The easiest way to remember what the risk-neutral measure is, or to explain it to a probability generalist who might not know much about finance, is to realize that it is: It is also worth noting that in most introductory applications in finance, the pay-offs under consideration are deterministic given knowledge of prices at some terminal or future point in time.
Therefore, today's price of a claim on a risky amount realised tomorrow will generally differ from its expected value.
Most commonly, investors are risk-averse and today's price is below the expectation, remunerating those who bear the risk.
It turns out that in a complete market with no arbitrage opportunities there is an alternative way to do this calculation: Instead of first taking the expectation and then adjusting for an investor's risk preference, one can adjust, once and for all, the probabilities of future outcomes such that they incorporate all investors' risk premia, and then take the expectation under this new probability distribution, the risk-neutral measure.
The main benefit stems from the fact that once the risk-neutral probabilities are found, every asset can be priced by simply taking the present value of its expected payoff.
The absence of arbitrage is crucial for the existence of a risk-neutral measure.
In fact, by the fundamental theorem of asset pricing, the condition of no-arbitrage is equivalent to the existence of a risk-neutral measure.
The method of risk-neutral pricing should be considered as many other useful computational tools—convenient and powerful, even if seemingly artificial.
be a d-dimensional market representing the price processes of the risky assets,
is a random variable on the probability space describing the market.
Then today's fair value of the derivative is where any martingale measure
If no equivalent martingale measure exists, arbitrage opportunities do.
In markets with transaction costs, with no numéraire, the consistent pricing process takes the place of the equivalent martingale measure.
There is in fact a 1-to-1 relation between a consistent pricing process and an equivalent martingale measure.
, consider a single-period binomial model, denote the initial stock price as
we find that the risk-neutral probability of an upward stock movement is given by the number Given a derivative with payoff
when it goes down, we can price the derivative via Suppose our economy consists of 2 assets, a stock and a risk-free bond, and that we use the Black–Scholes model.
In the model the evolution of the stock price can be described by Geometric Brownian Motion: where
Utilizing rules within Itô calculus, one may informally differentiate with respect to
and rearrange the above expression to derive the SDE Put this back in the original equation: Let
-martingales we can invoke the martingale representation theorem to find a replicating strategy – a portfolio of stocks and bonds that pays off
It is natural to ask how a risk-neutral measure arises in a market free of arbitrage.
For simplicity, consider a discrete (even finite) world with only one future time horizon.
Now it remains to show that it works as advertised, i.e. taking expected values with respect to this probability measure will give the right price at time 0.
Consider a portfolio P consisting of Ci amount of each Arrow security Ai.
The lack of arbitrage opportunities implies that the price of P and C must be the same now, as any difference in price means we can, without any risk, (short) sell the more expensive, buy the cheaper, and pocket the difference.
If the interest rate R were not zero, we would need to discount the expected value appropriately to get the price.
In particular, the portfolio consisting of each Arrow security now has a present value of
In a complete market, every Arrow security can be replicated using a portfolio of real, traded assets.
In a more realistic model, such as the Black–Scholes model and its generalizations, our Arrow security would be something like a double digital option, which pays off $1 when the underlying asset lies between a lower and an upper bound, and $0 otherwise.