Lets consider a pricing model like Vasicek.

Apparently, if you calibrate a derivatives pricing model to market prices this gives you risk neutral parameters. Its not clear to me as to WHY this will definitely be risk neutral. See here atmif.com/papers/rn.pdf under Market Calibration. "Many authors in the literature simply claim that they are using the "risk neutral" pricing measure when doing market calibrations." How can they just claim this?

If you need your pricing model to be arbitrage free to give a risk neutral measure, then is it just you need model which is arbitrage free and then you just calibrate it to the market and you are done? When you look at the literature, you start with some model of the underlying dynamics and then you apply Girsanov's theorem in order to become risk neutral. Are you calibrating the risk neutral version of your model or the original one to the market?

I thought that for something to be risk neutral it had to drift at the risk free rate $r dt$ (and plus a $\sigma dW$ term). However, derivatives pricing models almost always have non-zero drifts.

What am I not understanding here?

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    $\begingroup$ Copy of quant.stackexchange.com/questions/63725/… ? $\endgroup$
    – river_rat
    May 15, 2021 at 21:33
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    $\begingroup$ Non-dividend paying assets in the Heston model do have the short rate as instantaneous return, see this answer. $\endgroup$
    – Kevin
    May 15, 2021 at 23:11
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    $\begingroup$ Can you elaborate on "We only calibrate on one path" as I do not understand what you mean by that. $\endgroup$ May 20, 2021 at 13:46
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    $\begingroup$ Consider a diffusion model for a stock, no dividends. You can use a single historical path to estimate volatility, but that has nothing to do with risk neutrality. The risk neutral condition for the model dynamics under the risk neutral measure is that the stock price drift is always equal to the instantaneous rate, nothing more. $\endgroup$ May 20, 2021 at 14:05
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    $\begingroup$ In calibration, you seek model parameters to minimise the weighted sum of squared pricing errors, the difference between market prices and model prices (it's a bit like GMM, in case you have an econometrics background). You compute these model prices using risk-neutral pricing: expected payoff under $\mathbb{Q}$ and discounted at $r$. Thus, you find the parameters corresponding to that pricing approach (i.e. the risk-neutral parameters). They are useless to predict future price trajectories, but are more than sufficient for pricing (exotic) derivatives . $\endgroup$
    – Kevin
    May 21, 2021 at 11:26

1 Answer 1


There are two parts to your question which I try to answer separately. The first one is about what calibration actually is whereas the second question deals with risk-neutral pricing.

As an example, we can use any model. I continuously refer to the stochastic volatility model from Heston (1993) as an example for equity options. Any thoughts equally apply to other models or asset classes (think of interest rate derivatives and short rate models). The Heston model for a stock price $S_t$ reads as \begin{align*} \text{d}S_t&=\mu S_t\text{d}t+\sqrt{v_t}S_t\text{d}W_S, \\ \text{d}v_t&= \kappa(\theta-v_t)\text{d}t+\xi\sqrt{v_t}\text{d}W_v, \end{align*} where $\text{d}W_S\text{d}W_v=\rho\text{d}t$. Thus, there are five model parameters ($\mu,\kappa,\theta,\xi,\rho$).

Estimation and Calibration

Estimation finds the real-world distribution of stock returns; calibration aims to parametrise the risk-neutral distribution. To this end, estimation uses historical observations whereas calibration uses observed market prices. Estimation is commonly used for risk management (e.g., value-at-risk) whereas calibration is used for pricing/hedging/trading derivatives.

Let's begin with estimation: Take historical returns of the S&P 500 index and compute the (log-)likelihood function from the Heston model to find the parameters that are most likely to have given rise to the given sample of stock returns (This is called maximum likelihood estimation). A generalisation is GMM which compares sample moments from the data with moments implied by the model and finds the parameters that minimise the difference between the two. Intuitively, you seek parameters such that simulations of the model yield sample paths that look very similar to the actual data you observe in financial markets. In particular, this includes that asset prices grow at same rate $\mu$ that reflects the systematic risk of that asset. You could such a model for forecasting.

