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Random Processes and the Binomial and Poisson distributions.

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Presentation on theme: "Random Processes and the Binomial and Poisson distributions."— Presentation transcript:

1 Random Processes and the Binomial and Poisson distributions

2 Random Processes and Events A random process may result in a number of different outcomes. Each outcome is an event and has a probability assigned to it. Random Process Event D Event A Event B Event C Each event occurs with a probability: P(A), P(B), P(C) and P(D).

3 For example there are six possible outcomes of the throw of a die. A throw of a die Let event A = {“a six”} P(A) = 1/6

4 Probability If the random process is repeated n times (i.e. there are n trials) and event A occurs n(A) times then the probability of event A is defined as a relative frequency below: number of occurrences of event A number of trials A basic property of probability is: n(A)=0n(A)=n no outcomesall outcomesare event A P(A) approaches the true probability as n approaches infinity. Example experiment where a die is thrown n times (n trials) and event A = {“a six”}. nn(A)n(A)P(A) = n(A)/n ,676, ≈ ⅙

5 // // Computer simulation of the statistics of throwing a die. // #include int main() { const int n = ; // number of trials int m = 0; // number of sixes for (int i=0; i

6 Distribution of Outcomes Rearranging the frequency definition of probability, we can write: i.e. the number of events of type A approaches P(A) n as n approaches infinity. In reality we always have a finite number of trials, and sometimes the number is very small. For example, throw 5 dice and count the number of “sixes”. In this case P(A) n = ⅙  5 = 5/6. i.e. we expect about one “six” to appear but it may sometimes be “no sixes”, or “two sixes”, or even “five sixes” is possible. How can we quantify this?

7 Experiment A die is thrown 5 times and the number, k, of outcomes “six” is recorded. This is repeated n = 100,000,000 times to obtain the probability of k-outcomes. kn(k)n(k)n(k) / nBinomial 040,195, ,177, ,073, ,218, , , total100,000, The probability of k = 2 sixes is ≈ 16,073,542 / 100,000,000 ≈ P(k)P(k) k The distribution of k-outcomes follows the Binomial pmf. We will derive the formula for this next.

8 Binomial p.m.f The probability of obtaining the outcome A k-times in n random processes can be calculated as follows. One possible outcome is: A.A.A....A. B.B.B....B k terms n-k termsHere B is the event that is “not A”. The probability of this outcome is P(A.A.A....A. B.B.B....B)= P(A).P(A).P(A)....P(A). P(B).P(B).P(B)...P(B) = P(A) k. (1-P(A)) n-k =  k (1-  ) n-k where P(A) =  and P(B) = 1-  If the order of the sequence of outcomes of A is not important, then we can have a number of such outcomes...

9 The number of “k-combinations” is given by the Binomial Coefficient: k-terms And so the total probability of k outcomes of A is the Binomial pmf : For example, for n = 5 and k = 2 we can have the following 10 combinations: AABBB, BAABB, BBAAB, BBBAA, ABABB, BABAB, BBABA, ABBAB, BABBA, ABBBA. and the Binomial Coefficient gives: If event A is a “six” in the throw of a die, then  = 1/6 and so the probability of 2 sixes in 5 throws is That is the same number we found in the table two pages back.

10 Properties mean = n  variance  2 = n  (1-  ) (  =  2 )skewness = (1-2  )/ 

11 Binomial Examples

12 Special case: A large number of rare random processes We often experience random events that are the result of a large number of processes each with a small probability of the event. For example in radioactive decay a nucleus may have a very long half life (years) [  is very small] but a body containing a large number of nuclei [n is very large] may have a decay rate  1 s -1  n   1. This special case of the binomial pmf gives rise to the Poisson pmf: Where = n  represents the mean number of events in a given period. Example application: - The number of radioactive decays per minute. - The number of web hits per second. - The number of phone calls to a help line per hour.

13 Derivation of the Poisson pmf from the Binomial pmf In the Binomial pmf set n large, and  small  k << n Binomial k terms (small) Poisson Binomial

14 Consider a gain the throw of 5 dice (n=5) and the probability of 2 sixes (k=2,  = 1/6 ) Binomial pmf: Poisson pmf with = n  = 5/6: Given that n is not very large and  is not very small, the approximation is still not too bad. The Poisson pmf is often used in place of the binomial pmf for convenience, but remember it is just an approximation.

15 Properties (the Binomial pmf properties with the limit   0, and =n  ) mean = variance  2 = (  =  2 =  )skewness = 1/ 

16 Poisson Examples


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