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Using the second derivative of the Call-Option-Price one can try to recover the pricing density.

Formally: Assuming a constant interst rate $r$ and also not making any assumptions on the model used to evolve $S_t$

$C(t,S_t,K,r,T)=e^{-r(T-t)}\int_0^{\infty}(S_T-K)^+f(S_T|S_t)dS_T$

The density is then recovered via
$p(S_T|S_t)=e^{r(T-t)}\frac{\partial^2 C(t,S_t,K,r,T)}{\partial K^2}|_{K=S_T}$

As a follow-up to my last question:

  1. What are the applications of this recovered density ?

  2. How can we Interpret it ? (can it be considered the "real" probability density ? - seeing how it is used in a pricing context it should still be risk neutral)

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1 Answer 1

up vote 2 down vote accepted

1) A straigthforward application is to price any complex payoff at maturity using this.

By that I mean a payoff that is such that the price of the option is

$$P = e^{-r(T-t)}E[f(S_T)] $$

Which you can then calculate by integrating $f(S_T)$ w.r.t. to your density.

One of the challenges though is to have a proper marks and inter/extrapolation for the implied vols of the wings (i.e. far away from current forward) so that your density does not become negative.

Therefore another application for this density is that it is a good way to check for no arbitrage on a term option volatility curve.

2) Yes, it is just the probability which you can hedge against by using derivatives on the market. It's a bit like if a horse has 1 against 4 odds, that does not mean that there is a 20% chance that he will win, it just (sort of) means that if you have some product whose price needs to know that probability, the right number to put in the price is 20% because the hedging instrument you will use (i.e. better 1 against 4 on that horse, or against) will cost exactly that.

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so instead of making assumptions on the pricing model and perhaps using monte carlo - one uses this shortcut. Could you please elborate on your statement on the density not becoming negative :) –  Probilitator Mar 14 at 10:55
    
also I have to admit the part with the no-arbitrage test on the option vol curve is not really clear to me :( –  Probilitator Mar 14 at 10:57
    
1) you can use a MonteCarlo, but then you need to sample using this density. In fact the density might not be practical because to sample a distribution it's easier to start from the inverse of the cumulative distribution fn instead. Besides, if you only have 1 dimension an integral is more efficient than MC 2) the density is closely related to the price of a butterly, i.e. call(K-delta)-2*Call(K)+call(K+delta) (just make delta->0). Now clearly the payoff of a butterfly is always positive, so because of that the density should also be positive otherwise there's an arb. –  joelhoro Mar 14 at 18:11
    
thank you for the explanation. Could you perhaps also add more detail regarding "one of the challanges is to have a proper marks and ..." –  Probilitator Mar 20 at 9:23
1  
you're welcome. The issue is that traders will usually only get vol marks for a few strikes, and you need to in/extrapolate. If your vols are constant, you will get the lognormal density, but if you start moving some of the vols arbitrarily, or if your in/extrapolation are not well chosen then you will quickly end up with negative density. –  joelhoro Mar 21 at 12:42

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