f(...


1

$N_t$ process comes with its own Poisson law (probability measure) $P$ defined via intensity $\lambda$. Under it, $N_t-\lambda t$ is a martingale wrt ${\cal F}_t =\sigma(N_u | u\in [0,t])$ (as $E^P[N_t]=\lambda t$ and $N_t-\lambda t$ has independent increments). Any other equivalent Poisson law, $Q$, defined via a given intensity $\gamma$, can be built using ...


1

Consider a radon nikodym derivative, the Random variable: $Z(w)= 1_{N(T,w)=1}+1-Pr(N(T)=1)$. It is admissible since it is always positive and has an expectation 1. This will lead us to the formation of an equivalent measure, which I will denote by $'$. We start with the easy theorem that $E'(X)=E(XZ)$ for any random variable X, and RND $Z$ $E'(1_{N(T)=1})=E(...


1

This is easy to answer with the meta theorem given in the same chapter. Here you have two sources of randomness (W and N), and one risky asset. Q1: Arbitrage generally happens when you have more assets than the number of random sources, but here it is the other way around, so the answer is yes. Q2: You have one risky asset so you can delta hedge one ...


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