Why the Euler Equation Is Zero for Excess Returns
Summary
The discussion explains why a pricing equation with an expectation of one can appear as an equation with an expectation of zero. For a payoff priced today, the Euler equation states that its price equals the expected discounted payoff under the real-world probability measure, using a pricing kernel. Expressing the payoff as a gross return gives an expectation equal to one. For excess returns, the initial cost is zero by construction, so the corresponding pricing-kernel-weighted expectation is zero.
The key distinction is between gross and excess returns, not a switch to risk-neutral probabilities. The response also notes that notation can vary: one source uses R for returns, while another uses it for excess returns. The exchange offers a conceptual clarification rather than an empirical result or a full derivation of a risk-neutral measure.
Key ideas
- The Euler equation for a gross return equates its pricing-kernel-weighted expectation to one.
- The corresponding expectation for an excess return is zero because the excess payoff has zero initial price.
- The expectation in the stated Euler equation is taken under the real-world measure.
- Return notation must be checked because sources may use the same symbol for different quantities.
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Full text
# What is the equation $\mathbb{E}[mR]=0$?
# What is the equation $\mathbb{E}[mR]=0$?
https://economics.stackexchange.com/questions/16115/what-is-the-equation-mathbbemr-1
The above post asks what $$\mathbb{E}[mR]=1$$ means and gets some great answers. From the first answer is seems that the physical rather than risk neutral probabilities are used to take the expectation.
I'm reading the paper "Characteristics are Covariances" on p2 they introduce the Euler equation as:
$$\mathbb{E}[mR]=0$$
I have not seen this formulation from the paper anywhere else. They say that the only assumption used was that of "no arbitrage". The paper does not state which measure was used to take the above expectation, but given the prices are arbitrage free I am assuming it is a risk neutral one.
- A I correct to think that the expectation in the paper is taken under the risk-neutral measure?
- Is it the switch from the physical to risk neutral measure that causes the difference on the rhs of the two expressions, if so can someone run me through the derivation?
- Intuitively the first formulation seems to be a martingale but is calculated under the physical measure so I am assuming it is not arbitrage free can a price process be a martingale but still not be arbitrage free?
Thanks
Baz
## Answer by Bob Jansen (score 4, accepted)
https://quant.stackexchange.com/a/74263
I don't have the paper, I think you mean this text.
On the Economics Stack Exchange $R$ denotes the asset return. In the linked paper, $r$ denotes the excess return, i.e. $R = 1 + r$. I believe all the rest is the same.
## Answer by Kevin (score 4)
https://quant.stackexchange.com/a/74264
$\mathbb{E}[mR]=0$ is the Euler equation under $\mathbb{P}$ for excess returns.
In general, the Euler equation is $$P_t=\mathbb{E}_t[M_{t,t+1}X_{t+1}],$$ where $P_t$ is today's price, $M_{t,t+1}$ the pricing kernel and $X_{t+1}$ tomorrow's payoff. The expectation is conditional on information available at time $t$ and under the real world probability measure $\mathbb{P}$.
The Euler equation is often stated for returns rather than prices. Then, with $R_{t+1}=\frac{X_{t+1}}{P_t}$, we have $$1=\mathbb{E}_t[M_{t,t+1}R_{t+1}].$$
For excess returns ($R^e_{t+1}$), the price is zero by construction ($P_t=0$). The Euler equation is thus $$0=\mathbb{E}_t[M_{t,t+1}R^e_{t+1}].$$
Here are more examples from John Cochrane's fantastic book (chapter 1).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.