What is the Risk Neutral Measure?

I don't believe this has been answered on the internet well and with all the parts connecting.


  1. What is the risk neutral measure/pricing?

  2. Why do we need it?

  3. How we calculate the risk neutral measure or probabilities in practice?

  4. What connection has risk neutral pricing to the drift of a SDE? Does this help with 3)?


3 Answers 3



Great answer given by Kevin. I would like to contribute an additional perspective. My experience with and my understanding of the Risk Neutral measure is entirely based on "no arbitrage" and "replication / hedging" arguments.

The way I would like to explain this view is via the following three-step construction:

(i) First, I want to build the intuition with a one-period discrete model: only a single stock and a risk-free account, no derivatives. The aim is to show that even without trying to price derivatives, one can create a mathematical object called a "risk-neutral probability measure", just by assuming no arbitrage in the model.

(ii) I then want to show that pricing a derivative by replication of its payoff with the underlying instrument and the risk-free rate instrument is equivalent to taking the expectation of the derivative pay-off under the risk-neutral measure & discounting it.

(iii) I then want to highlight that the discrete model converges to the well-known continuous Black-Scholes model.

Part 1: Discrete single-period model:

I assume that today's stock price is $S_0$, and one period from now, the stock price can be $S_0 * u=S_u$ or $S_0 * d=S_d$, with $u$ and $d$ being "up" and "down" multiplicative factors. I assume that the risk-free rate is $r$.

Without imposing some conditions on $u$, $d$ and $r$, there might be some arbitrage opportunities. If for example $e^r>u$, I could short the stock at time $t_0$ and invest the proceeds $S_0$ into the risk-free account: in both future states at time $t_1$, I could then buy the stock back for less than my proceeds $S_0e^r$ from the risk free (because $S_u$ and $S_d$ would both be less than $S_0e^r$).

If, on the other hand, $e^r<d$, I could borrow exactly $S_0$ money at $t_0$ to buy the stock, and in both future states of the world at time $t_1$, the stock value would be higher than what I would have to repay ($S_0e^r$): so there would be arbitrage again.

Imposing $d \leq e^r \leq u$, will ensure no arbitrage in the one-period model.

Now I am going to perform the following algebraic manipulation:

$$ S_0 = \frac{S_0(u-d)}{(u-d)}= \\= \frac{1}{e^r}\frac{S_0(u-d)e^r}{(u-d)}= \\ =\frac{1}{e^r}\frac{S_0(u-d)e^r+(S_0ud - S_0ud)}{(u-d)}=\\= \frac{1}{e^r}\left( \frac{S_0ue^r -(S_0ud)}{u-d} + \frac{-S_0de^r+(S_0ud)}{u-d} \right)=\\=\frac{1}{e^r}\left(S_0u \left( \frac{e^r -d}{u-d} \right) + S_0d \left(\frac{u-e^r}{u-d} \right) \right)$$

The no arbitrage condition $d \leq e^r \leq u$ will result in the following bounds:

$$0 \leq \frac{e^r -d}{u-d} \leq 1$$

$$0 \leq \frac{u-e^r}{u-d} \leq 1$$


$$ \frac{e^r -d}{u-d} + \frac{u-e^r}{u-d} = 1 $$

Let's call $\frac{e^r -d}{u-d}:=p_u$ and $\frac{u-e^r}{u-d}:=p_d$. In the one period model, the stock going up and the stock going down are two different states of the world, i.e. there is no "intersection" between these states in the probabilistic sense. Therefore $p_u$ and $p_d$ are additive over disjoint sets and they are within the zero-one range, therefore mathematically, these parameters qualify as a probability measure.

Rewriting the algebraic manipulation above in terms of $p_u$ & $p_d$ yields the following:

$$ S_0 = \frac{S_u p_u + S_d p_d}{e^r} = \frac{1}{e^r}\mathbb{E} [S_1] $$

Also notice that in the entire construction above, we did not talk about the probabilities of the stock going up or down. Every market participant might have his or her Bayesian view of the world with different probabilities assigned to the stock going up or down. But the risk-neutral measure is agreed upon by the market as a whole as a consequence of no arbitrage.

