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

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Unformatted text preview: 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 ....
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This note was uploaded on 02/23/2012 for the course STAT 4382 taught by Professor Krwaw during the Fall '11 term at King Mongkut's Institute of Technology Ladkrabang.

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

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