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.


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

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.