This also brings up an interesting point: in my view, the risk neutral probabilities are probabilities only in the "mathematical object" sense. They do not actually represent "likelihoods", in the sense that we human beings like to interpret probabilistic events with.

Part 2: Pricing derivatives:

Let's assume we want to price a derivative on the stock with pay-off function $V(S_t)$ (could be a forward, option, whatever). The derivative pay-off in the two states will trivially be $V(S_u)$ and $V(S_d)$. We have two states, two underlying instruments: let's try to replicate the derivative pay-off in both states ($x$ is the number of stocks and $y$ is the amount invested in the risk-free account: I want to replicate the derivative pay-off in both states with $x$ stocks and $y$ risk-free investment):

$$ (i) x S_u + ye^r = V(S_u) $$ $$ (ii) x S_d + ye^r = V(S_d) $$

Solving gives:

$$ x = \frac{V(S_u)-V(S_d)}{S_0(u-d)} $$

$$ y = \frac{uV(S_d)-dV(S_u)}{(u-d)} \frac{1}{e^r} $$

Therefore the derivative price at time $t_0$ is the $x$ amount of the stock + $y$ amount invested in the risk-free account:

$$ V(S_0,t_0) = x*S_0 + y*1 = \\ = \frac{V(S_u)-V(S_d)}{S_0(u-d)}*S_0 + \frac{uV(S_d)-dV(S_u)}{(u-d)} \frac{1}{e^r}*1$$.

The above evaluates to:

$$\frac{1}{e^r}\left(V(S_u) \left( \frac{e^r -d}{u-d} \right) + V(S_d) \left(\frac{u-e^r}{u-d} \right) \right) $$

Notice that again we can write $\frac{e^r -d}{u-d}:=p_u$ and $\frac{u-e^r}{u-d}:=p_d$, where notably $p_u$ and $p_d$ are the same as in Part 1 above, Therefore, instead of having to compute the replication portfolio weights $x$ and $y$, the derivative can be priced as:

$$ V(S_0,t_0) = \frac{1}{e^r}\left(V(S_u) p_u + V(S_d) p_d \right) = \\ = \frac{1}{e^r} \mathbb{E}[V(S_1,t_1)]$$

Hopefully, by now you can see where I am going with this: the risk-neutral measure pricing technique has the following features:

(A) Is a consequence of no-arbitrage assumptions in the model

(B) Taking the expectation of a derivative pay-off and discounting it to today is the equivalent of: computing "replication portfolio" weights at each time-step, and pricing the derivative using these replicating weights at time $t_0$.

Part 3: Continous-time models:

Extending the one-period model leads to a multi-period "binomial tree" discrete model. Pricing a derivative on a multi-period tree would require working "backwards" from the terminal pay-off and computing the replicating portfolio pay-off at each node. Alternatively, the more convenient way is to use the risk-neutral expectation of the terminal pay-off and discounting it to "today": as that will produce the same result (as shown above) and will save us having to worry about the replicating portfolio weights.

There are multiple papers online showing how the binomial tree model converges to the Black-Scholes formula when the number of steps tends to infinity as $\delta t$ tends to zero (for example a great paper by John Hull here or alternatively this paper here).

As a sketch, we can show that the multi-period Binomial model for the stock converges to the well-known continuous Geometric Brownian Motion (GBM) model (which in turn can be used to derive the Black-Scholes formula directly by applying the risk-neutral expectation to the option pay-off at maturity where the stock process is simulated with GBM).

For the multi-period Binomial model, we have:


with $k\sim Bin(n,p)$. The "Cox-Ross-Rubinstein" parameters for "up" and "down" are then set as follows: $u:=e^{\eta\frac{T}{n}+\sqrt{\frac{T}{n}}\sigma}$, $d:=e^{\eta\frac{T}{n}-\sqrt{\frac{T}{n}}\sigma}$.

