# Physics 114: Lecture 9 Probability Density Functions Dale E. Gary NJIT Physics Department.

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Physics 114: Lecture 9 Probability Density Functions Dale E. Gary NJIT Physics Department

February 12, 2010 Binomial Distribution  If you raise the sum of two variables to a power, you get:  Writing only the coefficients, you begin to see a pattern: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1

February 12, 2010 Binomial Distribution  Remarkably, this pattern is also the one that governs the possibilities of tossing n coins :  With 3 coins, there are 8 ways for them to land, as shown above.  In general, there are 2 n possible ways for n coins to land.  How many permutations are there for a given row, above, e.g. how many permutations for getting 1 head and 2 tails? Obviously, 3.  How many permutation for x heads and n  x tails, for general n and x ? 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 n 2 n 0 1 1 2 2 4 3 8 4 16 Number of combinations in each row: (n choose x)

February 12, 2010 Probability  With fair coins, tossing a coin will result in equal chance of 50%, or ½, of its ending up heads. Let us call this probability p. Obviously, the probability of tossing a tails, q, is q = (1  p).  With 3 coins, the probability of getting any single one of the combinations is 1/2 n = 1/8 th, (since there are 8 combinations, and each is equally probable). This comes from (½) (½) (½), or the product of each probability p = ½ to get a heads.  If we want to know the probability of getting, say 1 heads and 2 tails, we just need to multiply the probability of any combination (1/8 th ) by the number of ways of getting 1 heads and 2 tails, i.e. 3, for a total probability of 3/8.  To be really general, say the coins were not fair, so p ≠ q. Then the probability to get heads, tails, tails would be (p)(q)(q) = p 1 q 2.  Finally the probability P(x; n, p) of getting x heads given n coins each of which has probability p, is  With 3 coins, there are 8 ways for them to land, as shown above.  In general, there are 2 n possible ways for n coins to land.  How many permutations are there for a given row, above, e.g. how many permutations for getting 1 head and 2 tails? Obviously, 3.  How many permutation for x heads and n  x tails, for general n and x ?

February 12, 2010 Binomial Distribution  This is the binomial distribution, which we write P B :  Let’s see if it works. For 1 heads with a toss of 3 fair coins, x = 1, n = 3, p = ½, we get  For no heads, and all tails, we get  Say the coins are not fair, but p = ¼. Then the probability of 2 heads and 1 tails is:  You’ll show for homework that the sum of all probabilities for this (and any) case is 1, i.e. the probabilities are normalized. Note: 0!  1

February 12, 2010 Binomial Distribution  To see the connection of this to the sum of two variables raised to a power, replace a and b with p and q :  Since p + q = 1, each of these powers also equals one on the left side, while the right side expresses how the probabilities are split among the different combinations. When p = q = ½, for example, the binomial triangle becomes  In MatLAB, use binopdf(x,n,p) to calculate one row of this triangle, e.g. binopdf(0:3,3,0.5) prints 0.125, 0.375, 0.375, 0.125. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 1 / 2 1 / 2 1 / 4 2 / 4 1 / 4 1 / 8 3 / 8 3 / 8 1 / 8 1 / 16 4 / 16 6 / 16 4 / 16 1 / 16

February 12, 2010 Binomial Distribution  Let’s say we toss 10 coins, and ask how many heads we will see. The 10 th row of the triangle would be plotted as at right.  The binomial distribution applies to yes/no cases, i.e. cases where you want to know the probability of something happening, vs. it not happening.  Say we want to know the probability of getting a 1, rolling five 6-sided dice. Then p = 1/6 (the probability of rolling a 1 on one die), and q = 1 – p = 5/6 (the probability of NOT rolling a 1). The binomial distribution applies to this case, with P B (x,5,1/6). The plot is shown at right. >> binopdf(0:5,5,1/6.) ans = 0.4019 0.4019 0.1608 0.0322 0.0032 0.0001

February 12, 2010 Binomial Distribution Mean  Let’s say we toss 10 coins N = 100 times. Then we would multiple the PDF by N, to find out how many times we would have x number of heads.  The mean of the distribution is, as before:  For 10 coins, with p = ½, we get  = np = 5.  For 5 dice, with p = 1/6, we get  = np = 5/6. 40 30 20 10 0  

February 12, 2010 Binomial Standard Deviation  The standard deviation of the distribution is the “second moment,” given by the variance:  For 10 coins, with p = ½, we get  For 5 dice, with p = 1/6, we get 40 30 20 10 0    

February 12, 2010 Summary of Binomial Distribution  The binomial distribution is P B :  The mean is  The standard deviation is

February 12, 2010 Poisson Distribution  An approximation to the binomial distribution is very useful for the case where n is very large (i.e. rolls with a die with infinite number of sides?) and p is very small—called the Poisson distribution.  This is the case of counting experiments, such as the decay of radioactive material, or measuring photons in low light level.  To derive it, start with the binomial distribution with n large and p << 1, but with a well defined mean  = np. Then  The term because x is small, so most of the terms cancel leaving a total of x terms each approximately equal to n.  This gives

February 12, 2010 Poisson Distribution  Now, the term (1 – p)  x  1, for small p, and with some algebra we can show that the term (1 – p) n  e .  Thus, the final Poisson distribution depends only on x and , and is defined as  The text shows that the expectation value of x (i.e. the mean) is  Remarkably, the standard deviation is given by the second moment as  These are a little tedious to prove, but all we need for now is to know that the standard deviation is the square-root of the mean.

February 12, 2010 Example 2.3  Some students measure some background counts of cosmic rays. They recorded numbers of counts in their detector for a series of 100 2-s intervals, and found a mean of 1.69 counts/interval. They can use the standard deviation formula from chapter 1, which is to get a standard deviation directly from the data. They do this and get s = 1.29. They can also estimate the standard deviation by  Now they change the length of time they count from 2-s intervals to 15-s intervals. Now the mean number of counts in each interval will increase. Now they measure a mean of 11.48, which implies while they again calculate s directly from their measurements to find s = 3.39.  We can plot the theoretical distributions using MatLAB poisspdf(x,mu), e.g. poisspdf(0:8,1.69) gives ans = 0.1845 0.3118 0.2635 0.1484 0.0627 0.0212 0.0060 0.0014 0.0003

February 12, 2010 Example 2.3, cont’d  The plots of the distributions is shown for these two cases in the plots at right.  You can see that for a small mean, the distribution is quite asymmetrical. As the mean increases, the distribution becomes somewhat more symmetrical (but is still not symmetrical at 11.48 counts/interval).  I have overplotted the mean and standard deviation. You can see that the mean does not coincide with the peak (the most probable value).    

February 12, 2010 Example 2.3, cont’d  Here is the higher-mean plot with the equivalent Gaussian (normal distribution) overlaid.  For large means (high counts), the Poisson distribution approaches the Gaussian distribution, which we will describe further next time.

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