slides10 - Lecture Stat 302 Introduction to Probability...

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Lecture Stat 302 Introduction to Probability - Slides 10 AD Feb. 2010 AD () Feb. 2010 1 / 23
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Discrete Random Variables A discrete r.v. X takes at most a countable number of possible values f x 1 , x 2 , ... g with p.m.f. p ( x i ) = P ( X = x i ) where p ( x i ) ° 0 and i = 1 p ( x i ) = 1 . Expected value/mean μ = E ( X ) = i = 1 x i p ( x i ) . Variance Var ( X ) = E ° ( X ± μ ) 2 ± = E ² X 2 ³ ± μ 2 . AD () Feb. 2010 2 / 23
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Bernoulli and Binomial Distributions Assume you have an experiment w.p. success p and w.p. failure ( 1 ± p ) . Bernoulli r.v.: You set X = 1 if success and X = 0 if failure so P ( X = 1 ) = p ( 1 ) = p , P ( X = 0 ) = p ( 0 ) = 1 ± p . Assume now you have n independent experiments, each w.p. success p and w.p. failure ( 1 ± p ) . Binomial r.v.: Set X = number of successes among the n experiments, then X 2 f 0 , 1 , 2 , ..., n g X = X 1 + X 2 + ² ² ² + X n where X k is the Bernoulli r.v. associated to experiment i and P ( X = k ) = ´ n k µ p k ( 1 ± p ) n ± k . AD () Feb. 2010 3 / 23
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Example Example : A survey from Teenage Research Unlimited (Northbrook, Ill.) found that 30% of teenage consumers receive their spending money from part-time jobs. If °ve teenagers are selected at random, °nd the probability that at least three of them will have part-time jobs. Answer : Let X be the number of teenagers having a part-time job among 5 teenagers, then X is Binomial of parameters n = 5, p = 0 . 3 so P ( X ° 3 ) = P ( X = 3 ) + P ( X = 4 ) + P ( X = 5 ) = 0 . 132 + 0 . 028 + 0 . 002 = 0 . 162 Example : What is the probability of obtaining 45 or fewer heads in 100 tosses of a fair coin? Solution : Let X be the number of heads, then X is binomial of parameters n = 100, p = 0 . 5 and P ( X ³ 45 ) = 45 k = 0 ´ n k µ p k ( 1 ± p ) k = 0 . 184. AD () Feb. 2010 4 / 23
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Mean of the Binomial Random Variable The mean/expected value of X is given by E ( X ) = np Proof 1 . We have X = X 1 + X 2 + ² ² ² + X n where X i is a Bernoulli r.v. w.p. p so E ( X ) = E ( X 1 + ² ² ² + X n ) = E ( X 1 ) + ² ² ² + E ( X n ) = np . Proof 2 . We have E ( X ) = n k = 0 k P ( X = k ) = n k = 0 k ´ n k µ p k ( 1 ± p ) n ± k = n k = 1 k n ( n ± 1 ) ! k ( k ± 1 ) ! ( n ± k ) ! p p k ± 1 ( 1 ± p ) n ± k = np n k = 1 ( n ± 1 ) ! ( k ± 1 ) ! ( n ± k ) ! p k ± 1 ( 1 ± p ) n ± k = np m l = 0 m ! l ! ( m ± l ) ! p l ( 1 ± p ) m ± l ( m n ± 1 , l k ± 1 ) = np AD () Feb. 2010 5 / 23
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Variance of the Binomial Random Variable The variance of the binomial random variable is Var ( X ) = np ( 1 ± p ) . Proof 1 . We have Var ( X ) = E ² X 2 ³ ± E ( X ) 2 = E ² X 2 ³ ± ( np ) 2 where X 2 = ( X 1 + ² ² ² + X n ) 2 = n k = 1 X 2 k + n k , l = 1 ; k 6 = l X k X l so E ² X 2 ³ = n k = 1 E ² X 2 k ³ + n k , l = 1 ; k 6 = l E ( X k X l ) . We have E ² X 2 k ³ = 1 2 ´ p + 0 2 ´ ( 1 ± p ) = p . To compute E ( X k X l ) where k 6 = l , we note that X k X l 2 f 0 , 1 g is a r.v. such that P ( X k X l = 1 ) = P ( X k = 1 ) P ( X l = 1 ) = p 2 , P ( X k X l = 0 ) = 1 ± P ( X k X l = 1 ) = 1 ± p 2 . AD () Feb. 2010 6 / 23
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Variance of the Binomial Random Variable As a safety check, you have indeed P ( X k X l = 0 ) = P ( X k = 0 , X l = 0 ) + P ( X k = 0 , X l = 1 ) + P ( X k = 1 , X l = 0 ) = ( 1 ± p ) 2 + ( 1 ± p ) p + p ( 1 ± p ) = 1 ± p 2 . Hence we have E ( X k X l ) = 1 ´ p 2 + 0 ´ ² 1 ± p 2 ³ = p 2 thus E ² X 2 ³ = n k = 1 E ² X 2 k ³ + n k = 1 , l > k E ( X k X l ) = n ´ p + n ( n ± 1 ) ´ p 2 Finally we have Var ( X ) = np + n ( n ± 1 ) ´ p 2 ± ( np ) 2 = np ( 1 ± p ) .
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