Lecture 11cs485-lec11-notes-2009-02-11

Lecture 11cs485-lec11-notes-2009-02-11 - CS 485:...

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CS 485: Mathematical Foundations for the Information Age Lecture 11 ▪ February 11, 2009 Jihai Chen (jc628) and Jeff Pankewicz (jhp36) A Review on Probability Distributions : 1) Binomial Distribution: The Binomial distribution models the probability of k successes in n independent trials, with success in each trial having probability p. Pr(k successes) = C(n,k) p k (1-p) n-k (we want exactly k successes, and the rest fail) Deriving the formula for expected value (E[k]) and variance ( s 2 ) of a Binomial distribution: E[k] = k*Prob(k) for all values of k = S k=0~n k*C(n,k) p k (1-p) n-k To simplify, consider (p+q) n = S k=0~n C(n,k)p k q n-k . Taking derivative of both sides w.r.t “p”, we get: n(p+q) n-1 = (1/p) S k=0~n k*C(n,k) p k (1-p) n-k = (1/p)*E[k] now if we set q=1-p, then (p+1-p) n-1 =1, so the LHS becomes just n, and we have: n=(1/p)E[k]=> E[k]= np Now, to get the variance, we take the second derivative of the equation (p+q) n = S k=0~n C(n,k)p k q n-k : n(n-1)(p+q) n-2 = S k=0~n k(k-1) C(n,k) p k-2 q n-k sub in q=1-p, and multiply both sides by p 2 , we get p 2 n(n-1) = S k=0~n k(k-1) C(n,k) p k q n-k Factor out the right hand side with k 2 and -k, we have p 2 n(n-1) = S k=0~n k 2 C(n,k) p k q n-k - S k=0~n kC(n,k) p k q n-k We realize that the first term of the RHS is E[k 2 ] , and the second term is E[k] = np, moving the second term on the right to the left, we receive: E[k 2 ]= p 2 n(n-1)+np Now we also note that
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Lecture 11cs485-lec11-notes-2009-02-11 - CS 485:...

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