It is well known that the Binomial distribution converges to Normal, in fact, using CLT, we get:


Going back to the equation for $S_n$, taking a log and substituting for $u$ and $d$, we get:

$$ln \left( \frac{S_n}{S_0} \right) = k \left( 2 \sqrt{\frac{T}{n}} \sigma \right) + n\left(\eta \frac{T}{n}-\sqrt{\frac{T}{n}}\sigma\right)$$

Which gives:

$$\mathbb{E}\left[ln \left( \frac{S_n}{S_0} \right)\right]=\sqrt{Tn}\sigma(2p-1)+\eta T$$

$$Var\left(ln \left( \frac{X_n}{X_0} \right) \right)=4T\sigma^2p(1-p))$$

So we see that:

$$\lim_{n\to\infty}\left( \frac{X_n}{X_0} \right)\xrightarrow{d}N(\sqrt{Tn}\sigma(2p-1)+\eta T,2\sqrt{T}\sigma\sqrt{p(1-p))}$$

Taking $p=0.5$, we get:

$$\lim_{n\to\infty}ln\left(\frac{X_n}{X_0}\right)\xrightarrow{d} N\left(\eta T,\sigma \sqrt{T}\right)$$

If we now set $\eta:=\mu -0.5\sigma^2$, we see that the discrete multiperiod Binomial model for the stock price converges exactly to the continuous Geometric-Brownian-Motion model in distribution.

Another interesting fact to note is that the replicating weight of the stock in part 1, i.e. $x$, converges to $N(d_1)$, i.e. the instantaneous option Delta.

I will conclude by producing the same summary as Kevin, but with the following additional comments:


  • Risk-neutral probability measures are artificial measures (agreed) made up of risk-aversion (SDF) and real-world probabilities (disagree here: don't think risk-aversion comes into it. I see it as an artificial measure entirely created by assuming the existence of no-arbitrage and completeness).

  • Derivatives can be priced relative to underlying assets. This hedging price can be computed as expectation with respect to the risk-neutral probability measure (agreed). Equivalent martingale measures are deeply related to the absence of arbitrage and completeness (agreed: I would say they are not just deeply related to these, they are the consequence of these).

  • The risk-neutral density can be estimated from observed market data (agreed: i.e. twice differentiating the Implied Vol surface with respect to strike). The risk-neutral framework connects many different approaches to derivatives pricing

  • $\begingroup$ So to price a new product what would you do? Find a replicating portfolio and then? Do you fit a model to the data for related products and then back out the parameters and then use these parameters to compute the expected value for the new product? $\endgroup$
    – Trajan
    Commented Jun 28, 2020 at 9:25
  • $\begingroup$ The risk neutral measure is created in "part 1", just as a consequence of assuming no arbitrage between the stock and the risk-free account. In the second part, I show that you first price the derivative by trying to calculate the replicating weights. Then I show that this is mathematically identical to taking the expectation of the derivative pay-off under the risk-neutral measure. You don't "create" the measure to price the derivative, the measure already exists in "part 1" and you just use the same measure in part 2. $\endgroup$ Commented Jun 28, 2020 at 9:25
  • $\begingroup$ ok thanks helps a lot, ill think about this a bit more $\endgroup$
    – Trajan
    Commented Jun 28, 2020 at 9:26
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    $\begingroup$ @JanStuller I just had to respond and add to my answer to defend the financial economics approach. :) You're right: the mere existence of $\mathbb{Q}$ is independent of any equilibrium notion. However, it's much easier to understand (in my opinion) how $\mathbb{P}$ and $\mathbb{Q}$ differ, when we know how $M$ deforms $\mathbb{P}$ and what the SDF $M$ is in the first place. Why is $\mathbb{Q}$ called ``risk-neutral''? It simply removes the risk, i.e. the $M$ from $\mathbb{E}^\mathbb{P}[MX]$ and instead writes $e^{-rT}\mathbb{E}^\mathbb{Q}[X]$. But $\mathbb{Q}$ simply absorbs (hides) the risk. $\endgroup$
    – Kevin
    Commented Jan 5, 2021 at 16:54
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    $\begingroup$ @Kevin: PS: I think the two answers can complement each other. Or at least mine can add some additional points to yours (which is already self-contained and is an excellent answer). $\endgroup$ Commented Jan 5, 2021 at 17:38

