Expected returns vs expected prices?

Question

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 1

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 $$P_t = E\left[\frac{D_{t+1}}{R_{t+1}}\right] = \frac{E(D_{t+1})}{E(R_{t+1})}=\frac{E(D_{t+1})}{E(D_{t+1})/P_t} = P_t$$