Suppose that I have an option on a single stock expiring at time $T$ and I replicate the payoff of this derivative by investing in the stock market and the money market. So this condition reads $$X(T) = V(T) \quad \text{almost surely}$$ where $X(T)$ is the value of my portfolio and $V(T)$ is the payoff of the derivative.

This condition holding under the actual probability measure is equivalent to it holding under the risk neutral measure, which I assume to exist and to be unique.

We then price the option by saying $$D(t)X(t) = \widetilde{E}[D(T)X(T)|F(t)] = \widetilde{E}[D(T)V(T)|F(t)]$$

I have a problem with the last equality. If the "almost surely" condition holds, then the last equality is implied. However, that equality does not necessarily imply the "almost surely" condition. Am I missing something here or is the fact that the price that comes of this method is unique and that it is a necessary condition for the almost surely condition to hold good enough for our purposes?


2 Answers 2


The theorem which justifies the equality of the expectations is Radon-Nikodym theorem. It says:


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 expectations:

$$E^P(DX)=\int DX \cdot dP=\int X \frac{dQ}{dP}\cdot dP=\int X dQ=E^Q(X)$$

  • $\begingroup$ I have no objection to the equality. It is crystal clear. What I am asking is how/if it implies the condition $$X(T) = V(T) \quad \text{almost surely}$$. The equality of expectations is implied by this condition. But I don't see how/if the equality of the expectations implies this condition. $\endgroup$
    – Calculon
    Commented Aug 23, 2014 at 15:16
  • $\begingroup$ @L'universo This is then the deeper reason, that changing the measure from $P$ to $Q$ means to price $X$ by its replicating portfolio aswell. Its the "Fundamental Theorem of Asset Pricing". Note that $Q$ is defined as the special measure $Q\sim P$ such that $X$ becomes martingale. $\endgroup$
    – emcor
    Commented Aug 23, 2014 at 15:22
  • 1
    $\begingroup$ @emcor: I'd suggest editing the initial answer rather than posting another one. $\endgroup$
    – olaker
    Commented Aug 23, 2014 at 15:24
  • $\begingroup$ @L'universo When you change the measure under an equivalent measure, if $P(V_T=X_T)=1$, you must also have $Q(V_T=X_T)=1$ because the measures are equivalent, meaning that $P=1\leftrightarrow Q=1$ (otherwise there would be some event with $P(\omega)=0$ and $Q(\omega)>0$. $\endgroup$
    – emcor
    Commented Aug 23, 2014 at 15:32

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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