Why Expected Returns Cannot Be Inverted into Expected Prices
Summary
The question examines whether a security’s price can be recovered by taking expectations of its dividend-to-price return definition and the rearranged price equation. The answer explains that the apparent conflict with the rule that the expectation of a ratio generally differs from the ratio of expectations comes from treating the return as random in the rearranged equation.
Here, the current price is known at the valuation time, so the return is the random future dividend divided by a constant. Its expectation is therefore the expected dividend divided by the current price. Rearranging that expected-return relation recovers the current price. This is an algebraic explanation for a single-period dividend-paying security; it does not establish a general method for pricing securities when returns, discount rates, or future prices are random.
Key ideas
- The current price is fixed at the valuation time, while the future dividend is uncertain.
- The return in this setup is random because it depends on the future dividend.
- The expectation of a ratio generally differs from the ratio of expectations when both quantities are random.
- Rearranging the expected-return definition recovers the known current price in this single-period setup.
Tags
Full text
# Expected returns vs expected prices?
# Expected returns vs expected prices?
This may be the most stupid question ever asked here, so sorry in advance for asking it.
Suppose we have a single period security which gives dividend $D_{t+1}$ and has current price $P_t$. By definition we have: $$R_{t+1}\equiv\frac{D_{t+1}}{P_t}\ \ (1)$$ Here is the first trouble:I believe I’m making no assumption by saying that by definition it’s still true that: $$P_{t}\equiv\frac{D_{t+1}}{R_{t+1}} \ \ (2)$$ So if we take expectations of (1) and (2) we have: $$E[R_{t+1}]=E\left[\frac{D_{t+1}}{P_t}\right]= \frac{E[D_{t+1}]}{P_t}\ \ (1^E)$$ $$E[P_{t}]=P_t=E\left[\frac{D_{t+1}}{R_{t+1}}\right] \ \ (2^E)$$ So if we solve $(1^E)$ for price we have the following equivalence : $$P_t=E\left[\frac{D_{t+1}}{R_{t+1}}\right]=\frac{E[D_{t+1}]}{E[R_{t+1}]} \ \ (3)$$ The reason why I’m puzzled is that, in general, $E[\frac{X}{Y}]\neq\frac{E[X]}{E[Y]}$. I can’t see my mistake given that I believe I’ve just worked with definitions (which should hold both ex-post and ex-ante) and taken expectations of them.
## Answer by Neeraj (score 1)
https://quant.stackexchange.com/a/24755
You are right in saying that $$E\bigg[\frac{X}{Y}\bigg] \ne \frac{E(X)}{E(Y)}$$ Here, both $X$ and $Y$ are random variable and hence their ratio is by definition is random variable too. If $Z$ is RV such that $$Z=\frac{X}{Y} \Rightarrow E[Z] = E[\frac{X}{Y}] \neq \frac{E(X)}{E(Y)}$$
But in your case, $Z$ is not a random variable but known and constant. See, $$R_{t+1}= \frac{D_{t+1}}{P_t} \Rightarrow P_t = \frac{D_{t+1}}{R_{t+1}}$$ where $P_t$ is known at time $t$ or constant. That is why, we have $$P_t = E\left[\frac{D_{t+1}}{R_{t+1}}\right] = \frac{E(D_{t+1})}{E(R_{t+1})}=\frac{E(D_{t+1})}{E(D_{t+1})/P_t} = P_t$$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.