# XYZ Corporation announces its earnings four times per year. Based on historical data, you estimate..

XYZ Corporation announces its earnings four times per year. Based on historical data, you estimate that in any given quarter the probability that XYZ Corporation’s earnings will exceed consensus estimates is 30%. Also, the probability Sample Problem Question: Assume we have a mixture distribution with two independent components with equal variance. Prove that the variance of the mixture distribution must be greater than or equal to the variance of the two component distributions. Answer: Assume the two random variables, X1 and X2, have variance s2. The means are µ1 and µ2, with corresponding weights w and (1 – w). The mean of the mixture distribution, X, is just the weighted average of the two means: µ µ = = E X[ ] w E1 1 [ ] X w+ = 2 2 E X[ ] w w 1 2 + – ( ) 1 µ The variance is then: E X[( – = µ µ ) )] [ w E ( ) X w – + )] ( ) – – E X[( µ) ] 2 1 1 2 2 2 1 First, we solve for one term on the right-hand side: E X E X w w E X [( ) ] [( ( ) ) ] [( ( 1 2 1 1 2 2 1 1 1 1 – = – – – = – – µ µ µ µ – – = – – – – – w E X X w )( )) ] [( ) ( )( )( µ µ µ µ µ 2 1 2 1 1 2 2 1 1 1 2 µ µ µ s µ µ 1 2 2 1 2 2 2 1 2 2 1 1 ) ( ) ( ) ] ( ) ( ) + – – = + – – w w Similarly for the second term: E X[( ) ] w ( ) 2 2 2 2 1 2 2 – = µ s + – µ µ Substituting back into our original equation for variance: E X[( – = µ s ) )] ( + – w w)(µ µ – ) 2 2 1 2 2 1 Because w and (1 – w) are always positive and (µ1 – µ2) 2 has a minimum of zero, w(1 – w)(µ1 – µ2) 2 must be greater than or equal to zero. Therefore, the variance of the mixture distribution must always be greater than or equal to s2. c04.indd 86 11/11/13 6:56 PM Distributions 87 of exceeding the consensus in any one quarter is independent of the outcome in any other quarter. What is the probability that XYZ Corporation will exceed estimates three times in a given year?