1
$\begingroup$

I'm reading Hull's Options, Futures and other Derivatives and it intrigues me that the distribution of the continuously compounded rate of return x is: $x \sim \phi(\mu - \frac{\sigma^2}{2}, \frac{\sigma^2}{T})$

This happens when: $\frac{\Delta S}{S} \sim \phi(\mu \Delta t, \sigma^2\Delta t)$

My question is: what are the practical implications of $\mu - \frac{\sigma^2}{2} < \mu$? Does it mean that although most of the times your return will be less than $\mu$, your expected return is $\mu$?

What does Ernest P. Chan means when he says:

Suppose a certain stock exhibits a true (geometric) random walk, by which I mean there is a 50-50 chance that the stock is going up 1% or down 1% every minute. If you buy this stock, are you most likely, in the long run, to make money, lose money, or be flat? ... Most traders will blurt out the answer “Flat!”, and that is wrong. The correct answer is you will lose money, at the rate of 0.005% every minute!

Does it mean that although most of the times your return will be less than μ, your expected return is μ?

$\endgroup$
  • 1
    $\begingroup$ That must be a typo, you lose money at the rate 0.005% every minute. ($\mu=0,\sigma =0.01$ so $\mu-\sigma^2/2=0.00005$ or 0.005%) $\endgroup$ – Alex C Apr 14 at 1:03
  • 3
    $\begingroup$ Hi: the rough idea behind that is the same as the idea that, if a stock's price was originally 1 dollar and it went up 50 percent the first day and down 50 percent the next day, then the final stock price at the end of the second day is 75 cents. The point is that if the price has an equal chance of going up or down during some period, then the expected log return over the same period is less than zero. There are papers on this concept that show this WAY more rigorously. If I can dig up a good one, I'll send a link. $\endgroup$ – mark leeds Apr 14 at 7:11
  • 1
    $\begingroup$ Hi: Google s your friend because there are so many relared articles and papers on the net. But I have read the one at the link below and remember it being an interesting read at the time. I can't really recall the contents now but it might be worth taking a look at it for a more formal answer to your question. papers.ssrn.com/sol3/papers.cfm?abstract_id=2027471 $\endgroup$ – mark leeds Apr 14 at 18:42
1
$\begingroup$

The key word in your question is compounded. The expected arithmetic return for each $\Delta t$ is $\mu$, but the growth rate is $\mu - \frac{\sigma^2}{2}$. As others mentioned, volatility reduces the growth rate.

This is similar to the area of a rectangle. If one rectangle has sides 3 and 1, its average side length is 2, and its area is 3. If another is a square with length 2, its average side length is also 2, but the area is 4. Variability reduces area relative to the average side length.

$\endgroup$
  • 1
    $\begingroup$ Charles made an interesting analogy but just to relate it to finance: When you do geometric rets, you are essentially calculating the product of rets ( due to re-investment ). When you do arithmetic, you're adding rets. So, in example earlier, if you just added the 50 percent up and 50 percent return down, you get zero percent which means you still have a dollar. So, it's the re-investment ( second day we assume that we are playing with 1 dollar and 50 cents ) assumed by the geometric return calculation which causes the drain on the expected return. $\endgroup$ – mark leeds Apr 15 at 17:47
  • $\begingroup$ Good point @markleeds. To match the rectangle analogy better: 2% returns twice gives you 1.02*1.02 -1 = 4.04%. 3% return followed by 1% returns 1 basis point less (1.03*1.01 -1 = 4.03%). This difference is smaller than the rectangle example because of scaling (.02 difference relative to 1 vs 2 difference relative to 2). The rectangle example is more like 200% return twice vs 100%, 300%. $\endgroup$ – Charles Fox Apr 15 at 18:03
  • $\begingroup$ Gotcha Charles. I tend not to think geometrically but it's interesting to try. Sort of like a brain exercise. Thanks. $\endgroup$ – mark leeds Apr 16 at 2:25

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.