Someone plans to use x dollars buying some stock share, the stock price is $a$ dollars per share now. One year later, the stock price will possibly increase to $ra$ or decrease to $a/r$ (r>1) which the probability is $p$, or remain the price now which probability is $1-2p$. If he want to own more shares, should he buy the stock share or one year later?

Solution I:

The expected value of the share price one year later is $a(rp+\frac{p}{r}+1-2p)$.

So one year later he can buy $\frac{x}{a(rp+\frac{p}{r}+1-2p)}<\frac{x}{a}$.

So buy the stock share now.

Solution II:

One year later, if price is $ra$, he can buy $\frac{x}{ra}$ shares; if price is $\frac{a}{r}$, he can buy $\frac{xr}{a}$ shares.

Then the expected value of the shares he can buy one year later is $$\frac{x}{ra}*p+\frac{xr}{a}*p+\frac{x}{a}*(1-2p)>\frac{x}{a}$$.

So buy the stock one year later.

What's wrong with Solution II.

  • $\begingroup$ Do you mean what's wrong with solution I? $\endgroup$ – André Nicolas May 28 '12 at 14:10
  • $\begingroup$ @André Nicolas But I think solution I is right. $\endgroup$ – Charles Bao May 28 '12 at 14:13

The second solution is right. The first solution is not.

Let random variable $W$ be the number of shares we will have if we adopt a waiting strategy. We want to compare $E(W)$ with $\frac{x}{a}$. By the correct calculation you did, we have $$E(W)=\frac{x}{a}\left(1+ r+\frac{1}{r}-2\right).$$ But in various ways one can see that $r+\frac{1}{r}\ge 2$, with equality only if $r=1$.

As to why the first solution is not correct, the simple answer is that the second solution is based on a concrete calculation of the expectation of $W$, which is the random variable we are interested in. So the second solution is correct. The first solution gives a different answer, so cannot be right.

The problem is that the first calculation does not correctly compute $E(W)$.

If $Z$ is a random variable, and $f$ is a function, then (usually) $E(f(Z))\ne f(E(Z))$.

In solution I, $Z$ is the price, and $W=\frac{x}{Z}$. Solution I implicitly assumes that $E(\frac{x}{Z})=\frac{x}{E(Z)}$. There is no reason to assume that this equation will hold, and in fact it does not.


There is nothing wrong with your second solution. In fact, I think it is better than the first solution.

In the first case you computed the expected price one year from now, and say that the number of shares you can buy is inversely related to this. In the second case you compute the expected number of shares you can buy one year from now directly.

I'd argue that the second solution is actually better than the first, as it is more directly related to the quantity you're interested in (the number of shares you can buy).

Mathematical interlude:

You're running into Jensen's inequality, which says that for a convex function $f$ of a random variable $S$, you have

$$f(ES) \leq Ef(S)$$

where $E$ is the expectation operator. Since the reciprocal function is convex, you have

$$1/ES \leq E(1/S)$$

for any random variable $S$. Here the left hand side is the inverse of the expected stock price, and the right-hand side is proportional to the expected amount of stock you can buy.


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