# Calculating expected return

Assume I have a trading system that I'm evaluating over a three-year period. The returns are 25%, -40% and 25%. Empirically, I can see that this system loses because at the end of three years, I have less than when I started.

Wikipedia defines expected return as follows:

E(R)= Sum: probability (in scenario i) * the return (in scenario i)


If we insert our values into this formula, we get the following:

E(R) = (.33 * .25) + (.33 * -.40) + (.33 * .25) = .033


So a positive expected return for a system that loses every time, no matter if you re-arrange the order of the returns.

What's wrong here?

To further explain how the game works, consider an initial start value of 100. What can you expect at the end of the game? There are no re-investments, no withdrawls, no dividends, no broker fees nor SEC fees. It is a simple game. Here is some R code to illustrate the game.

first  <- c(.25, .25, -.4)
second <- c(-.4, .25, .25)
third  <- c(.25, -.4, .25)


Pass any of the above sequence of returns into this function:

game <- function(x){

start <- 100

for(i in 1:NROW(x))

start <- start + start*x[i]

return(start)
}


NOTE: I asked a similar question on quantexchange, but I'm interested here in the math behind the expected return equation.

The difference between the two ways to look at the return is whether you are reinvesting your gains. If you start with 100 dollars and reinvest all of the money the next year, your balances will be: 125, 75, 93.75 dollars - you lost money. However if you invest 100 dollars every year, then you get +25 - 40 +25 = +10 dollars gain. That's one way you could think about the Wikipedia formula. A better way to interpret it, though, is that you are investing in one of the three years chosen at random, and it gives you the expected return.

• So my error is not re-upping during losing years and taking profits on winning years? Apr 21 '11 at 14:23
• @Milk It's not that simple. Suppose you invest 100 dollars. After year 1, you take 25 out, leaving 100. After year 2, you have lost 40, so you put the \$25 back in plus another 15. After year 3, you have 125. In toto you were forced to have 115 in cash (not 100), on which you made only 10 after 3 years. All this assumes you cannot earn interest on the unused cash sitting around and you cannot borrow money. This only begins to hint at the mistakes that will be made when using oversimplified formulas for expectation. Reading a good text on financial calculus will serve you well.
– whuber
Apr 21 '11 at 18:48
• @Milk An excellent starting point is Stochastic Calculus for Finance; Volume I: The Binomial Asset Pricing Models by Steven E. Shreve, Springer-Verlag, New York, 2005.
– whuber
Apr 21 '11 at 18:50
• I'm accepting this but need to say the wiki equation is awkward at best. It's application to trading is limited. There are other expected return formulae that are better suited to examine trade system characteristics. Apr 22 '11 at 0:18

Wikipedia is right. So are your calculations: this system should win.

However: it also assumes that your investments in each part are equal in size (because you give equal probability 1/3 to each). If this is not true, that may explain the difference with your empirical observations (perhaps you should share the numbers with us on those). e.g. If you invested twice as much in the part that has a negative return than in the two other parts, the probabilities become 1/4,1/2,1/4 and you'd have a losing system.

This is also ignoring any extra costs, so if these extra costs are bigger than your meager (?) 1/30 return, this could be another explanation (I'm not familiar with the practicalities of trading).

• I've included some R code by way of explanation. So you're saying I'm not applying the wiki formula correctly? Presumably by assigning the probabilities, no? Apr 21 '11 at 14:20
• Now I see: you did not divide your initial amount into three parts (that is what the Wikipedia formula would be for), the three returns are just the returns each year. In your case, the formula is simply ((100(1+0.25))(1-0.4))(1+0.25). There is no probability involved in this way. Apr 21 '11 at 14:29