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-1

I guess that the unclarity comes from the fact that hedge pricing is an incomplete model, as it does not take the FVA into account. Not from flaws of the mathematics used in the model.


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In our lecture, we were told to omit the proof because it was too difficult. Maybe it will help you though if you can read it here:


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After applying the theorem you can take an $Y$ orthogonal to the separation plane, pointing in the right direction, you will easily have the property $X.Y>0$ for all $X$.


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The theorem which justifies the equality of the expectations is Radon-Nikodym theorem. It says: $$E^P(DX)=E^Q(X)$$ where $D=dQ/dP$ is a change of measure process with $E(D)=1$, $D>0$ and $Q\sim P$. Note that $Q$ is further specified as the special riskneutral measure under which $X$ becomes a martingale. You can see it easily by writing out the ...


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"Almost surely" is actually like a "surely" condition, it holds under $P$ and under $Q$ measure because $Q\sim P$. "Almost surely" is just to say that $0$-probability events are not included. So for example, in a binomial model the price can only go up or down with probability $p$ and $1-p$ respectively, both $>0$. When you perfectly replicated the claim ...


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Essentially, if there is no arbitrage, then for any ratio of prices of assets $X_t/N_t =: M_t$, there must be a measure (depending on $N$ only) such that the ratio is a martingale (under that measure): $E_N[ X_T/N_T \,| F_t ] = X_t/N_t$. ...which is of course what you wrote, I'm just highlighting that it holds for any two assets. An asset here is ...


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The solution of your SDE is known as the Stochastic Exponential: $$X_t=X_0e^{\sigma Z_t-\sigma^2t^2/2}$$ (You can check this solution by applying Ito to the function $f(t,Z_t)=\ln X_t$.) Taking the expectation of $X_t$ to check its martingale property, since $Z_t\sim N(0,t)$ then $E(e^{\sigma Z_t})=e^{\sigma^2t^2/2}$: $$E(X_t)=E(X_0e^{\sigma ...


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Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t). However, we can't conclude a driftless SDE is always a martingale. ...



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