3.53.6 Expected values and variance Prof. Tesler Math 186 Winter - PowerPoint PPT Presentation

3.5–3.6 Expected values and variance Prof. Tesler Math 186 Winter 2019 Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 1 / 24

Expected Winnings in a Game Setup A simple game: Pay $1 to play once. 1 Flip two fair coins. 2 Win $5 if HH, nothing otherwise. 3 � $ 5 with probability 1 / 4 ; The payoff is X = $ 0 with probability 3 / 4 . The net winnings are � $ 5 − $ 1 = $ 4 with probability 1 / 4 ; Y = X − 1 = $ 0 − $ 1 = − $ 1 with probability 3 / 4 . Playing the game once is called a trial. Playing the game n times is an experiment with n trials. Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 2 / 24

Expected Winnings in a Game Average payoff If you play the game n times, the payoff will be $5 about n / 4 times and $0 about 3 n / 4 times, totalling $ 5 · n / 4 + $ 0 · 3 n / 4 = $ 5 n / 4 The expected payoff (long term average payoff) per game is obtained by dividing by n : E ( X ) = $ 5 · 1 / 4 + $ 0 · 3 / 4 = $ 1 . 25 For the expected winnings (long term average winnings), subtract off the bet: E ( Y ) = E ( X − 1 ) = $ 4 · 1 / 4 − $ 1 · 3 / 4 = $ 0 . 25 That’s good for you and bad for the house. A fair game has expected winnings = $0. A game favors the player if the expected winnings are positive. A game favors the house if the expected winnings are negative. Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 3 / 24

Expected Value of a Random Variable (Technical name for long term average) The expected value of a discrete random variable X is � E ( X ) = x · p X ( x ) x The expected value of a continuous random variable X is � ∞ E ( X ) = x · f X ( x ) dx − ∞ E ( X ) is often called the mean value of X and is denoted µ (or µ X if there is more than one random variable in the problem). µ doesn’t have to be a value in the range of X . The previous example had range X = $ 0 or $ 5 , and mean $ 1 . 25 . Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 4 / 24

Expected Value of Binomial Distribution Consider a biased coin with probability p = 3 / 4 for heads. Flip it 10 times and record the number of heads, x 1 . Flip it another 10 times, get x 2 heads. Repeat to get x 1 , · · · , x 1000 . Estimate the average of x 1 , . . . , x 1000 : 10 ( 3 / 4 ) = 7 . 5 (Later we’ll show E ( X ) = np for the binomial distribution.) An estimate based on the pdf: About 1000 p X ( k ) of the x i ’s equal k for each k = 0 , . . . , 10 , so 1000 10 � � k · 1000 p X ( k ) x i 10 � i = 1 k = 0 average of x i ’s = = k · p X ( k ) 1000 ≈ 1000 k = 0 which is the formula for E ( X ) in this case. Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 5 / 24

Interpretation of the word “Expected” Although E ( X ) = 7 . 5 , this is not a possible value for X . Expected value does not mean we anticipate observing that value. It means the long term average of many independent measurements of X will be approximately E ( X ) . Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 6 / 24

Function of a Random Variable Let X be the value of a roll of a biased die and Z = ( X − 3 ) 2 . z = ( x − 3 ) 2 p X ( x ) p Z ( z ) x 1 q 1 4 2 q 2 1 p Z ( 0 ) = q 3 3 q 3 0 p Z ( 1 ) = q 2 + q 4 4 q 4 1 p Z ( 4 ) = q 1 + q 5 5 q 5 4 p Z ( 9 ) = q 6 6 q 6 9 pdf of X : Each q i � 0 and q 1 + · · · + q 6 = 1 . pdf of Z : Each probability is also � 0 , and the total sum is also 1 . Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 7 / 24

Expected Value of a Function Let X be the value of a roll of a biased die and Z = ( X − 3 ) 2 . z = ( x − 3 ) 2 p X ( x ) p Z ( z ) x 1 q 1 4 2 q 2 1 p Z ( 0 ) = q 3 3 q 3 0 p Z ( 1 ) = q 2 + q 4 4 q 4 1 p Z ( 4 ) = q 1 + q 5 5 q 5 4 p Z ( 9 ) = q 6 6 q 6 9 E ( Z ) , in terms of values of Z and the pdf of Z , is � E ( Z ) = z · p Z ( z ) = 0 ( q 3 ) + 1 ( q 2 + q 4 ) + 4 ( q 1 + q 5 ) + 9 ( q 6 ) z Regroup it in terms of X : 6 � ( x − 3 ) 2 p X ( x ) = 4 q 1 + 1 q 2 + 0 q 3 + 1 q 4 + 4 q 5 + 9 q 6 = x = 1 Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 8 / 24

Expected Value of a Function Let X be a discrete random variable, and g ( X ) be a function, such as ( X − 3 ) 2 . The expected value of g ( X ) is � E ( g ( X )) = g ( x ) p X ( x ) x For a continuous random variable, � ∞ E ( g ( X )) = g ( x ) f X ( x ) dx − ∞ Note that if Z = g ( X ) then E ( Z ) = E ( g ( X )) . Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 9 / 24

