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## New answers tagged risk-neutral-measure

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(... 3 We have \begin{align} V(t) &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B} (S(T_2)-K)^+)] \\ &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K} (S(T_2)-K))] \\ &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K} S(T_2)]-K\mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K}] \\ \end{align} The second term ... 7 There are two parts to your question which I try to answer separately. The first one is about what calibration actually is whereas the second question deals with risk-neutral pricing. As an example, we can use any model. I continuously refer to the stochastic volatility model from Heston (1993) as an example for equity options. Any thoughts equally apply to ... 1 With EMMQ$, associated$Q$-Brownian motion$W$, filtration${\cal F}$, and $$d\beta_t = r_t \beta_t dt, \; \beta_t ={\rm e}^{\int_0^t r_u du},$$ consider martingale: $$M_t =E\left[{\rm e}^{-\int_0^T r_u du} C_T | {\cal F}_t\right].$$ By martingale representation theorem, there is a process$N_t$such that $$M_t = M_0 + \int_0^t N_u dW_u,$$ where$\$ M_0=...

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