solhw43823 - SOLUTION FOR HOMEWORK 3 STAT 4382 Well welcome...

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SOLUTION FOR HOMEWORK 3, STAT 4382 Well, welcome to your third homework. 1. Problem 1.3.1. For a Poisson RV with mean λ , Pr( X is odd) = summationdisplay i =0 Pr( X = 2 i + 1) = summationdisplay i =0 e - λ λ 2 i +1 / [(2 i + 1)!] . Now remember Taylor’s formulas e λ = summationdisplay i =0 λ i /i ! and e - λ = summationdisplay i =0 ( - λ ) i /i ! . Visualize these equations and get that Pr( X is odd) = e - λ [(1 / 2)( e λ - e - λ )] = (1 - e - 2 λ ) / 2 . 2. Problem 1.3.5. Here Y = n - X where X is Binomial ( n,p ). Note that Y is the number of failures in n trials which is also Binomial ( n, 1 - p ). Then E ( XY ) = E ( X ( n - X )) = nE ( X ) - E ( X 2 ) = n ( np ) - [ np (1 - p ) + ( np ) 2 ] = np ( n - 1 + p - np ) = np ( n - 1)(1 - p ) . Next, Cov ( X,Y ) = E ( XY ) - E ( X ) E ( Y ) = np ( n - 1)(1 - p ) - ( np )[ n (1 - p )] = - np (1 - p ) . Note that the covariance is negative as it should be, because if X increases then Y decreases, and vice versa. 3. Problem 1.3.8(a). Here X is Binomial ( n,p ) and Y is also Binomial ( m,p ) and in- dependent of X . Set Z = X + Y . Because p is the same, we can write (remember the discussion in class) that X = n i =1 X i where X i are independent Bernoulli ( p ), and similarly Y = m i =1 Y i where Y i are independent Bernoulli ( p ). Then we get that Z = n + m summationdisplay i =1 Z i where Z i are independent Bernoulli ( p ). This yields that Z is Binomial ( m + n,p ). Note that you can also prove this via moment generating function which is M X ( t ) = [ pe t + (1 - p )] n for the earlier defined Binomial X . 1
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Remark: The sum of several Binomial RVs is Binomial only if the probability of success ( p ) is the same! 4. Problem 1.3.12. Well, here X is Poisson ( λ ), λ = 4, and the question is to find the probability Pr( X > 7). The answer is Pr( X > 7) = summationdisplay i =7 e - 4 4 i / [ i !] .
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