Expected Value of a Continuous distribution Consider the dartboard of radius 3 example, with pdf � Probability density with 31 bins if 0 � r � 3 ; 2 r / 9 f R ( r ) = 0.8 otherwise. 0 Probability density Throw n darts and make a 0.6 histogram with k bins. 0.4 r 1 , r 2 , . . . are representative values of R in each bin. 0.2 The bin width is ∆ r = 3 / k , the height is ≈ f R ( r i ) , 0 and the area is ≈ f R ( r i ) ∆ r . 0 1 2 3 The approximate number of darts in bin i is n f R ( r i ) ∆ r . Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 10 / 24

Expected Value of a Continuous Distribution The estimated average radius is � i r i · n f R ( r i ) ∆ r � = r i · f R ( r i ) ∆ r n i As n , k → ∞ , the histogram smoothes out and this becomes � 3 r · f R ( r ) dr 0 Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 11 / 24

Mean of a continuous distribution Consider the dartboard of radius 3 example, with pdf � if 0 � r � 3 ; 2 r / 9 f R ( r ) = otherwise. 0 The “average radius” (technically the mean radius or expected value of R ) is � 3 � 3 � ∞ 2 r 2 r · 2 r µ = E ( R ) = r · f R ( r ) dr = 9 dr = 9 dr 0 0 − ∞ 3 = 2 ( 3 3 − 0 3 ) 2 r 3 � � = = 2 � 27 27 � 0 Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 12 / 24

Expected Values — Properties The gambling slide earlier had E ( X − 1 ) = E ( X ) − 1 . Theorem where a , b are constants. E ( aX + b ) = aE ( X ) + b Proof (discrete case). � E ( aX + b ) = ( ax + b ) · p X ( x ) x � � = x · p X ( x ) + b p X ( x ) a x x = a · E ( X ) + b · 1 = aE ( X ) + b � Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 13 / 24

Expected Values — Properties The gambling slide earlier had E ( X − 1 ) = E ( X ) − 1 . Theorem where a , b are constants. E ( aX + b ) = aE ( X ) + b Proof (continuous case). � ∞ E ( aX + b ) = ( ax + b ) · f X ( x ) dx − ∞ � ∞ � ∞ = x · f X ( x ) dx + b f X ( x ) dx a − ∞ − ∞ = a · E ( X ) + b · 1 = aE ( X ) + b � Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 14 / 24

Expected Values — Properties These properties hold for both discrete and continuous random variables: for any constants a , b . E ( aX + b ) = a E ( X ) + b E ( aX ) = a E ( X ) E ( b ) = b � � g ( X ) + h ( X ) = E ( g ( X )) + E ( h ( X )) E Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 15 / 24

Variance These distributions both have mean=0, but the right one is more spread out. 0.1 0.1 pdf pdf 0.05 0.05 0 0 ! 20 0 20 ! 20 0 20 x x The variance of X measures the square of the spread from the mean: σ 2 = Var ( X ) = E ( X − µ ) 2 � � � The standard deviation of X is σ = Var ( X ) and measures how wide the curve is. Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 16 / 24

Variance � ( X − µ ) 2 � The same definition Var ( X ) = E is used for both the discrete and continuous cases, but expected value is computed differently in the two cases. � � Why don’t we use E ( X − µ ) or E | X − µ | to measure the spread? E ( X − µ ) = E ( X ) − µ = µ − µ = 0 , so it doesn’t measure the spread. Both | X − µ | and ( X − µ ) 2 are nonnegative. � ( X − µ ) 2 � We will see that E leads to useful properties. � � It turns out that E does not have nice properties. | X − µ | Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 17 / 24

Variance formula σ 2 = E � ( X − µ ) 2 � Consider the dartboard of radius 3 example, with pdf � if 0 � r � 3 ; 2 r / 9 f R ( r ) = otherwise. 0 µ = 2 from earlier slide. σ 2 = Var ( R ) � ( R − µ ) 2 � � ( R − 2 ) 2 � = E = E � 3 � ∞ ( r − 2 ) 2 · 2 r ( r − 2 ) 2 f R ( r ) dr = = dr 9 0 − ∞ � r 4 � 3 2 r 3 − 8 r 2 + 8 r 3 18 − 8 r 3 27 + 4 r 2 �� � = dr = � 9 9 � 0 � 3 4 0 18 − 8 ( 3 3 ) + 4 ( 3 2 ) � − 0 = 1 = 27 9 2 Variance: σ 2 = 1 � Standard deviation: σ = 1 / 2 2 Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 18 / 24

Variance — Second formula There are two equivalent formulas to compute variance. In any problem, choose the easier one: σ 2 = Var ( X ) ( X − µ ) 2 � � (Definition) = E E ( X 2 ) − µ 2 (Sometimes easier to compute) = Proof. ( X − µ ) 2 � � Var ( X ) = E E ( X 2 − 2 µ X + µ 2 ) = E ( X 2 ) − 2 µ E ( X ) + µ 2 = E ( X 2 ) − 2 µ 2 + µ 2 = E ( X 2 ) − µ 2 = � Prof. Tesler 3.5–3.6 Expected values and variance Math 186 / Winter 2019 19 / 24