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The well-known risk-neutral pricing formula goes as follows (extracted from Shreve's Volume 2, section $5.2.4$ (Pricing Under the Risk-Neutral Measure)):

Given any $T>0$ and any $t\in[0,T],$ if $V(T)$ denotes the payoff of a derivative security at time $T$ which is $\mathscr{F}(T)$-measurable. Let $R(t)$ be the interest rate process. Then Steven Shreve indicates that $$V(t)=\tilde{\mathbb{E}}[e^{-\int_t^TR(s)ds}V(T)|\mathscr{F}(t)], \quad 0\leq t\leq T.$$

In the proof, it seems that he does not use any assumption at all. But from this post, it seems that we need to remove arbitrage opportunity for the formula above to hold.

Question: What assumptions do we need to fulfill when applying the risk-neutral pricing formula above?

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Shreve is a bit naughty here but, of course, he is right. When you have the risk-neutral measure $\mathbb{Q}$ or $\tilde{\mathbb{P}}$, you can price derivatives as discounted expectation by the very definition of the risk-neutral measure (better called: equivalent martingale measure). So indeed, once you have $\tilde{\mathbb{P}}$, you can price derivatives without any further assumptions.

But Shreve has not (yet) told you how do you find a risk-neutral measure? When does such a measure exist? The first Fundamental Theoream of Asset Pricing states that the absence of arbitrage possibilities is equivalent to the existence of (at least) one risk-neutral measure. The second FTAP states that a complete market can have at most one risk-neutral measure.

Thus, the main assumption is merely the absence of arbitrage (strategies with zero initial cost, positive probability of a positive payoff and zero probability of loosing money).

Note that you do not need further assumptions such as continuous trading (it applies equally to discrete models (Cox-Ross-Rubinstein, see Shreve's first book) and to time continuous models. The risk-neutral pricing formula also applies to jumps models which have discontinuous sample paths (see jump-diffusions like Merton and Kou or more advanced models like Variance Gamma, Normal Inverse Gaussian, CGMY etc.). In general, you don't need any assumptions on the dynamics of the stock price or the interest rate.

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  • $\begingroup$ I agree that it holds in discrete models, but such models can be seen as "toy models" to understand the concepts. If you take CRR, even with a very time step, you have only two possible states which is highly unrealistic; and if you scale your tree nodes with volatility and timestep as they advocate, you may find that this does mean continuous stock price :) $\endgroup$ – siou0107 Dec 4 '19 at 10:26
  • $\begingroup$ About stock prices with jumps, it is also true, RN pricing holds for derivatives. But it is much less straightforward, and the range of NA prices is typically very large. What banks do is that they "pick one risk-neutral measure" (refer to Guyon & Henry-Labordère Nonlinear Option pricing), but you should then be careful with the expression "Risk-neutral pricing works". Option trading remains a risky business under such wider assumptions. $\endgroup$ – siou0107 Dec 4 '19 at 10:29
  • $\begingroup$ Mathematically, discrete time models may be ``toy''` models but they do carry a lot of economic intuition and are indeed used to price options. After all, you need to discretise every time continuous model which does have a closed-form solution (so pretty much most of them). So, as the OP asked about where and when risk-neutral pricing holds, I think it is important to point out that time continuous is absolutely not required at all. $\endgroup$ – KeSchn Dec 4 '19 at 12:06
  • $\begingroup$ In response to the jump models. You’re right. Whenever the market is incomplete (more risk sources than tradable assts), you have multiple EMMs (see 2nd FTAP). The same applies e.g. to stochastic volatility models. Still, risk-neutral pricing does apply here which is what the OP asked. So continuous sample paths are not required either. Assuming that jump (or volatility) risk are diversifiable, we arrive at results like Merton (1976) or Heston (1993). $\endgroup$ – KeSchn Dec 4 '19 at 12:06
  • $\begingroup$ I just think it is very dangerous to link risk-neutral pricing to the Black Scholes world if it so much broader. Risk-Neutral Pricing is inherently related to the PDE approach (which is numerally dealt with finite differences) by Feynman Kac and to Monte Carlo simulations via the Law of Large Numbers. The QUAD method and Fourier tools stem directly from the idea that option prices are discounted expectations. And these are standard tools which are used for much more than just the Black Scholes world. So the assumption for risk-neutral pricing is really only the absence of arbitrage. $\endgroup$ – KeSchn Dec 4 '19 at 12:06
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For Risk-neutral Pricing to “work”, you need assumptions where risk elimination by trading financial instruments is possible : no counterparty risk, no transaction costs, continuous trading, continuous asset paths.

If such assumptions are not fulfilled (which is the case in real markets ; however, for large banks they are sufficiently near from reality), risk-neutral pricing cannot work.

A great reference on this topic is Joshi's The Concepts and Practice of Mathematical Finance. And, of course, the books of Paul Wilmott, who is always very keen to explain where finance theory breaks down ;)

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  • $\begingroup$ They look like assumptions in Black-Scholes models.... $\endgroup$ – Idonknow Dec 4 '19 at 7:40
  • $\begingroup$ Yes they do, since Black-Scholes is the merest example of such 'perfect' market. Basically, you can price by risk-neutral expectation BECAUSE you can diminish risk by trading the underlying (or eventually, derivatives on the underlying as in fixed income or in SV models). $\endgroup$ – siou0107 Dec 4 '19 at 8:16
  • $\begingroup$ I see. Am I right to say that since the risk-neutral pricing formula involves risk-neutral probability measure, the $V(T)$ must involve risk-neutral probability measure as well? For example, If $V(T) = (S(T) - K)^+$ is the payoff of a standard European call option, by assuming Black-Scholes world, the Brownian motion in the Geometric Brownian motion's SDE satisfied by $S(t)$ must be with respect to risk-neutral probability measure? $\endgroup$ – Idonknow Dec 4 '19 at 9:29
  • $\begingroup$ I am not sure to understand your comment, but indeed you must calculated the expected value of your payoff under the risk-neutral measure. $\endgroup$ – siou0107 Dec 4 '19 at 10:22

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