Why Option-Implied Price Distributions Are Risk-Neutral

# What the implied distribution really is?


# What the implied distribution really is?












From volatility surfaces we have a implied distribution of $S_T$. This distribution is the real world distribution or this is a risk neutral distribution?

## Answer by Quantuple (score 4, accepted)

https://quant.stackexchange.com/a/31481

In this related question How to derive the implied probability distribution from B-S volatilities?, it is shown how to infer the implied probability density of the future prices of a risky asset from a continuum of call prices written on that asset (Breeden-Litzenberger identity).

The developments, which I invite you to read, basically rely on the fact that the call price writes $$ C=e^{-rT} \int_0^\infty (S-K)^+ p(S) dS $$ or equivalently $$ C = e^{-rT} \Bbb{E} \left[ (S_T-K)^+ \right] \tag{1} $$ under some measure where $S_T$ is a random variable with probability density function: $$ d\Bbb{P}(S_T \leq S)/dS = p(S) $$

Now, the fundamental theorem of asset pricing tells us that equation $(1)$ holds under the so-called risk-neutral measure (a measure equivalent to the real-world measure but under which the $t$-value of any self-financing strategy is a martingale when expressed with respect to the risk-free money market account numéraire).

Consequently, the implied density $p(S)$ you compute by evaluating $$ p(S) = e^{rT} \frac{\partial^2 C}{\partial K^2}(K=S) $$

is indeed a risk-neutral pdf (because it relies on the risk-neutral expression of the call price $(1)$.)

## Answer by Will Gu (score -1)

https://quant.stackexchange.com/a/31480

I'm not a probabilist but I tend to think real-world distribution in financial world doesn't exist (or at least is not a proper term). None of the financial events are really repeatable or IID.

And to your question, it should be risk-neutral density function.