Expected values and variance

Transcription

1 Expected values and variance Prof. Tesler Math 186 January 27, 214 Prof. Tesler Expected values and variance Math 186 / January 27, / 23

2 Expected Winnings in a Game Setup A simple game: 1 Pay $1 to play once. 2 Flip two fair coins. 3 Win $5 if HH, nothing otherwise. The payoff is X = { $5 with probability 1/4; $ with probability 3/4. The net winnings { are $5 $1 = $4 with probability 1/4; Y = X 1 = $ $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 Expected values and variance Math 186 / January 27, / 23

3 Expected Winnings in a Game Average payoff If you play the game n times, the payoff will be $5 about n/4 times and $ about 3n/4 times, totalling $5 n/4 + $ 3n/4 = $5n/4 The expected payoff (long term average payoff) per game is obtained by dividing by n: E(X) = $5 1/4 + $ 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 = $.25 That s good for you and bad for the house. A fair game has expected winnings = $. A game favors the player if the expected winnings are positive. A game favors the house if the expected winnings are negative. Prof. Tesler Expected values and variance Math 186 / January 27, / 23

4 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 x p 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 = $ or $5, and mean $1.25. Prof. Tesler Expected values and variance Math 186 / January 27, / 23

5 Expected Value of Binomial Distribution Consider a biased coin with probability p = 3/4 for heads. Flip it 1 times and record the number of heads, x 1. Flip it another 1 times, get x 2 heads. Repeat to get x 1,, x 1. Estimate the average of x 1,..., x 1 : 1(3/4) = 7.5 (Later we ll show E(X) = np for the binomial distribution.) An estimate based on the pdf: About 1p X (k) of the x i s equal k for each k =,..., 1, so average of x i s = 1 i=1 x i 1 1 k= k 1 p X (k) 1 which is the formula for E(X) in this case. = 1 k= k p X (k) Prof. Tesler Expected values and variance Math 186 / January 27, / 23

6 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 Expected values and variance Math 186 / January 27, / 23

7 Function of a Random Variable Let X be the value of a roll of a biased die and Z = (X 3) 2. x p X (x) z = (x 3) 2 p Z (z) 1 q q q 3 p Z () = q 3 4 q 4 1 p Z (1) = q 2 + q 4 5 q 5 4 p Z (4) = q 1 + q 5 6 q 6 9 p Z (9) = q 6 pdf of X: Each q i and q q 6 = 1. pdf of Z: Each probability is also, and the total sum is also 1. Prof. Tesler Expected values and variance Math 186 / January 27, / 23

8 Expected Value of a Function Let X be the value of a roll of a biased die and Z = (X 3) 2. x p X (x) z = (x 3) 2 p Z (z) 1 q q q 3 p Z () = q 3 4 q 4 1 p Z (1) = q 2 + q 4 5 q 5 4 p Z (4) = q 1 + q 5 6 q 6 9 p Z (9) = q 6 E(Z), in terms of values of Z and the pdf of Z, is E(Z) = z z p Z (z) = (q 3 ) + 1(q 2 + q 4 ) + 4(q 1 + q 5 ) + 9(q 6 ) Regroup it in terms of X: = 4q 1 + 1q 2 + q 3 + 1q 4 + 4q 5 + 9q 6 = 6 (x 3) 2 p X (x) x=1 Prof. Tesler Expected values and variance Math 186 / January 27, / 23

9 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)) = x g(x) p 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 Expected values and variance Math 186 / January 27, / 23

10 Expected Value of a Continuous distribution Probability density Probability density with 31 bins Consider the dartboard of radius 3 example, with pdf { 2r/9 if r 3; f R (r) = otherwise. Throw n darts and make a histogram with k bins. r 1, r 2,... are representative values of R in each bin. The bin width is r = 3/k, the height is f R (r i ), and the area is f R (r i ) r. The approximate number of darts in bin i is n f R (r i ) r. Prof. Tesler Expected values and variance Math 186 / January 27, / 23

11 Expected Value of a Continuous Distribution The estimated average radius is i r i n f R (r i ) r n = i r i f R (r i ) r As n, k, the histogram smoothes out and this becomes 3 r f R (r) dr Prof. Tesler Expected values and variance Math 186 / January 27, / 23

12 Mean of a continuous distribution Consider the dartboard of radius 3 example, with pdf { 2r/9 if r 3; f R (r) = otherwise. The average radius (technically the mean radius or expected value of R) is µ = E(R) = 3 = 2r3 27 r f R (r) dr = 3 = 2(33 3 ) 27 r 2r 9 dr = = 2 3 2r 2 9 dr Prof. Tesler Expected values and variance Math 186 / January 27, / 23

13 Expected Values Properties The gambling slide earlier had E(X 1) = E(X) 1. Theorem E(aX + b) = ae(x) + b where a, b are constants. Proof (discrete case). E(aX + b) = x (ax + b) p X (x) = a x x p X (x) + b x p X (x) = a E(X) + b 1 = ae(x) + b Prof. Tesler Expected values and variance Math 186 / January 27, / 23

