jan 13 IndicatorExample

# jan 13 IndicatorExample - 6 = 6 E[X i = 6(5/6 n Also we can...

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Example from class, Jan 13 Roll a 6-sided die n times. Let X = the number of unrolled faces. Find E[X] and Var(X) (in terms of n ) Solution Define indicator variables X i = {1 if the i th face is unrolled after n rolls, {0 otherwise; for i = 1, 2, …, 6 Then X = X 1 + … + X 6 Using the results for indicator variables, E[X i ] = P( i th face is unrolled after n rolls) = (5/6) n Var(X i ) = (5/6) n (1 – (5/6) n ) Cov(X i , X j ) = E[X i X j ] – E[X i ]E[X j ] = P(both faces i and j are unrolled after n rolls) – (5/6) n (5/6) n = (4/6) n – (5/6) 2 n Using rules for linear combinations of random variables, E[X] = E[X 1 + … + X 6 ] = E[X 1 ] + … + E[X
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Unformatted text preview: 6 ] = 6 E[X i ] = 6 (5/6) n Also we can find the number of distinct faces rolled, 6 – X Var(X) = Var(X 1 + … + X 6 ) = Var(X 1 ) + … + Var(X 6 ) + 2 Σ i<j Σ Cov(X i , X j ) = 6 Var(X i ) + 2*15 Cov(X i , X j ) since there are (6C2) = 15 pairs i < j = 6(5/6) n (1 – (5/6) n ) + 30 ((4/6) n – (5/6) 2 n ) You could simplify, but you don’t have to. The neat thing to look at is what these expressions look like as functions of n . Can you think of a logical explanation why the mean and variance behave that way as n increases?...
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