Life Without a Risk-Neutral Measure

How would we price assets without the measure $\mathbb Q$? Well, we would start with some version of the Euler equation $P_t=\mathbb{E}_t[M_{t+1}P_{t+1}]$, where $M$ is the stochastic discount factor (SDF). This equation holds under very weak assumptions (law of one price) and uses real-world probabilities. So, we take the return in every future state of nature multiply with the SDF to account for riskiness and weight this product by real world probabilities. This gives the current price of any asset (underlying and derivative).

The main problem: What is the SDF? In order to find the SDF, we need a general equilibrium model (such as C-CAPM or CAPM) which requires us to make assumptions about investor's utility function (simple CRRA or recursive Epstein-Zin?) etc. There are many different proposed SDFs in the asset pricing literature. And there is disagreement which models fits the data best.


Recalling the Euler equation $P_t=\mathbb{E}_t[M_{t+1}P_{t+1}]$, the idea is to merge the SDF into the expectation (i.e. changing the probabilities associated to the expected value) which allows us to write $$P_t=\mathbb{E}_t[M_{t+1}P_{t+1}]=\frac{1}{1+r}\mathbb{E}_t^Q[P_{t+1}].$$ Here, I kind of assume discrete time steps and discrete compounding at a fixed rate $r$ but it of course equally applies in a time continuous setting.

The benefits are clear. We can now compute today's price of an asset by computing the expectation of the future price (or payoff) and discount this expectation at the risk-free rate. Importantly, we do not need to specify a utility function and an SDF anymore. We only need to find these new probabilities (the artificial measure $\mathbb Q$) and compute the expectation with respect to this new measure. In particular, the derivatives price will not depend on real-world probabilities.

What's the intuition about $\mathbb Q$? Well, it's a merger of real-world probabilities with the stochastic discount factor. This means $\mathbb Q$ adjusts the probabilities of outcomes by joining them with risk preferences (state prices). The new probabilities would then correspond to a world where all investors are risk-neutral (in which investors do not ask for risk premia and discount every cash flow at the risk-free rate $r$). This explains why $\mathbb Q$ is also called risk-neutral measure (i.e. $M$ represents risk and it vanished, what's left as discount factor is the risk-free rate of return, $r$). Risk-neutrality means being indifferent between any gamble and its expected payoff.

In line with the SDF interpretation, risk-neutral probabilities of bad events (declining stock prices) are higher than their corresponding real-world probabilities. On the other hand, risk-neutral probabilities deflate the likelihood of good events (increasing stock prices). As a result, the risk-neutral density is skewed to the left (negative skewness).


Another common term for the risk-neutral measure is equivalent martingale measure. Equivalent simply means that both measures agree on which events have zero probability. A martingale is a (integrable and adapted) stochastic process which models a fair game, i.e. $\mathbb{E}[X_t|\mathcal{F}_s]=X_s$ (the best prediction for the future value $X_t$ given the knowledge $\mathcal{F}_s$ at time $s$ is the value at time $s$ itself, $X_s$.

Stock prices have some real world drift $\mu$ which rewards investors for holding this risky assets. Typically, $\mu>r$. In a risk-neutral world, agents do not care about risk and do not ask for such a risk premium. Thus, in a market with risk-neutral investors, stocks (and every other assets) return the risk-free rate $r$. If we now discount stock prices using the risk-free asset as numéraire, we eliminate the drift and the discounted stock price becomes a $\mathbb Q$-martingale. Importantly, this only holds in the artificial risk-neutral world. In the real world, stocks do bear an equity risk premium and their price is not a martingales.

By the tower law, discounted derivative prices are martingales as well. They are defined by $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[X|\mathcal{F}_s]$, where the random variable $X$ is the future payoff (we actually assume the absence of arbitrage here for $\mathbb{Q}$ to exist, see below).

