tag:blogger.com,1999:blog-35364652.post116168701261755658..comments2023-02-09T02:30:59.837-05:00Comments on Quantitative Trading: Maximizing Compounded Rate of ReturnErnie Chanhttp://www.blogger.com/profile/02747099358519893177noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-35364652.post-21430808931110370892014-10-26T14:42:44.916-04:002014-10-26T14:42:44.916-04:00Hi Jason,
Excellent demonstration that the exponen...Hi Jason,<br />Excellent demonstration that the exponential of the expected log return gives the mode of the future profit, while the expected value of the exponential of the log return gives the mean of that same profit!<br /><br />ErnieErnie Chanhttps://www.blogger.com/profile/02747099358519893177noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-49325612244313548412014-10-26T12:09:51.014-04:002014-10-26T12:09:51.014-04:00Hi Ernie,
Perhaps they are amused because this il...Hi Ernie,<br /><br />Perhaps they are amused because this illustrates that one needs to be careful in interpreting the expectation value of the log return. <br /><br />If the Stock is S at time 0, and S' at time 1, let R= S'/S, then as you said:<br /><br />exp(Eln(R)) !=E[R].<br /><br />So if the exponential of Eln(R) is not the expected price ratio, then what is it?<br /><br />With your example, following on from my comment above:<br /><br />1+r=exp(ElnR).<br /><br />So we can interpret exp(ElnR) <i>as</i> the compounded growth factor for the <i> most likely </i> outcome in the long run.<br /><br />This is technically the thing that was asked for in your question. So you are right. <br /><br />It made me wonder how general is this connection between exp(ELog[R]) and the mode. For example, is it also true for a geometric random walk? <br /><br />Recall that for gbm:<br />S'=S*exp((m-s^2/2)t+s*sqrt(t) Z), <br />where Z is a standard normal variable, m is the drift and s is the standard deviation. <br /><br />So, <br />ln(R)=(m-s^2/2)t+s*sqrt(t) Z,<br />which implies that<br />exp(ElnR)=exp((m-s^2/2)t).<br /><br />On the other hand, the mode occurs when Z=0 (for a standard normal) i.e., if R* is is the mode then,<br /><br />R*=exp((m-s^2/2)t)<br /> =exp(ElnR).<br /><br />So indeed the mode is related to the exponential of the expectation value of the log return. <br /><br />However, note a difference to before. Now the mode of R <i>is</i> exp(ElnR). In the binomial model you had originally, it was the equivalent (constant) compound growth factor of the mode of R, r+1, which was equal to the exp(ElnR). They are both related to the mode of R, but with slight differences to account for the differences in the models. Jasonhttps://www.blogger.com/profile/14357219983580399572noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-56616512636952193862014-10-26T10:05:53.153-04:002014-10-26T10:05:53.153-04:00Jason,
Interesting analysis!
You may be right tha...Jason,<br />Interesting analysis!<br /><br />You may be right that the mode is money-losing, while the mean is flat. <br /><br />However, there is another angle to this paradox. If we compute the expected value (mean) of the profit, it is actually 0 too (assuming that the one-period returns has 0 mean). On the other hand, if we compute the expected log return (i.e. the expected compounded rate of return), it is negative, as I stated in my post.<br /><br />The difference is due to the fact that the expected value of the exponential function of a normal variable is not equal to the exponential of the expected value of a normal variable. The former is the expected value of our net worth, which is flat (unchanged), and the exponent of the latter is our expected compound rate of return, which is negative.<br /><br />To a trader, profit is what counts. So yes, you are right that in this case there is no profit. But to financial professors (such as Prof. John Hull who wrote the widely read Options textbook and who computed the compound rate of return for normal 1-period returns), they are amused by the fact that this compounded return is actually negative.<br /><br />ErnieErnie Chanhttps://www.blogger.com/profile/02747099358519893177noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-46025472120634501222014-10-26T08:41:22.527-04:002014-10-26T08:41:22.527-04:00I think this blog post is highly misleading.
