Expectation MATH Expectation. Benjamin V.C. Collins, James A. Swenson MATH 2730

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1 MATH 2730 Expectation Benjamin V.C. Collins James A. Swenson

2 Average value Expectation Definition If (S, P) is a sample space, then any function with domain S is called a random variable. Idea Pick a real-valued random variable X : S R. If we repeat our experiment many times, what would we expect to be the average value of X over all trials?

3 Expectation of a paradise Example Let X denote the sum shown on two fair dice. Roll the dice 36 times, recording X each time. Estimate the most likely average value of X. Solution. Let D = {1, 2, 3, 4, 5, 6}, S = D D, and P(a, b) = 1 36 for all (a, b) S. Let X : S R be defined by X (a, b) = a + b. Out of 36 rolls, we expect X = 2 once, X = 3 twice, etc. So our expected average value is = = 7.

4 Generalizing from the example Summary Let a S. If N is large, we expect a to occur about N P(a) times, which adds N P(a) X (a) to our sum. Thus the average value is expected to be 1 NP(a)X (a), or just P(a)X (a). N a S a S In the example, we grouped together outcomes where X has the same value. In general, for any t R, we expect X to equal t about N P(X = t) times, which adds N P(X = t) t to our sum. Thus the average value is expected to be t R tp(x = t).

5 Formulæ Expectation Definition Let (S, P) be a sample space and X : S R a random variable. The expectation (or expected value, or mean) of X is µ = E(X ) = a S P(a)X (a). Proposition If (S, P) is a sample space and X : S R is a random variable, then E(X ) = tp(x = t). t R

6 Face value of a card Expectation Example Let (S, P) model the draw of one card from a 52-card deck, and let Z : S N give the face value of the card. The expectation of Z is E(Z) = t R = 10 t=1 tp(x = t) tp(x = t) = P(X = 1) + 2P(X = 2) + + 9P(X = 9) + 10P(X = 10) = = 1 85 ( ) =

7 Center of mass Expectation Example Suppose we have n numbered weights on a seesaw. Let S = {1,..., n}, set P(k) = 1 n for all k S, and define X : S R so that X (k) is the position of the k th weight, treating the seesaw as a number line. If the system balances, where is the fulcrum? k X (k)

8 Center of mass Expectation Solution. Fix units so that each weight has mass P(k). Let c R be the location of the fulcrum. The moment of the k th weight is P(k)[X (k) c], the mass times the signed distance from the fulcrum. The system balances when its net moment is zero: P(k)[X (k) c] = 0 k S [P(k)X (k) cp(k)] = 0 k S P(k)X (k) k S k S cp(k) = 0 P(k)X (k) = c k S k S E(X ) = c P(k)

9 Center of mass Expectation Example In our example, c = E(X ) = 2 8 ( 2) (1) (3) (7) = = = 3 2. k X (k)

10 Center of mass Expectation Example In our example, c = E(X ) = 2 8 ( 2) (1) (3) (7) = = = 3 2. Even if not all outcomes are equally likely, we can picture E(X ) as the balancing point of a seesaw with mass P(X = t) at each t R. We don t even have to change our proof!

11 Working with multiple random variables Proposition If (S, P) is a sample space and X : S R and Y : S R are random variables, then there is a random variable X + Y defined by (X + Y )(a) = X (a) + Y (a), and its expectation is Proof. E(X + Y ) = E(X ) + E(Y ). E(X + Y ) = a S P(a)[X (a) + Y (a)] = (a) + P(a)Y (a)] a S[P(a)X = P(a)X (a) + P(a)Y (a) a S a S = E(X ) + E(Y ).

12 Working with multiples of random variables Proposition If (S, P) is a sample space, X : S R is a random variable, and c R, then there is a random variable cx defined by (cx )(a) = c X (a), and its expectation is E(cX ) = ce(x ). Proof. E(cX ) = a S P(a)[cX (a)] = c a S P(a)X (a) = ce(x ).

13 Working with multiples of multiple random variables Theorem (expectation is linear) If (S, P) is a sample space, X : S R and Y : S R are random variables, and c 1 and c 2 are real numbers, then E(c 1 X + c 2 Y ) = c 1 E(X ) + c 2 E(Y ). Proof. E(c 1 X + c 2 Y ) = E(c 1 X ) + E(c 2 Y ) = c 1 E(X ) + c 2 E(Y )

14 Working with products of random variables Proposition If (S, P) is a sample space and X : S R and Y : S R are random variables, then there is a random variable XY defined by (XY )(a) = X (a)y (a). The expectation of XY is E(XY ) = E(X )E(Y ) if X and Y are independent random variables.

15 Working with products of random variables Proof. E(XY ) = t R = t t R tp(xy = t) = t R = = u R (u,v) R R (u,v) R R:uv=t (u,v) R R:uv=t P(X = u and Y = v) uvp(x = u)p(y = v) uvp(x = u)p(y = v) uvp(x = u)p(y = v) = E(X )E(Y ). v R

16 Pay attention to details! Expectation Warning This is not an theorem! It is possible that E(XY ) = E(X )E(Y ) even when X and Y are not independent!

