Risk-Neutral and Real-World Probabilities in Asset Pricing

# What is the relationship between the risk-neutral and real-world probability measure for a random payoff?


# What is the relationship between the risk-neutral and real-world probability measure for a random payoff?












Say there's some financial security that gives you 1 dollar with probability $p$ and zero otherwise. We know that the value of this is not the expected value $p$ but rather the expected value under the risk-neutral measure (chosen by the market) $q$.

However, $q$ ought to at least depend on $p$, i.e. $q = q(p)$.

What is the nature of this function, $q(p)$? Do we know anything about it? Could we say it's linear? $q(p) = a + bp$? Or would that be an assumption we impose as a model?

## Answer by Attack68 (score 4)

https://quant.stackexchange.com/a/84107

> However, $q$ ought to at least depend on $p$, i.e. q = q(p)

Why? I think that you are suggesting that because there is a (known) $p$ then $q$ should be directly relatable to it, since that will ultimately be the realized probability distribution. I would counter that since $q$ exists and it is not equal to $p$, there must be some independent, structural component that is driving $q$. And since it is independent it is not relatable to $p$ in any defined manner.

In financial markets $p$ is often latent and unknowable, anyway, (i.e what is the real world probability of Apple Shares closing up tomorrow, versus the option implied probability of Apple shares closing up tomorrow), whereas $q$ is often calculable from market pricing. I would suggest that if one is able to confidently model $p$ from independent data, then, by comparing one's model with $q$, trading opportunities should present themselves if one has the risk and margin framework to run the trade to realisation.

Regarding your deleted comment, the probability measures $\mathbb{P}$ and $\mathbb{Q}$ on $(\Omega, F)$ are equivalent which means that they agree on which sets in $F$ have zero probability.

## Answer by Rylan (score 2)

https://quant.stackexchange.com/a/84110

What you're describing is very closely related to Arrow-Debreu securities, if you're looking for more reading on the matter. There generally doesn't exist an easy relationship. I agree in principle with @dm63's comment of when $q > p$; if memory serves me correctly it's a little more general though.

For example, if you think of home insurance as a binary option paying off in the state that your house burns down and with only a very small market, you're likely willing to pay a premium such that $q > p$. This example not being a traded security might not be satisfying, but loosely, $q>p$ often in cases where the security "pays you when you need the money most".

Returning to the world of tradeable securities, it's still important to be cautious. A risk neutral probability distribution is basically a tool to calculate the amount of money you'd need to hedge the security, and it can look very different from the "real world" probability distribution. For example, suppose we have a stock with the following dynamic in the real world measure: $$\frac{dS_t}{S_t} = -c(log(S_t) - \log(S_0)) dt + \sigma dW_t$$

Its dynamic in the risk neutral measure will be $$\frac{dS_t}{S_t} = rdt + \sigma dW_t$$

The variance $log(S_T) - log(S_t)$ will asymptotically reach some fixed value in the real world, while in the risk neutral measure it will increase linearly with $T-t$.

## Answer by dm63 (score 1)

https://quant.stackexchange.com/a/84109

In this case q is the market price of the instrument (assuming the payment is immediate, or equivalently that interest rates are zero). So then we are in the world of asset pricing. If I’m not mistaken, if the process for p is correlated with the market then q<p. A zero correlation should imply q=p and a negative correlation q>p.

If I were presented with this personally and p was based on a biased coin flip, I would pay $q_1 <p $ and if I had to sell it to someone I would need to receive $ q_2 >p$

## Answer by GratisGuru (score 1)

https://quant.stackexchange.com/a/84147

Look into radon nikodym derivatives and change of measures for better clarity. When you implement a change of measure, it is assumed that there is a known ratio of p(x)/q(x), the probabilities under measure p and q for event x (may be discrete or continuous).