Expected Value & VarianceLectures 5 & 6(Lectures 5 & 6) 1 / 23Expected ValueE(X ) =Xall kk P(X = k )Interpretation1A probability-weighted average.2Long-run average of X.3The fair value of a gamble.4The balance point for a probability histogram/bargraph.(Lectures 5 & 6) 2 / 23How to compute E(X)k 2 5P(X = k ) 1/3 2/3E(X ) = 2 · P(X = 2) + 5 ·P(X = 5)= 2 (1/3) + 5 (2/3)= 2/3 + 10/3 = 12/3 = 4It is a weighted average of the values of X and the weights are theprobabilities that correspond to these values.(Lectures 5 & 6) 3 / 23A BetYou pay $100 for a bet.If you win you get $210 and if you lose you get $0.X = your profit/lossX = $110, if you winX = −$100, if you loseAssume that the probabilities to win/lose areP(win) = 0.55P(lose) = 0.45Is this a fair bet?(Lectures 5 & 6) 4 / 23A BetThe pdf of X isk $110 −$100P(X = k ) 0.55 0.45To check whether the bet is fair or not, compute the expectation ofyour profit/loss:E(X ) = 110 P(X = 110) − 100 P(X = −100)= $110 (0.55) −$100 (0.45)= $15.5Is it fair?It is not a fair bet, since the expectation is not 0... but should youplay?(Lectures 5 & 6) 5 / 23Fair Bet(Lectures 5 & 6) 6 / 23(Lectures 5 & 6) 7 / 23Who wants to be a Millionaire?Quit = $100,000Guess right X = $250,000Guess wrong X = $32,000Two cases:Choose at random among all 4 possible answers.P(right) = 1/4IE(X ) = (250, 000)(1/4) + 32, 000(3/4) = $86, 500Get the 50-50 option. So, you choose at random between 2answers.P(right) = 1/2IE(X ) = (250, 000)(1/2) + 32, 000(1/2) = $141, 000(Lectures 5 & 6) 8 / 23Adding a ConstantE(X + α) = E(X ) + α(Lectures 5 & 6) 9 / 23Functions of a Random Variablek 10 12 15P(X = k ) 0.4 0.5 0.1E(X ) = 10(0.4) + 12(0.5) + 15(0.1) = 11.5Double the value of Xk 20 24 30P(2 X = 2 k ) 0.4 0.5 0.1E(2 X ) = 20(0.4) + 24(0.5) + 30(0.1) = 23E(2 X )= 2 E(X )= 2(11.5) = 23(Lectures 5 & 6) 10 / 23Multiplying by a ConstantE(b X ) = b E(X )(Lectures 5 & 6) 11 / 23Properties of E(X )1E(X + α) = E(X ) + α.2E(b X) = b E(X ).3E(X + Y ) = E(X ) + E(Y )4If X and Y are independent, thenE(XY ) = E(X ) E(Y )(Lectures 5 & 6) 12 / 23Example: Blue and green dieTwo fair dice are rolled, one blue and one green.Y = faces on the blue die, Z = face on the green dieX = sum of the two faces, W = product of the two faces(i) Determine the expected values of Y and Z .The sample space for each die blue or green isΩ = {1, 2, 3, 4, 5, 6}The PDF of Y and Z isy Y = 1 Y = 2 Y = 3 Y = 4 Y = 5 Y = 6P(Y = y) 1/6 1/6 1/6 1/6 1/6 1/6(Lectures 5 & 6) 13 / 23Example: Blue and green diePDF:y Y = 1 Y = 2 Y = 3 Y = 4 Y = 5 Y = 6P(Y = y) 1/6 1/6 1/6 1/6 1/6 1/6Expectation:E(Y ) = 1 ·16+ 2 ·16+ 3 ·16+ 4 ·16+ 5 ·16+ 6 ·16=216.Similarly, E(Z ) =216.(Lectures 5 & 6) 14 / 23Example: Blue and green die(ii) What is the mean value of X ?X = Y + ZE(X ) = E(Y + Z ) = E(Y ) + E(Z ) =216+216=426= 7.(iii) What is the expectation of W ?W = YZISince the outcome of one die does not depend on the other one,the random variables Y and Z are independent.E(W ) = E(YZ ) = E(Y )E(Z ) =216·216= 12.25.(Lectures 5 & 6) 15 / 23Measuring the SpreadVariance = how spread out the pdf is.Variance = average squared distance from the mean:IMean: E(X )IDistance from the mean: X − E(X )ISquared distance from the mean: (X − E(X ))2IAverage squared distance from the mean:Var (X ) = E(X − E(X ))2(Lectures 5 & 6) 16 / 23Another way to compute the VarianceVar(X ) = E(X2) − (E(X))2Remark!E(X2) is not the same as (E(X ))2!!!!(Lectures 5 & 6) 17 / 23How to compute Var(X )k 1 2 3P(X = k ) 0.2 0.6 0.21E(X ) = 1(0.2) + 2(0.6) + 3(0.2) = 2.2E(X2) =Pk2P(X = k )k21 4 9P(X = k ) 0.2 0.6 0.2E(X2) = 1(0.2) + 4(0.6) + 9(0.2) = 4.43Compute the Variance:Var (X ) = E(X2) −(E(X ))2= 4.4 − 22= 0.4(Lectures 5 & 6) 18 / 23Standard DeviationStandard Deviation =√Varianceσ =rVar(X )(Lectures 5 & 6) 19 / 23PropertiesVar (αX ) = α2Var (X )For example,Var (3 · X ) = 32· Var (X ) = 9 · Var (X )Var (α + X ) = Var (X )For example,Var (3 + X) = Var (X )If X and Y are independentVar (X + Y ) = Var (X ) + Var (Y )(Lectures 5 & 6) 20 / 23(Lectures 5 & 6) 21 / 23(Lectures 5 & 6) 22 / 23(Lectures 5 & 6) 23 /