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Review

Review. Discrete Distributions. Binomial distribution, Negative binomial distribution, Hypergeometric distribution, Poisson distribution. Expected Value. If X is a discrete rv and p(x) is the value of its probability distribution at x , the expected value of X is defined as. Example.

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Review

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  1. Review

  2. Discrete Distributions • Binomial distribution, • Negative binomial distribution, • Hypergeometric distribution, • Poisson distribution.

  3. Expected Value • If X is a discrete rv and p(x) is the value of its probability distribution at x, the expected value of X is defined as

  4. Example • Toss a coin 4 times. X = number of heads. What’s E(X) ? • The pmf of X is • x: 0 1 2 3 4 • p(x): 1/16 4/16 6/16 4/16 1/16 • So,

  5. Example • Let X be a Bernoulli rv with pmf • Then E(X) = 0p(0) + 1p(1) = p. So the expected value of X is just the probability that X takes on the value 1.

  6. Example • X = number of children born up to and including the first boy. The pmf of X is • Then

  7. Expected Value of a Function of a RV • If a rv X has a pmf p(x), then the expected value of any function h(X) is computed by • Special case: h(x) = a·x + b. • E(a X + b) = a·E(X) + b. • Why?

  8. Variance • The expected value measures the center of a probability distribution. • Variance measures the variability of a pmf.

  9. Variance • Let X have pmf p(x) and expected value . Then the variance of X, denoted by • The standard deviation (SD) of X is

  10. Example • If X has pmf : • x 1 2 6 8 • p(x) .4 .1 .3 .2 • Then •  = 1×.4 + 2×.1 + 6×.3 + 8×.2 = 4 . • 2 = (1 - 4)2×.4 + (2 - 4)2 × .1 + (6 - 4)2 ×.3 • + (8 - 4)2 ×.2 = 8.4. • and  = 2.90.

  11. A Shortcut Formula • Proof:

  12. Rules of Variance • In particular,

  13. Moments • The kth moment about the origin of a rv X, denoted by µk’ , is the expectedvalue of Xk, , symbolically, • µk’ = E(Xk) = x xk · p(x). • The kth moment about the mean of a rv X, denoted by µk, is the expectedvalue of (X - µ)k, , symbolically, • µk = E[(X - µ)k] = x (x - µ)k · p(x).

  14. Special Cases • The expectation, or the mean, is the 1st moment about the origin. • µ = µ1’ = E(X) = x x · p(x). • The variance is the 2nd moment about the mean • 2 = µ2 = E[(X - µ)2] = x (x - µ)2 · p(x).

  15. The Binomial Distribution

  16. Binomial Distribution • For X ~ Bin(n,p), the cdf will be denoted by

  17. Mean & Variance • If X ~ Bin(n, p), then • E(X) = np, • V(X) = npq (where q = 1-p.)

  18. Example(Cont) • n = 5, p = 11/32 . Then • E(X) = n · p = 5 · 11/32 = 1.72. • V(X) = n · p · q = 5 · 11/32 · 21/32 = 1.13. •  = (1.13)1/2 = 1.06.

  19. Hypergeometric and Negative Binomial Distribution

  20. Introduction • The hypergeometric and negative binomial distribution are both closely related to the binomial distribution.

  21. Introduction • The negative binomial distribution arises from fixing the number of S’s and letting the number of trials to be random. • The hypergeometric distribution is the exact probability model for sampling without replacement from a finite dichotomous (S,F) population.

  22. Negative Binomial Dist’n • The experiment consists of a sequence of independent trials. • Each trial results in either S or F. • The probability of success, p, is constant from trial to trial. • Trials are performed until a total of ssuccesses have been observed, where s is a prespecified positive integer.

  23. Negative Binomial RV • X = the number of F’s that precede the rth success, is called a negative binomial rv. • Possible values of X are 0, 1, 2, …

  24. pmf • Denote by nb(x; r, p) the pmf of X. Then • Why? • Total # of trials = x; The last trial must be a success. Among the first (x-1) trials, there are (s - 1) successes & x-s failures.