14 Expected Values Properties The gambling slide earlier had E(X 1) = E(X) 1. Theorem E(aX + b) = ae(x) + b where a, b are constants. Proof (continuous case). E(aX + b) = = a (ax + b) f X (x) dx x f X (x) dx + b = a E(X) + b 1 = ae(x) + b f X (x) dx Prof. Tesler Expected values and variance Math 186 / January 27, / 23

15 Expected Values Properties These properties hold for both discrete and continuous random variables: E(aX + b) = a E(X) + b for any constants a, b. E(aX) = a E(X) E(b) = b ( ) E g(x) + h(x) = E(g(X)) + E(h(X)) Prof. Tesler Expected values and variance Math 186 / January 27, / 23

16 Variance These distributions both have mean=, but the right one is more spread out..1.1 pdf.5 pdf.5!2 2 x!2 2 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 Expected values and variance Math 186 / January 27, / 23

17 Variance formula σ 2 = E((X µ) 2 ) Consider the dartboard of radius { 3 example, with pdf 2r/9 if r 3; f R (r) = otherwise. µ = 2 from earlier slide. σ 2 = Var(R) = E((R µ) 2 ) = E((R 2) 2 ) = (r 2) 2 (r 2) 2 2r f R (r) dr = dr 9 3 2r 3 8r 2 ( ) + 8r r 4 = dr = r r2 3 ( ) = 18 8(33 ) (32 ) = Variance: σ 2 = 1 2 Standard deviation: σ = 1/2 3 Prof. Tesler Expected values and variance Math 186 / January 27, / 23

18 Variance Second formula There are two equivalent formulas to compute variance. In any problem, choose the easier one: σ 2 = Var(X) = E((X µ) 2 ) (Definition) = E(X 2 ) µ 2 (Sometimes easier to compute) Proof. Var(X) = E((X µ) 2 ) = E(X 2 2µX + µ 2 ) = E(X 2 ) 2µE(X) + µ 2 = E(X 2 ) 2µ 2 + µ 2 = E(X 2 ) µ 2 Prof. Tesler Expected values and variance Math 186 / January 27, / 23

19 Variance formula σ 2 = E(R 2 ) µ 2 Consider the dartboard of radius 3 example, with pdf { 2r/9 if r 3; f R (r) = otherwise. µ = E(R) = 2 E(R 2 ) = 3 r 2 2r 9 dr = 3 2r 3 9 dr = 2r = 2(81 ) 36 = 9/2 Variance: σ 2 = E(R 2 ) µ 2 = = 1 2 Standard deviation: σ = 1/2 Prof. Tesler Expected values and variance Math 186 / January 27, / 23

20 Variance Properties Var(aX + b) = a 2 Var(X) density pdf µ µ±σ density pdf µ µ±σ x y=2x+2 Adding b shifts the curve without changing the width, so b disappears on the right side of the formula. Multiplying by a dilates the width a factor of a, so variance goes up a factor a 2. For Y = ax + b, we have σ Y = a σ X. Example: Convert measurements in C to F: F = (9/5)C + 32 µ F = (9/5)µ C + 32 σ F = (9/5)σ C Prof. Tesler Expected values and variance Math 186 / January 27, / 23

21 Variance Properties Var(aX + b) = a 2 Var(X) density pdf µ µ±σ density pdf µ µ±σ x y=2x+2 Proof of Var(aX + b) = a 2 Var(X). E((aX + b) 2 ) = E(a 2 X 2 + 2ab X + b 2 ) = a 2 E(X 2 ) + 2ab E(X) + b 2 (E(aX + b)) 2 = (ae(x) + b) 2 = a 2 (E(X)) ( 2 + 2ab E(X) + b 2 Var(aX + b) = difference = a 2 E(X 2 ) (E(X)) 2) = a 2 Var(X) Prof. Tesler Expected values and variance Math 186 / January 27, / 23

22 Mean and Variance of the Binomial Distribution For the binomial distribution, Mean: µ = np Standard deviation: σ = np(1 p) At n = 1 and p = 3/4: µ = 1(3/4) = 75 σ = 1(3/4)(1/4) 4.33 pdf Binomial distribution µ µ±σ Binomial: n=1, p= x Approximately 68% of the probability is for X between µ ± σ. Approximately 95% of the probability is for X between µ ± 2σ. More on that in Chapter 4. Prof. Tesler Expected values and variance Math 186 / January 27, / 23

23 Mean of the Binomial Distribution Proof that µ = np for binomial distribution. E(X) = k k p X (k) = n k= k (n) k p k q n k Calculus Trick: Differentiate: Times p: Evaluate left side: (p + q) n = n ( n ) k= k p k q n k p (p + q)n = n k= k( n k ) p k 1 q n k p p (p + q)n = n k= k( n p p (p + q)n = p n(p + q) n 1 k) p k q n k = E(X) = p n 1 n 1 = np since p + q = 1. So E(X) = np. We ll do σ = np(1 p) later. Prof. Tesler Expected values and variance Math 186 / January 27, / 23