You...I think this blog post is highly misleading. <br /><br />You say that "If you buy this stock, are you most likely, in the long run, to make money, lose money, or be flat?"<br /><br />Well it is clear to the traders that if you buy this stock you don't expect to make any money. And <i>they</i> are correct. <br /><br />Usually, profit is what one means by "making money." The profit when the stock goes up is 1% (of investment), and when it goes down the profit is a loss of -1%. Both occur with equal probability. So forget about the stock market or geometric random walk etc this is a very simple game, with expected profit of 0. <br /><br />The key word here is "most likely." This means that you are asking about the event that has the highest probability of occurring or more precisely <i>the mode</i> - not the expectation value or mean. The entire confusion results from the fact that the mode and mean do not have to have the same value. <br /><br />When there are an even number of rounds, say 2N, the mode is unique (when there are an odd number of rounds there are two such outcomes having equal highest probability of occurrence). Since we will take N to infinity we can work with an even N without loss of generality.<br /><br />The most likely event is the event in which there have been an equal number of up and down moves. Or when the stock has moved from the price S to:<br /><br />S(1+s)^N(1-s)^N<br />=S(1-s^2)^N<br /><br />We can find the average compounded growth rate, r, for this <i>most likely</i> outcome, over the (2N) periods by letting<br /><br />(1+r)^(2N)=(1-s^2)^N<br />which implies: <br />r=-s^2/2+O(s^4).<br /><br />So indeed one is most likely, in the long run, to lose money at the said rate. But somewhat confusing, one also expects no profit or loss (as mentioned by Per above). <br /><br />It is basically a trick question about the difference between the mode (the most likely outcome) and the mean (the expected outcome).Jasonhttps://www.blogger.com/profile/14357219983580399572noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-22865197316856137972011-05-15T09:24:41.187-04:002011-05-15T09:24:41.187-04:00Hi sjev,
You are right. It was also pointed out by...Hi sjev,<br />You are right. It was also pointed out by various readers before and I posted a correction some time ago here: http://epchan.blogspot.com/2006/11/correction-maximizing-compounded-rate.html<br /><br />ErnieErnie Chanhttps://www.blogger.com/profile/02747099358519893177noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-14219093507885483582011-05-14T14:05:17.604-04:002011-05-14T14:05:17.604-04:00...Ernie, I think you've made a mistake statin......Ernie, I think you've made a mistake stating expected .5% per minute loss.<br /><br />Let's redo the math: your expected return per time step is 0.5*log(1.01)+0.5*log(0.99), which is -5.0003e-005 or rather 0.005% . You are 1e3 off.sjevhttps://www.blogger.com/profile/17452562180989360928noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-68021710523196319592010-07-04T23:09:55.064-04:002010-07-04T23:09:55.064-04:00Regarding: "The correct answer is you will lo...Regarding: "The correct answer is you will lose money, at the rate of 0.5% every minute!"<br /><br />I think it is worth mentioning that the expected profit from that investment is 0.<br /><br />For example, for a two minute run, there are four equally likely scenarios: Lose-Lose, Lose-Win, Win-Lose, Win-Win. The respective returns from a 1 unit investment would be 0.9801, 0.9999, 0.9999 and 1.0201, which average 1 - even if most outcomes are losses.Perhttps://www.blogger.com/profile/12545623419387288000noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-31972884169112203952006-11-09T09:25:00.000-05:002006-11-09T09:25:00.000-05:00Yes, m is the short run mean return, s is the shor...Yes, m is the short run mean return, s is the short run mean s.d. See John Hull's book on options also.Ernie Chanhttps://www.blogger.com/profile/02747099358519893177noreply@blogger.comtag:blogger.com,1999:blog-35364652.post-29238040178496220222006-11-09T09:05:00.000-05:002006-11-09T09:05:00.000-05:00EC-
In m – s2/2, does m always mean rate of retur...EC-<br /><br />In m – s2/2, does m always mean rate of return and s2/2 always mean standard deviation in the short run?<br /><br />Sorry for so many Qs, I'm quite new to quantitative trading. thanks!NAhttps://www.blogger.com/profile/09308905056511377295noreply@blogger.com