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# Advanced Math/Time Series and Conditional Probability

Question
QUESTION: Hi Scott,

I recently stumbled across a discussion on a financial radio show that raised some questions about the probability of consecutive up/down days in the S&P. One of the speakers had tallied the occurrences of consecutive up/down days in the market from 1950 to the present and noted that the data set was normally distributed. Using the total number of 3 consecutive up-days divided by the total number of points, he estimated the probability of a 3 consecutive up-days at approximately 0.162. He did this for each of the discrete values of streaks to calculate the probability of a streak appearing in the time series of any recorded size.

Using Bayes' theorem, he wanted to show, for example, that the probability of 4 consecutive up days given 3 consecutive up-days was not significantly different from the probability of a single up-day (which was 0.53). In other words, given the existence of a streak in consecutive up-days, the probability of another up-day was no different than the probability of a single up-day given no prior streak. He did the following calculation using Bayes' theorem:

p(4 up-days | 3 up-days) =  p(4 up-days)/p(3 up-days)

= 0.086 / 0.162

= 0.531

He did this calculation for each of the discrete values for up/down-days. For example, p(8 up-days | 7 up-days), p(8 down-days | 7 down-days), ect. Using Bayes' theorem, he demonstrated that the probability for a continuation of each of the streaks was approximately 0.53.

My question is, does this negate the original distribution that assigned a probability of 4 consecutive up-days as 0.086? I am struggling to understand the use of conditional probability in this context. Given the fact that the original data set of consecutive up/down-days was normally distributed, can one claim that that a streak of 8 consecutive up-days, for example, is less probable than a new streak, which would be a down-day? If so, given a streak of 7 up-days, would it not be less probable for the series to take on a streak of 8 up-days rather than reverting to the mean, that is, produce a down-day? Does the fact that, according to the previous calculations, the probability of an up-day does not change despite the existence of a streak of any size negate the the application of the '68-95-99.7% Rule' with a normally distributed data set? In other words, according to the application of Bayes' theorem above, is the probability of the continuation of a 7 up-day streak really 0.53 despite the low probability of an 8-day streak?

Thanks for the insight.

ANSWER: The chance of getting N up-days is smaller depending upon the size of N.
The chance of the next day being an up-day, however, is not dependent on how many occurred so far.

As an example, consider drawing a face card out of a deck, and then putting the card back in and  reshuffling after each draw.  Since, in each suit, there are three face cards and ten cards that are not, this makes the chances of drawing a face card as 3/13.  It makes no difference how many have been drawn in the past - the chance is still 3/13.  The chance of getting two in a row 9/169, which is a lot smaller.  However, given that one has already been drawn, the chance of the next one being a face card is independent of the last draw since each card is reshuffled back into the deck after each draw.

That is similar to the problem given. The chance of a streak may be low, but the chance of having an up-day is independent of how many up-days have occurred previously.

---------- FOLLOW-UP ----------

QUESTION: Thanks for the clarification. As a follow up question, wouldn't the fact that a streak of 3 up-days MUST precede a streak of 4 up-days imply that given the fact that today's close completed a 3 day streak, the probability of tomorrow closing as an up-day is less than the probability of tomorrow closing as a down-day given the normal distribution of streaks? Is the idea that the continuation of streaks is random despite the clearly defined probability of a streak of any particular size occurring? It seems that if the probability of a streak of 4 up-days is 0.086, given that today's close completed a 3 day streak, shouldn't the probability that tomorrow closes an up day still be 0.086?

This is an added comment I just remembered from somewhere in my past.

If there are 10 people in the room, the chance of all of them having different birthdays is 88%.
If there are 20 people in the room, the chance of all of them having different birthdays is only 59%.
If there are 30 people in the room, the chance of all of them having different birthdays is only 29%.
If there are 40 people in the room, the chance of all of them having different birthdays is only 11%.

That is because for only 2 people, there is only one pair to look at.
If there are 3 people, A, B, and C, there are 3 pairs to look at - AB, AC, or BC.
If there are 4 people, A, B, C, and D, there are 6 pairs to look at - AB, AC, AD, BC, DB, and CD.

To get the number of pairs to look at, it is (n-1)!.
That is, 2-1 = 1, and 1! = 1; 3-1 = 2, and 2! = 2; 4-1 = 3, and 3! = 6.
Now when we get up to 10 people, there are 9!, and that is 9*8*7*6*5*4*3*2*1 = 362,880.
Even though there are only 10 people in the room, there are far more than only 10 pairs.

To actually compute that chance of all the birthdays being different, for two it is only 1/365.
See, the 1st person is born on someday, and whatever day that is, that leaves 364 days for the 2nd person.  When we just keep multiplying by 366-1 on top and dividing by 365, our number quickly becomes small.  Granted, for the first 36 people each of the number turns out to be at least 90% of being different, but when we multiply all of them together, it becomes small quickly.

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#### Scott A Wilson

##### Expertise

I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?

##### Experience

Experience in the area; I have tutored people in the above areas of mathematics for over two years in AllExperts.com. I have tutored people here and there in mathematics since before I received a BS degree back in 1984. In just two more years, I received an MS degree as well, but more on that later. I tutored at OSU in the math center for all six years I was there. Most students offering assistance were juniors, seniors, or graduate students. I was allowed to tutor as a freshman. I tutored at Mathnasium for well over a year. I worked at The Boeing Company for over 5 years. I received an MS degreee in Mathematics from Oregon State Univeristy. The classes I took were over 100 hours of upper division credits in mathematical courses such as calculus, statistics, probabilty, linear algrebra, powers, linear regression, matrices, and more. I graduated with honors in both my BS and MS degrees. Past/Present Clients: College Students at Oregon State University, various math people since college, over 7,500 people on the PC from the US and rest the world.

Publications
My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.

Education/Credentials
Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

Awards and Honors
I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.

Past/Present Clients
My clients have been students at OSU, people who live nearby, friends with math questions, and several people every day on the PC. I would guess that you are probably going to be one more.