The above derivative price does not depend on the drift of the stock price. That's an important insight. In praxis this is great because we do not need to estimate the mean rate of return of a stock. Furthermore, even if two agents had completely different estimates for this drift, they could still agree on a fair price of an option written on this stock because the drift does not matter. In some extreme case, if $K=0$, a call option would just pay the stock and that could be replicated by just buying the stock. No need to consider the drift. Alternatively, at the money (forward), put and call options have the same value. If the drift mattered, either put or call options would increase in value, violating this relationship.

Fundamental Theorems of Asset Pricing

The fundamental theorems of asset pricing are key results in finance linking the market properties arbitrage and completeness to risk-neutral probability measures.

  • The first theorem of asset pricing states that the absence of arbitrage is equivalent to the existence of at least one equivalent martingale measure.
  • The second theorem of asset pricing states that a market is complete if and only if at most one equivalent martingale measure exists.

In simple terms, an arbitrage (free lunch) is a self-financing trading strategy which does not require any funds at inception and yields a non-negative payoff in the future with a positive probability of an actual gain (strictly positive payoff). A complete market is a market in which any reasonable payoff can be replicated.

Because the set of all equivalent martingale measures is convex, either no EMM exists (there exist arbitrage strategies), or precisely one EMM exists (market is free of arbitrage and complete) or infinitely many EMMs exist (market is free of arbitrage but incomplete). It is impossible to only have 2, 42 or 3141592 EMMs.

The assumption that no arbitrage exists is an easy one to accept. We thus know that at least one equivalent martingale measure exists. The question is completeness. This is debatable (Is volatility risk tradable? What about jump risk? etc). If infinitely many EMMs exist, then option prices are not unique, i.e. there exist pricing intervals whose elements all represent valid, arbitrage-free derivative prices. In these cases, one either neglects some risk sources (pretending the market is complete) by choosing a corresponding model or one implements some minimum variance hedging strategy, accepting that there's not one true price.

Unfortunately, absence of arbitrage and completeness are a bit at odds. The more asset payoffs there are, the more complete the market is but arbitrage possibilities are also likelier to exist.

Relationship to Hedging Prices

A key element in no-arbitrage pricing theory are linear pricing functionals which extend hedging prices and are monotone (higher payoffs lead to higher prices). By Riesz' Representation Theorem, such a linear functional can be represented by an inner product, i.e. there exists some random variable $M$ such that the pricing functional $\pi$ can be written as $\pi(X)=\mathbb{E}^\mathbb{P}[MX]$ for any payoff (contract) $X$. This $M$ is of course the SDF.

So, if no arbitrage strategy exists, we have (at least) one linear pricing functional. This functional gives rise to an SDF and an SDF can be merged with real-world probabilities to give a risk-neutral measure. This chain directly links hedging prices to EMMs. In fact, there’s an one-to-one relationship between pricing functionals and EMMs. Thus, just like EMMs, the set of pricing functionals is convex. If markets are complete, then the SDF and EMM are unique. As a result, they have to coincide with the simple hedging price.

Agreeing with hedging prices further motivates why individual risk preferences do not enter the risk-neutral pricing framework. You price assets relative to each other. You assume you know the prices of some basic (primitive) assets (from some equilibrium model) and then you price further assets (derivatives) by trading the basic options. This is possible by the absence of arbitrage. The prices of the original assets already incorporate risk-aversion etc. So, by relative pricing (hedging), you do not need to include preferences again.

Estimating the Risk-Neutral Density

Breeden and Litzenberger (1978) show that $$\mathbb{Q}[\{S_T\geq \kappa\}] = -e^{rT}\frac{\partial C(S_0,K,T)}{\partial K}\bigg|_{K=\kappa},$$ which in turn means that the risk-neutral density can be extracted as follows $$q_T(\kappa) = e^{rT}\frac{\partial^2 C(S_0,K,T)}{\partial K^2}\bigg|_{K=\kappa}.$$

We can thus use observed option prices, $C(S_0,K,T)$, to estimate the risk-neutral density $q_T$. Clearly, the risk-neutral density changes over time and with option maturity.

