2
$\begingroup$

Suppose we do not have a particular pricing model, we have just a frictionless market with constant interest rate (say $0$), and some traded stock $S$ which does not pay dividends. For any expiry $T$ to price options/contingent claims consistently we need a pricing rule, or equivalently a pricing (risk-neutral) probability measure. In particular, to price European call options we only need a marginal of such probability measure being the distribution of $S_T$.

Standard non-arbitrage arguments imply that the futures price $F(0,T) = \Bbb E^{\Bbb Q}S_T$ must satisfy $F(0,T) = S_T$, otherwise using a static replication strategy we can exploit mispricing in case of inequality. Hence, for the pricing distribution $\Bbb Q$ at least the first moment is fixed externally. What about the rest of the distribution? Say, we only focus on distributions that can be completely recovered from their moments. The first one is fixed, what about others, are we completely free in choosing them given the conditions above?

$\endgroup$
1
$\begingroup$

The answer is: yes. We can consider a model that assumes there is only one jump with distribution $p$, and otherwise the stock value does not change. Then for $p$ to be a martingale measure the only condition is on expectation of $p$. Hence, any distribution with desired expectation can be a marginal of some pricing measure.

| improve this answer | |
$\endgroup$
0
$\begingroup$

The other moments are not free.

Suppose we are in the standard BS environment with one stock and one bond and a single brownian motion. Suppose we have a derivative that at maturities pays: $V_T=S^2_T$ and we want to price it. Under the martingale measure we know that: $E_t^Q[S_T]=S_t e^{r(T-t)}$ and $Var^Q_t(S_T)=S_t^2e^{2r(T-t)}(e^{\sigma^2(T-t)}-1)$. Using the fact that $Var(X)=E(X^2)-E(X)$ we can pin-down the value of the security $V$ at time $t$.

Hope this example shows that we are not free to pick other moments. I am not sure whether I fully understood your question though.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I assumed that we have not chosen a particular model. Of course Brownian motion would impose some additional constraints just by its special structure. To say the least, the resulting distribution is going to be lognormal. I think I found that the answer is yes. Will post it soon, thanks for the interest though. $\endgroup$ – Ilya Jul 13 '15 at 5:47
0
$\begingroup$

basically, you have very few constraints. The main other constraint you might consider is that if the real-world probability of lying in a given set is positive, the risk-neutral probability must be too.

So a point mass at the futures price is not valid unless you believe that this is the case in real-world too.

(see chapter 6 of my book "concepts etc" for further discussion.)

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.