Wednesday, May 12, 2010

How do you produce a 63 day winning streak?

Here is how I have been trying to figure this out. Suppose we wanted to figure out what a daily winning percentage had to be in order to observe a 50 percent probability of a 63 day winning streak. It would be (.5)^(1/63), because the probability of 63 straight wins would be Pr(one win)^(63). It turns out that (.5)^(1/63)=.989, which I will round to .99.

Now lets say a firm has a proprietary trading model that is correct 51 percent of the time. This means that on the average day, it will come out ahead (suppose all trades are $1 trades). But if a trader makes one trade a day, he will close the day ahead only 51 percent of the time. If he makes 100 trades a day, however, while his winning percentage per trade remains the same, put his winning percentage per day goes up a lot. Specifically, the standard error for a daily outcome goes down by 1/10, from sqrt(.51*.49) to sqrt(.51*.49/100), or from about .25 to .025. The chance of finishing the day losing on average is based on how many standard deviations away .5 is from .51. In this case, it does from .01/.25 (or not far at all) to .01/.025, or .4 standard deviations away. In a normally distributed world, this means there is a 65 percent chance of finishing the day ahead, assuming each trade has a .51 batting average and 100 trades per day.

To get to winning 99 percent of days, we need to get the standard error for the day to be sufficiently low that .5 is more 2.4 standard deviations away from .51, so the standard error needs to be .01/2.4 or about .004. So we need to find X such that sqrt((.49*.51)/X)=.004. or X=.25/(.004^2)=15,625 trades per day.

Three big assumptions go into this calculations. First, it assumes a stable model. Over the course of one quarter, this may be reasonable. Second, it assumes a model with a 51 percent winning percentage. This is a huge assumption (I do not know what a reasonable number might be). Third, it assumes normality. This is probably not too bad; we do know that Chebyshev's Inequality says that (1-1/k^2) share of any distribution must be within k standard deviations of the mean. This means that 99 percent of any distribution is within 10 standards deviations, but that is an extreme outcome.


nummy said...

this assumes the probability of winning is always 50/50 and stays constant. the probability of the market ending down or up every day (or making a winning trade) is not 50/50 ... it might be 60/40 one day and then 40/60 the next .. it is constantly changing.

rjs said...

the casino is rigged: their computers are the house, you place your bets but you always lose...they drive prices up and harvest the short sellers, then drive them back down and harvest the stop loss orders...

Maxine Udall (girl economist) said...

What rjs said.

rjs said...

richard; now you can try figuring out the probablility of all four big banks having the same winning streak:

‘Perfect Quarter’ at Four U.S. Banks Shows Fed-Fueled Revival (Bloomberg) -- Four of the largest U.S. banks, including Citigroup Inc., racked up perfect quarters in their trading businesses between January and March, underscoring how government support and less competition is fueling Wall Street’s revival.Bank of America Corp., JPMorgan Chase & Co. and Goldman Sachs Group Inc., the first, second and fifth-biggest U.S. banks by assets, all said in regulatory filings that they had zero days of trading losses in the first quarter. Citigroup Inc., the third-largest, doesn’t break out its daily trading revenue by quarter. It recorded a profit on each trading day, two people with knowledge of the results said.“The trading profits of the Street is just another way of measuring the subsidy the Fed is giving to the banks,” said Christopher Whalen, managing director of Torrance, California- based Institutional Risk Analytics. “It’s a transfer from savers to banks.”

PD said...

I disagree. This run of positive trading days reflects 'getting closer to home' as goldman traders repeated over and over in their testimony. They are simply taking less risk. These positive returns reflect these companies are happy making markets and profiting on the bid/ask spread rather than on speculative trades. Thus, the 50/50 or even 60/40 probabilities are far off.

rjs said...

PD: i agree. they take no risk at all. in las vegas, they call that vigorish...

PD said...

to extend the casino analogy:

The 'house' has a wide variety of customers from the savvy card counters to the suckers playing the big six wheel. The banks have the same variety of customers and with the lack of pricing clarity in the market, wide spreads can be made. Unfortunately most of the big six wheel players are your pension funds or underdeveloped sovereigns. Hedge funds are the card counters. Michael Masters is on the leading edge to add more clarity to these OTC markets.

. said...

On the last point: Chebyshev works for you here only if the variance is finite. For distributions that look like power laws (which really match what equities look like) variances are "unbounded"--they don't go to a fixed value, but continue to increase as the sample size increases. That's why various modelers did not use power laws when modeling data: everything gets ugly when you your variance shoots up like that. It would be nice if the distributions in finance all mapped to the exponential family, but they don't.

devin said...

I disagree with the Vegas analogies. In casinos, some nights the house loses. Any casino that won every night would lose all its customers.

rjs said...

jon stewart does a better job of explaining how that happened than i could...

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