Pricing Assets Under Risk-Neutral and Physical Measures
Summary
The document explains how the risk-neutral measure prices a stock when its terminal value is discounted at the short rate. Under an equivalent martingale measure, discounted asset prices are martingales, so the stock value is the conditional expectation of its discounted future value. The same pricing relationship can be expressed under the physical measure, but it requires a state price density to adjust for risk.
The answer outlines how that density incorporates the asset’s excess return relative to the risk-free rate and its volatility. It distinguishes the bond’s discounted payoff from a risky asset’s payoff: ordinary physical-measure discounting alone does not generally give the asset price. The treatment assumes an appropriate state price density exists and simplifies its dynamics to a single-asset setting; it does not derive the density or address market incompleteness, multiple sources of risk, or practical estimation.
Key ideas
- Risk-neutral pricing expresses an asset value as the expected discounted payoff under an equivalent martingale measure.
- Under the physical measure, pricing requires a state price density to account for risk.
- The state price density multiplied by an asset price is a physical-measure martingale.
- Discounting a risky payoff under the physical measure alone generally does not produce its market value.
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# Confusion regarding the risk neutral and physical measures
# Confusion regarding the risk neutral and physical measures
I may be confused. I am looking at the risk neutral vs. physical measures. We know that knowing the short interest rate stochastic process $r$, a bond maturing at time $T$ can be considered as a derivative with payoff \$$1$. Its price at time $t$ is $$P(t) = \mathbf E^Q\big[e^{-\int_t^Tr}\big|\mathcal F_t\big],$$ in the risk neutral measure $Q$. Now if we want to compute the time $t$ value $V(t)$ of a stock at time $T$, should we compute the expectation in the risk neutral measure $Q$ $$V(t)=\mathbf E^Q\big[e^{-\int_t^Tr}S(T)\big|\mathcal F_t\big]$$ or the physical measure $P$ $$V(t)=\mathbf E^P\big[e^{-\int_t^Tr}S(T)\big|\mathcal F_t\big]$$ ?
## Answer by fni (score 1)
https://quant.stackexchange.com/a/41465
The equivalent martingale measure (EMM) $\mathbb{Q}$ is a measure under which all the asset prices discounted using a risk-free bond are martingales, i.e. given the bond price $B(t)$ and the asset price $S(t)$ we have $$\frac{S(t)}{B(t)} = E_t^{\mathbb{Q}}\left[\frac{S(T)}{B(T)}\right]$$ If the bond follows the following ODE $dB(t) = r(t)B(t)dt$ then $\frac{B(t)}{B(T)} = e^{-\int_t^T r(s)ds}$ and therefore we get the equation you have suggested under $\mathbb{Q}$
$$ S(t) = E_t^{\mathbb{Q}}\left[e^{-\int_t^T r(s)ds}S(T)\right]$$
If you want to price under the physical measure $\mathbb{P}$, then you know there exists a state price density $\xi(t)$ that (assuming only one asset) evolves according to the following process $d\xi(t) = -\xi(t)\left(r(t)dt + \frac{\mu(t)-r(t)}{\sigma(t)}dW(t)\right)$ and whose product with any asset is a $\mathbb{P}$ martingale, i.e.
$$ \xi(t)S(t) = E_t^{\mathbb{P}}\left[\xi(T)S(T)\right]$$
implying the following pricing equation under the physical measure:
$$ S(t) = E_t^{\mathbb{P}}\left[\frac{\xi(T)}{\xi(t)}S(T)\right]$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.