The biggest problem with this approach is that option prices are needed at every positive strike. Around ATM strikes, there are enough liquid options which can easily be interpolated but at extreme strikes (very OTM and very ITM), data becomes an issue. Using the put-call parity, one typically focusses on more liquid OTM options but estimating the tails of the risk-neutral density is difficult and one often uses semi-parametric approaches which assume some functional form for the tails.

Relationship to Numerical Methods

The risk-neutral pricing equation $P_t=\frac{1}{1+r}\mathbb{E}^\mathbb{Q}_t[P_{t+1}]$ is key to understand (almost) all numerical methods used in finance:

  • Finite differences: They are used to solve the PDEs which define derivative prices. By the Feynman-Kac theorem, the diffusion equations appearing in finance can be written as a conditional expectation - the risk-neutral price.
  • Binomial trees: You directly approximate the (risk-neutral) evolution of the underlying asset and compute expectations backwards through time. In particular, real-world probabilities do not enter this formula, just like in the risk-neutral framework.
  • Monte Carlo simulations: You simply simulate the future price (in a risk-neutral world), compute the average (expectation) of this price and discount back at the risk-free rate (you directly approximate the aforementioned risk-neutral pricing equation).
  • Fourier methods: You again start with the risk-neutral pricing equation and just change the integration domain: instead of integrating with respect to the risk-neutral density, you simply integrate in the Fourier domain using characteristic functions. This is fully equivalent and often easier.
  • Quadrature: You directly use the (risk-neutral) transition density of the underlying asset price and numerical integration to approximate the expectation (an integral) of the future price. So, the risk-neutral pricing framework is in the centre of quantitative finance and connects (almost) all methods used in quantitative finance.

Do we need to consider risk aversion for a risk-neutral measure?

I just saw that @JanStuller's answer disagrees with me relating risk aversion to the risk-neutral measure. He is, of course, absolutely right. We don't need to do this. If the law of one price holds, then a stochastic discount factor (and state prices) exists and prices are linear functionals. If the market is additionally free of arbitrage, then the SDF (and every state price) is strictly positive (in each state of nature, not just almost surely). In this case, we can also define a risk-neutral measure by merging SDF and real-world probabilities to obtain an equivalent martingale measure. The state price approach leads to the pricing by hedging/replication in Jan's answer.

So far the theory. Nothing of this depends on equilibrium, risk aversion or utility maximisation. In this sense, Jan is absolutely right and we can link SDF, risk-neutral measure and state prices without further assumptions. However, isn't the question how do we explain economically what's going on? What does the SDF measure? How do risk-neutral probabilities differ from real-world probabilities? What's the intuition behind this artificial measure? Why is the risk-neutral measure called risk-neutral if it has nothing to do with risk aversion? To answer these questions, we need to think deeper about the SDF. It turns out that $P=\mathbb{E}^\mathbb{P}[MX]$ is just the first-order condition of an investor who maximizes his utility. The SDF $M$ is then indeed a function of his risk-aversion. Ultimately, the SDF prices every asset -- and that price should reasonably be related to preferences, tastes and attitudes towards risk.

So the bottom line is: the existence of a risk-neutral measure is utterly independent of utility functions etc. However, in order to make sense of the measure change in economic terms and to build up intuition, I think it's wise to include some financial economics.