  25. Review of Chapter 3 • Hypergeometric distribution, • Poisson distribution.

  26. Example • What’s the probability that < 3 requests are received during a particular hour? • P( X < 3) = P(0) + P(1) + P(2) • = e-5 + 5· e-5 + 52 · e-5/2 • = 0.125.

  27. Example • What’s the probability that exactly 10 requests are received during a particular 2-hour period? • Rate = 2 × 5 = 10. • P(X = 10) = e-10 1010/10! = 0.125.

  28. Example • How many calls do they expect to get during a 45-min period? • E(X) = (3/4) · 5 = 3.75.

  29. Continuous RVs& Probability Distributions

  30. Continuous RV • An rv X is continuous if its set of possible values is an entire interval of numbers. • Example: • X = the pH of a random soil sample • X = the weight of a randomly selected • person.

  31. pdf • Let X be a continuous rv. Then a probability density function (pdf) of X is a function f(x) such that for any two numbers a and b with a b, • For f(x) to be a pdf, f(x) must satisfy: • f(x)  0 for all x, and

  32. Example • Waiting time at a bus station. A bus arrives every 10 minutes. So the waiting time is from 0 to 10. One possible pdf for waiting time X is • The probability of waiting between 3 to 5 minutes is:

  33. Uniform Distribution • A continuous rv X is said to have a uniform distribution on the interval [A, B] if the pdf of X is • Graphs of uniform distributions.

  34. Probability at a Point • When X is a discrete rv, each possible value is assigned positive probability. This is no longer true for continuous rv. • If X is a continuous rv, then for any number c, P(X = c) = 0. Consequently, P(a  X  b) = P(a < X  b) = P(a  X < b) • = P(a < X < b).

  35. Example • Let X = the “time headway” for two randomly chosen consecutive cars on a freeway during a period of heavy flow. Suppose the pdf of X is given by: • f(x) = 0.15 e-0.15( x - 0.5), x 0.5. • f(x) = 0 for x < .5 and f(x) decreases exponentially fast as x increase from .5.

  36. Example • First, it clear that f(x) 0. Now we verify • The probability that headway time is at most 5 seconds is

  37. CDFs & Expected Values

  38. cdf • The cumulative distribution function (cdf) F(x) for a continuous rv X is defined for every number x by • For each x, F(x) is the area under the density curve to the left of x. It is the probability of observing X a value smaller than or equal to x.

  39. Example • Let X have a uniform distribution on the interval [A, B]. Then • So, for x < A, F(x) = 0 and for x  B, F(x) = 1. For A  x  B,

  40. Example • The entire cdf is: • The graph of the cdf looks like:

  41. Propositions • Compute probabilities using F(x): P(a  x  b) = F(b) - F(a). • Obtaining pdf from cdf: • If X is a continuous rv with cdf F(x) differentiable at every point x, then the pdf f(x) =F ’(x).

  42. Example • For uniform distribution on [A, B], the cdf is • So, for example, if A < a < b < B, then P(a < X < b) = F(b)-F(a) = (b-a)/(B-A). • The pdf • f(x) = F ’(x) = 1/(B-A) for A < x < B.

  43. Expected Values • The expected value (or, mean) of a continuous rv X with pdf f(x) is • If X is a continuous rv with pdf f(x) and h(X) is any function of X, then

  44. Example • The pdf of the waiting time (in minutes) at a checkout is given by • f(x) = x/8 for 0  x < 4. • What’s the probability of waiting less than 3 min? • What’s the expectation of the waiting time?

  45. Example • What’s the probability of waiting less than 3 min? • What’s the expectation of the waiting time?

  46. Variance & S.D. • The variance of a continuous rv X with pdf f(x) and mean  is • The standard deviation (S.D.) of X is • V(X) = E(X2) - [E(X)]2.

  47. Linear Transformation • If h(X) = a X + b and V(X) = 2, then • V(h(X))=V(a X + b) = a 2 2 • and • aX+b = |a| .

  48. Example(Cont) • The pdf of the waiting time at a checkout: • f(x) = x/8 for 0  x < 4. • Find the variance of the waiting time. •  = E(X) = 2.667.

  49. Normal Distribution

  50. Introduction • The normal distribution is the most important distribution in all of probability and statistics. • Many numerical populations have distributions that can be approximated very well by a normal curve.

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