Let's turn to calibration: Take current market prices for S&P vanilla call and put options. Using the closed-form solution for option prices in the Heston model, you now seek model parameters which minimise the (squared) difference between observed market prices and implied model prices. This time, you don't care about historical return data. You only take yesterday's option prices and you match your model with these prices. If you calibrate a short rate model, you can use the prices of zero bonds, bond options, caps etc. Your choice of calibration instruments should match your intended application of the model. Because these instruments are priced using a risk-neutral framework (see below), the calibrated parameters correspond to the $\mathbb{Q}$ measure in which assets are assumed to grow at rate $r$ (and not $\mu$). Thus, these parameters must not be used for forecasting! These parameters, instead, are used for valuing other (complex) derivatives.

Two points are in order

  • There are two different time horizons: Estimation requires a (long) time series of returns. Calibration requires price data from one single day.

  • In practice, there are many subtitles when one wants to implement estimation or calibration in practice: How to discretise a continuous time model? What moments to match for GMM? How to weight different market prices? How to clean options data? Calibrate to prices or implied volatilities, etc.

Pricing and Risk-Neutral Distribution

Option pricing is (almost) always done under the risk-neutral ($\mathbb{Q}$) measure. The value of a European-style call option is computed as discounted expected payoff \begin{align*} C=e^{-rT}\mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}]. \end{align*}
The risk-neutral distribution ($\mathbb{Q}$) is necessary such that we can discount at the risk-free rate $r$ (instead of bothering with risk premiums embedded $\mu$). The idea is that the $\mathbb{P}$ distribution (i.e., real stock price movements) contains these risk premiums which are extremely difficult to identify. The risk-neutral pricing framework allows us to avoid all of this and arrive at the same option prices while pretending everyone is risk-neutral (we pretend that $\mu=r$).

The crux is that we do not need to know $\mu$ to price options. When we calibrate a model, we minimise the difference between market prices and model prices. These model prices are computed using the risk-neutral framework (all option prices are). Thus, you only recover the risk-neutral distribution from option prices, see also Breeden and Litzenberger (1978). But that's no problem. We only need that risk-neutral distribution to value derivatives.

The $\mathbb{P}$ and $\mathbb{Q}$ distribution are linked via the stochastic discount factor (or pricing kernel) which is just a scaled Radon-Nikodym derivative (change of measure via Girsanov's theorem). Thus, the $\mathbb{P}$ parameters and the $\mathbb{Q}$ parameters are also linked, using the risk premiums embedded in the pricing kernel, see this answer. Thus, if we knew the true SDF and the true $\mathbb{Q}$ parameters, we could recover the true $\mathbb{P}$ parameters.

  • $\begingroup$ This is a great answer. Can you suggest how I could rephrase my question in order to get it undownvoted? I feel your answer deserves better! $\endgroup$
    – Trajan
    May 24, 2021 at 14:45
  • $\begingroup$ @Trajan Thank you very much. Firstly, I'm glad you like the answer! I think a lot of the confusion in the comments came from the fact that when the term "calibration" was used, no one was thinking about the difference to estimation and the difference between fitting $\mathbb{P}$ and $\mathbb{Q}$ parameters. But I think it became clear in the comments (to me at least). $\endgroup$
    – Kevin
    May 24, 2021 at 14:56
  • $\begingroup$ None of the textbooks Ive read cover estimation or calibration $\endgroup$
    – Trajan
    May 24, 2021 at 15:05
  • $\begingroup$ @Trajan I think I've previously mentioned the book from Hirsa? I have seen many other textbooks discussing the topic. In the end, calibration is only a least squares fit. But you're right. Standard introductions like Shreve or Hull don't cover how to find model parameters. These books are more concerned with teaching the basic tools. But if you switch to applications, you find a lot of material about calibration and estimation. Please let me know if you have more questions/if I can add to my answer $\endgroup$
    – Kevin
    May 24, 2021 at 15:29

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