# Which are the practical implications that the continuously compounded rate of return can be smaller than the expected rate of return?

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?

• 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%) – Alex C Apr 14 '19 at 1:03
• 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. – mark leeds Apr 14 '19 at 7:11
• 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 – mark leeds Apr 14 '19 at 18:42

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.
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