17 Spread Expectation

18 What is the variance? Definition Let (S, P) be a sample space and X : S R a random variable; let µ = E(X ). The variance of X is σ 2 = Var(X ) = E[(X µ) 2 ]. [The number σ = Var(X ) is the standard deviation of X.] Idea Qualitatively: Var(X ) is large if X is often far from µ; Var(X ) is small if X is usually close to µ. That is, Var(X ) is a measure of spread in the values of X.

19 Rolling one die Example Roll a fair die; let X denote the number rolled. The expectation of X is E(X ) = t R tp(x = t) = 6 t/6 = 1 6 t=1 6 t = 1 ( ) 7 = = 7 2. t=1

20 Rolling one die Expectation Example The variance of X is Var(X ) = E [ (X µ) 2] = E = t R = 1 6 [ ( X 7 ) ] 2 2 ( t 7 2 P(t) 2) [ ( 1 7 ) 2 ( ) ] = = =

21 An easier formula Proposition If X is a real-valued random variable, then Var(X ) = E [ X 2] E[X ] 2. Proof. Var(X ) = E[(X µ) 2 ] = E(X 2 2µX + µ 2 1) = E(X 2 ) 2µE(X ) + µ 2 E(1) = E(X 2 ) 2E(X ) 2 + E(X ) 2 (1) = E(X 2 ) E(X ) 2.

22 Rolling one die Expectation Example Roll a fair die; let X denote the number rolled. The variance of X is Var(X ) = E(X 2 ) E(X ) 2 [ ] = t 2 P(X = t) t R [ 6 ] = 1 t t=1 ( ) = 1 6 6(7)(13) = = =

23 Gardening Expectation Example Given: the probability that a beet seed germinates is 0.8. I m planting a seed; let X = 1 if the seed germinates and X = 0 otherwise. [X is called a zero-one random variable, or an indicator random variable.] Find the expectation and variance of X. Remark My seed is a Bernoulli trial with p = 0.8.

24 Gardening Solution. E(X ) = k R kp(x = k) = P(X = 1) = p = 0.8. Var(x) = E(X 2 ) E(X ) 2 = E(X ) E(X ) 2 = p p 2 = p(1 p) = pq = = 0.16.

25 Moral of the example Summary If X is a zero-one random variable modeling a Bernoulli trial, then E(X ) = p; Var(X ) = pq = p(1 p).

26 Gardening Expectation Example Given: the probability that a beet seed germinates is 0.8. I will plant 9 beet seeds: let X denote the number of seeds that germinate. Find E(X ) and Var(X ). Remark Assuming that the seeds germinate independently, my garden is a binomial experiment with p = 0.8 and n = 9. Hence P(X = k) = ( ) n k p k q n k = ( 9 k) (0.8) k (0.2) 9 k.

27 Gardening Expectation Solution. E(X ) = = = 9 k=0 9 k=1 8 j=0 = 9(0.8) ( 9 k (0.8) k) k (0.2) 9 k k 9! k(k 1)!(9 k)! (0.8)k (0.2) 9 k 9! (j)!(9 (j + 1))! (0.8)j+1 (0.2) 9 (j+1) (j = k 1) 8 j=0 8 = 9(0.8) j=0 8! (j)!(8 j)! (0.8)j (0.2) 8 j ( 8 j ) (0.8) j (0.2) 8 j = 9(0.8)[ ] 8 = 9(0.8) = 7.2.

28 Gardening Remark By the same proof, the expectation of any binomial random variable is E(X ) = np. But this was too much work! Instead, let X i = 1 if the i th seed germinates and X i = 0 otherwise. Then: X = X 1 + X X 9 E(X ) = E(X 1 + X X 9 ) = E(X 1 ) + E(X 2 ) + + E(X 9 ) (linearity) = p + p + + p (Bernoulli example) = np = 9(0.8) = 7.2.

29 Gardening Expectation We can try to find Var(X ) in the same way, but we don t have a formula for Var(X + Y ). So... Lemma If X : S R and Y : S R are independent random variables on (S, P), then Var(X + Y ) = Var(X ) + Var(Y ).

30 Gardening Proof of lemma. Var(X + Y ) = E[(X + Y ) 2 ] E[X + Y ] 2 = E(X 2 + 2XY + Y 2 ) [E(X ) + E(Y )] 2 = E(X 2 ) + 2E(XY ) + E(Y 2 ) [E(X ) 2 + 2E(X )E(Y ) + E(Y ) 2 ] = E(X 2 ) E(X ) 2 + E(Y 2 ) E(Y ) 2 +2 [E(XY ) E(X )E(Y )] = Var(X ) + Var(Y ).

31 Gardening Solution to the beet problem. Let X i = 1 if the i th seed germinates and X i = 0 otherwise. We have assumed that these are independent variables, so: X = X 1 + X X 9 Var(X ) = Var(X 1 + X X 9 ) = Var(X 1 ) + Var(X 2 ) + + Var(X 9 ) (lemma) = pq + pq + + pq (Bernoulli example) = npq = 9(0.8)(0.2) = 1.44.

32 Moral of the example Summary If X is a binomial random variable, then E(X ) = np; Var(X ) = npq = np(1 p).