  • Risk-neutral probability measures are artificial measures made up of SDF (risk-aversion) and real-world probabilities.
  • The SDF can be linked to state prices which pave the way for pricing by replication. The SDF can be linked to the risk-neutral measure via a change of measure. What does the SDF measure? That's an economic question. In any equilibrium model, the SDF reflects agent's attitude towards risk.
  • Derivatives can be priced relative to underlying assets. This hedging price can be computed as expectation with respect to the risk-neutral probability measure.
  • Equivalent martingale measures are deeply related to the absence of arbitrage and completeness
  • The risk-neutral density can be estimated from observed market data
  • The risk-neutral framework connects many different approaches to derivatives pricing
  • 1
    $\begingroup$ "What is the SDF? In order to find the SDF, we need a general equilibrium model (such as C-CAPM or CAPM)" Why do we need a general equilibrium model? $\endgroup$
    – Trajan
    Commented Jun 26, 2020 at 19:13
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    $\begingroup$ @Permian You need a general equilibrium model for the SDF to be determined endogenously in the model. Of course, you can assume the SDF as a particular form and endow a (partial equilibrium) model exogenously with an SDF but for the model to give rise to an SDF, it needs to be a general equilibrium model. In the CAPM: $M_{t+1}=a-bR_{m,t+1}$ with $b>0$ and in the C-CAPM $M_{t+1}=\beta\left(\frac{c_{t+1}}{c_t}\right)^{-\gamma}$ where $c_t$ is aggregate consumption, $\beta<1$ is the subjective discount factor of an homogeneous agent and $\gamma$ the parameter of a CRRA utility function. $\endgroup$
    – Kevin
    Commented Jun 26, 2020 at 22:55
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    $\begingroup$ Very clear answer @KeSchn $\endgroup$
    – dm63
    Commented Jun 28, 2020 at 3:31
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    $\begingroup$ @Permian In response to “how to find risk neutral probabilities”, there are two ways: the formula I gave in the answer above to estimate the density (model free). Alternatively, you assume some model (Black Scholes, Heston, local vol, Merton, ...) and calibrate this model to market data. These models give you option prices and an EMM but you have to accept the corresponding model assumptions. But that’s what you’d do most likely in praxis: choose a model, estimate its parameters and use a (closed-form) option price formula. You normally don’t explicitly write down the risk neutral density. $\endgroup$
    – Kevin
    Commented Jun 28, 2020 at 10:35
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    $\begingroup$ @BlgKhalil Risk-neutral probabilities combine marginal utility (SDF) and real-world probabilities. The former is higher in bad states of nature (when people don't like risk). Geometrically, the mean of the risk-neutral density is $r$, the mean of the real-world density is $\mu$. Thus, the SDF has to shift the real-world density down somehow (by inflating the left tail and reducing the right tail). $\endgroup$
    – Kevin
    Commented Oct 11, 2021 at 10:13

I believe the other answers are nearly exhaustive; but here's a bit of intuition I'd like to add:

Think of the decision (= equilibrium price) of a market as:

Decision = f(probabilities, risk aversion)

where probabilities are the chances of various events happening, and risk aversion is the taste preference of the market.

Now it turns out that the 'iso-curve' always has a point where risk aversion is 0. That is, for some probabilities p and risk aversion a, I can find probabilities p' such that:


That is, I can always look at the market as if it was risk neutral (no risk aversion), provided I play around with the possibilities in a clever way (p changes to p').

For intuition into this, let us now try to have a risk neutral and a risk averse person arrive at the same 'decision' of investing:

The decision of a risk averse person at market probabilities of a up and down move (0.5,0.5) can be replicated by a risk neutral person at fictitious probabilities (0.4,0.6). [Ignore the specific values, just note that I've increased the probability of the down move].

This is because the averse person is much more 'sensitive' to the down move gives it a 'higher weight' in his decision making process. So, I need to incentivize him with a lower probability of a down move. However, the risk neutral person is much less sensitive and will make the same 'decision' at higher probability of a down move. This tells me that risk aversion and probability are 2 sides of the same coin, and can be inter-played with without affecting the 'decision (prices)'.

So, the risk averse man can be thought of as a risk neutral man by just accommodating the risk aversion in the probabilities.

This is what the risk neutral measure achieves: a change in probabilities of events, without compromising the market decision.

The above can always be done if there is no arbitrage in the market.

The derivative pricing thing comes into the picture if you impose completeness - that everything is replicable. That pins down p' - these probabilities are now 'unique'. This is why we use them in derivative pricing, because now pricing payoffs is trivial - as we only need to value a payoff using expectation (as a risk neutral person does).


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