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10-07 Random Variables (3)

# 10-07 Random Variables (3) - BTRY 4080 STSCI 4080 Fall 2010...

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BTRY 4080 / STSCI 4080 Fall 2010 210 Section 4.6: Bernoulli and Binomial Distributions Bernoulli random variable: A single trial is performed, with possible outcomes being “success” or “failure”. Suppose the probability of success is p . Let X denote the number of successes. Then, X Bernoulli ( p ) . is notational shorthand for “is distributed as” Probability mass function (parameters: p ) p (1) = P ( X = 1) = p ; p (0) = P ( X = 0) = 1 p. Mean & variance: E ( X ) = p and V ar ( X ) = p (1 p )

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BTRY 4080 / STSCI 4080 Fall 2010 211 Binomial random variable: Suppose n independent Bernoulli trials are performed, where each trial results in a success with probability p . Let X denote the number of successes in these n independent trials. Then: X Binomial ( n,p ) . Probability mass function (parameters: n , p ) p ( i ) = P ( X = i ) = parenleftbigg n i parenrightbigg p i (1 p ) n i , i = 0 , 1 ,...,n Mean & variance: E ( X ) = np and V ar ( X ) = np (1 p ) Proof that n i =0 p ( i ) = 1 is a direct application of the Binomial Theorem; formulas for mean and variance rely on similar calculations (p. 139).
BTRY 4080 / STSCI 4080 Fall 2010 212 Example (see Slide 199/205 – last time!): A coin is tossed 6 times. For a single coin toss, suppose S = { H,T } and P ( H ) = 1 / 3 . In addition, suppose the results for all tosses are independent. Let X denote the number of heads. Find E ( X ) and V ar ( X ) . Solution: As we now know: X is Binomial (6 , 1 3 ) (here, n = 6 and p = 1 3 ) with pmf p ( x ) = 8 < : ` 6 x ´` 1 3 ´ x ` 2 3 ´ 6 - x x ∈ { 0 , 1 , 2 , 3 , 4 , 5 , 6 } 0 otherwise So, we should have E ( X ) = 6 1 3 « = 2 and V ar ( X ) = 6 1 3 «„ 2 3 « = 4 3 , exactly the same results that we obtained before.

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BTRY 4080 / STSCI 4080 Fall 2010 213 Calculations in R: > x = 0:6 > p = function(i) choose(6,i) * (1/3)ˆi * (2/3)ˆ(6-i) > x [1] 0 1 2 3 4 5 6 > p(x) [1] 0.087791 0.263374 0.329218 0.219479 0.082305 0.016461 0.001372 > muX = sum(x * p(x)) [1] 2 > varX = sum(xˆ2 * p(x)) - (muX)ˆ2 [1] 1.333333
BTRY 4080 / STSCI 4080 Fall 2010 214 Example (Slide 122): Consider an urn with n 1 red balls and n 2 blue balls. We take a total sample of size n from this urn, where n n 1 + n 2 . What is the probability of seeing exactly x red balls if sampling is done with replacement? Solution: Let Y be the number of red balls. On each draw (trial), the probability of drawing a red ball is p = n 1 / ( n 1 + n 2 ) . Assume successive draws are independent. Then “exactly x red balls” is the same as the event { Y = x } , where Y Binomial ( n,p ) . Thus: P ( Y = x ) = parenleftbigg n x parenrightbigg p x (1 p ) n x , x = 0 , 1 ,...,n ; or, as we argued (much) earlier, P ( Y = x ) = parenleftbigg n x parenrightbiggparenleftbigg n 1 n 1 + n 2 parenrightbigg x parenleftbigg n 2 n 1 + n 2 parenrightbigg n x , x = 0 , 1 ,...,n.

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BTRY 4080 / STSCI 4080 Fall 2010 215 Section 4.7: Poisson Distribution Poisson random variable: A random variable X , taking on one of the values { 0 , 1 , 2 , ...... } , is a Poisson random variable with parameter λ > 0 if the pmf of X is given by p ( i ) = P ( X = i ) = e λ λ i i ! , i = 0 , 1 , 2 , 3 ,... Formula for pmf is an immediate consequence of the (Maclaurin) series expansion for e λ . In particular: since e λ = summationdisplay i =0 λ i i !
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