1 / 37

Lectures prepared by: Elchanan Mossel Yelena Shvets

Lectures prepared by: Elchanan Mossel Yelena Shvets. Binomial Distribution. Toss a coin 100, what’s the chance of 60 ?. hidden assumptions: n-independence; probabilities are fixed. ={ }. Magic Hat:.

arty
Download Presentation

Lectures prepared by: Elchanan Mossel Yelena Shvets

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lectures prepared by:Elchanan MosselYelena Shvets

  2. Binomial Distribution • Toss a coin 100, what’s the chance of 60 ? hidden assumptions: n-independence; probabilities are fixed.

  3. ={ } Magic Hat: Each time we pull an item out of the hat it magically reappears. What’s the chance of drawing 3 I-pods in 10 trials?

  4. Suppose we roll a die 4 times. What’s the chance of k ? Let . ={} Binomial Distribution k=0

  5. Suppose we roll a die 4 times. What’s the chance of k ? Let . ={} Binomial Distribution k=1

  6. Suppose we roll a die 4 times. What’s the chance of k ? Let . ={} Binomial Distribution k=2

  7. Suppose we roll a die 4 times. What’s the chance of k ? Let . ={} Binomial Distribution k=3

  8. Suppose we roll a die 4 times. What’s the chance of k ? Let . ={} Binomial Distribution k=4

  9. Suppose we roll a die 4 times. What’s the chance of k ? Let . ={} Binomial Distribution k=0 k=1 k=2 k=3 k=4

  10. Binomial Distribution How do we count the number of sequences of length 4 with 3 . and 1 .? .1 1. 1.. ..1 .2. ...1 ..3. .3.. 1... ....1 ...4. ..6.. .4... 1....

  11. Binomial Distribution This is the Pascal’s triangle, which gives, as you may recall the binomial coefficients.

  12. 1a 1b (a + b) = 1b2 2ab 1a2 (a + b)2 = 1a3 (a + b)3 = 3a2b 3ab2 1b3 1a4 6a2b2 4ab3 1b4 4a3b (a + b)4 = Binomial Distribution

  13. Newton’s Binomial Theorem

  14. Binomial Distribution For n independent trials each with probability p of success and (1-p) of failure we have This defines the binomial(n,p)distribution over the set of n+1 integers {0,1,…,n}.

  15. Binomial Distribution This represents the chance that in n draws there was some number of successes between zero and n.

  16. Example: A pair of coins • A pair of coins will be tossed 5 times. • Find the probability of getting • on k of the tosses, k=0 to 5. binomial(5,1/4)

  17. # =k Example: A pair of coins • A pair of coins will be tossed 5 times. • Find the probability of getting • on k of the tosses, k=0 to 5. To fill out the distribution table we could compute 6 quantities for k = 0,1,…5 separately, or use a trick.

  18. Binomial Distribution:consecutive odds ratio Consecutive odds ratio relates P(k) and P(k-1).

  19. Use consecutive odds ratio to quickly fill out the distribution table Example: A pair of coins for binomial(5,1/4): • k • 0 • 1 • 2 • 3 • 4 • 5

  20. Binomial Distribution We can use this table to find the following conditional probability: P(at least 3 | at least 1 in first 2 tosses) = P(3 or more & 1 or 2 in first 2 tosses) P(1 or 2 in first 2)

  21. bin(5,1/4): • k • 0 • 1 • 2 • 3 • 4 • 5 bin(2,1/4): • 2 • k • 0 • 1

  22. Stirling’s formula How useful is the binomial formula? Try using your calculators to compute P(500 H in 1000 coin tosses) directly: My calculator gave me en error when I tried computing 1000!. This number is just too big to be stored.

  23. The following is calledthe Stirling’s approximation. It is not very useful if applied directly : nn is a very big number if n is 1000.

  24. Stirling’s formula:

  25. Binomial Distribution Toss coin 1000 times; P(500 in 1000|250 in first 500)= P(500 in 1000 & 250 in first 500)/P(250 in first 500)= P(250 in first 500 & 250 in second 500)/P(250 in first 500)= P(250 in first 500) P(250 in second 500)/ P(250 in first 500)= P(250 in second 500)=

  26. Binomial Distribution: Mean Question: For a fair coin with p = ½, what do we expect in 100 tosses?

  27. Binomial Distribution:Mean • Recall the frequency interpretation: • p ¼ #H/#Trials • So we expect about 50 H !

  28. Binomial Distribution:Mean • Expectedvalue or Mean(m) • of a binomial(n,p) distribution m = #Trials £ P(success) = n pslip !!

  29. Binomial Distribution: Mode Question: What is the most likely number of successes? Mean seems a good guess.

  30. Binomial Distribution:Mode Recall that To see whether this is the most likely number of successes we need to compare this to P(k in 100) for every other k.

  31. Binomial Distribution:Mode • The most likely number of successes is called the mode of a binomial distribution.

  32. Binomial Distribution:Mode If we can show that for some m P(1) · … P(m-1) · P(m) ¸ P(m+1) … ¸ P(n), then m would be the mode.

  33. Binomial Distribution:Mode so we can use successive odds ratio: to determine the mode.

  34. > > > > > > Binomial Distribution:Mode successive odds ratio: If we replace · with > the implications will still hold. So for m=bnp+pc we get that P(m-1) · P(m) > P(m+1).

  35. mode = 50 Binomial Distribution:Mode - graph SOR k

  36. Binomial Distribution:Mode - graph distr k mode = 50

  37. Lectures prepared by:Elchanan MosselYelena Shvets

More Related