Topic 3: The Expectation of a Random Variable
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1 Topic 3: The Expectation of a Random Variable Course 003, 2017 Page 0
2 Expectation of a discrete random variable Definition (Expectation of a discrete r.v.): The expected value (also called the expectation or mean) of a discrete random variable whose distinct possible values are x 1, x 2,... is defined by E(X) = x j P(X = x j ) j=1 The expectation is simply a weighted average of possible outcomes of X, weighted by their probabilities. We can think of the expectation as a point of balance: if weights are placed on a weightless rod, where should a fulcrum be placed so that the rod balances? E(X) need not occur with positive probability. For example, consider the experiment of rolling a fair die and X = number of dots. E(X) = 6 x=1 x 6 I {1,2...6} (x) = 3.5 If X can take only a finite number of different values, this expectation always exists. With an infinite sequence of possible values, it will exist if high values of X have low probability weights. Page 1
3 Expectations: some useful results Result 1: If X and Y are r.v. s with the same distribution, then (if either side exists) E(X) = E(Y) The proof follows immediately from the definition of E(X). Note that the converse of the above statement is NOT true. This is a common mistake. Result 2: For any r.v. X and a constant c E(cX) = ce(x) Result 3 (linearity of expectation): For any r.v.s X, Y, E(X + Y) = E(X) + E(Y) Note: We do not require independence of these two r.v.s. It is instructive to consider the experiment of tossing two fair coins. Let X and Y be indicator variables for at least one head and at least one tail. Verify linearity for this case by constructing the distribution of X + Y. Let the events {s} denote the smallest events in the sample space over which X and Y are defined. Then E(X) + E(Y) = s X(s)P({s}) + s Y(s)P({s}) = s (X + Y)(s)P({s}) = E(X + Y) Page 2
4 The expectations of Bernoulli and Binomial r.v.s Bernoulli: Using the definition of the expectation, this is 1p + 0q = p Binomial: A binomial r.v. can be expressed as the sum of n Bernoulli r.v.s: X = I 1 + I I n. By linearity, E(X) = E(I 1 ) + E(I 2 ) +... E(I n ) = np. Page 3
5 The Geometric distribution Definition (Geometric distribution): Consider a sequence of independent Bernoulli trials, all with success probability p (0, 1). Let X be the number of failures before the first successful trial. Then X Geom(p) The PMF is given by P(X = k) = q k p for k = 0, 1, 2,... where q = 1 p. This is a valid PMF since the sum p q k = p 1 q = 1 The expectation of a geometric r.v. is defined as E(X) = kq k p We know that q k = 1 1 q. Differentiating both sides w.r.t. q, we get multiplying both sides by pq, we get 1 E(X) = pq (1 q) 2 = q p kq k 1 = 1 (1 q) 2. Now Page 4
6 The Negative Binomial distribution Definition (Negative Binomial distribution): Consider a sequence of independent Bernoulli trials, all with success probability p (0, 1). Let X be the number of failures before the rth success. Then X NBin(r, p) The PMF is given by for k = 0, 1, 2,... where q = 1 p. ( ) n + r 1 P(X = n) = p r q n r 1 We can write the Negative Binomial as a sum of r Geom(p) r.v.s: X = X 1 + X 2 + X r The expectation of a Negative Binomial r.v. is therefore E(X) = r q p Page 5
7 Expectation of a function of X Result: If X is a discrete random variable and g is a function from R to R, then E(g(X)) = x g(x)p(x = x) where the sum is taken over all possible values of X. Intuition: Whenever X = x, g(x) = g(x), so we can assign p(x) to g(x). An easy but illustrative example is g(x) = X 3 This is a very useful result, because it tells us we don t need the PMF of g(x) to find its expected value, and we are often interested in functions of random variables. Examples: expected revenues from the distribution of yields, earnings from a chance game... Page 6
8 Variance of a random variable Definition (variance and standard deviation): The variance of and r.v. X is Var(X) = E[(X EX) 2 ] The square root of the variance is called the standard deviation of X. Notice that by linearity we have E(X EX) = 0 SD(X) = Var(X) We can expand the above expression to rewrite the variance in a form that is often more convenient: Var(X) = E(X 2 ) (EX) 2 Page 7
9 Variance properties 1. Var(X) 0 with equality if and only if P(X = a) = 1 for some constant a. 2. Var(aX + b) = a 2 Var(X) for any constants a and b. It follows that Var(X) = Var( X) Proof: Var(aX + b) = E[(aX + b (aex + b)) 2 ] = E[(a(X EX)) 2 ] = a 2 E[(X EX) 2 ] = a 2 Var(X) 3. Var(X X n ) = Var(X 1 ) + + Var(X n ) for independent random variables X 1,..., X n. Proof: Denote EX by µ. For n = 2, E(X 1 + X 2 ) = µ 1 + µ 2 and therefore Var(X 1 + X 2 ) = E[(X 1 + X 2 µ 1 µ 2 ) 2 ] = E[(X 1 µ 1 ) 2 + (X 2 µ 2 ) 2 + 2(X 1 µ 1 )(X 2 µ 2 )] Taking expectations, we get E[(X 1 µ 1 ) 2 + (X 2 µ 2 ) 2 + 2(X 1 µ 1 )(X 2 µ 2 )] = Var(X 1 ) + Var(X 2 ) + 2E[(X 1 µ 1 )(X 2 µ 2 )] But since X 1 and X 2 are independent, E[(X 1 µ 1 )(X 2 µ 2 )] = E(X 1 µ 1 )E(X 2 µ 2 ) = (µ 1 µ 1 )(µ 2 µ 2 ) = 0 It therefore follows that Var(X 1 + X 2 ) = Var(X 1 ) + Var(X 2 ) Using an induction argument, this can be established for any n Note that this result is only true with independent random variables Page 8
10 The Poisson distribution Definition (Poisson distribution): An r.v. X has the Poisson distribution with parameter λ > 0 if the PMF of X is P(X = k) = e λ λ k for k = 0, 1, 2,.... We write this as X Pois(λ). This is a valid PMF because the Taylor series 1 + λ + λ2 f(k) = e λ λ k = e λ e λ = 1. k As λ gets larger, the PMF becomes more bell-shaped. 2! + λ3 3! +... converges to e λ so Page 9
11 The Poisson mean and variance The expectation E(X) = e λ k λk = e λ k=1 k λk = λe λ The variance is also λ. To see this let s start with finding E(X 2 ): Since e λ λ k = 1, since this vanishes): Differentiate again to get k=1 λ k 1 (k 1)! = λe λ e λ = λ λ k = e λ. Now differentiate both sides w.r.t λ (and ignore the k = 0 term k=1 k λk 1 = e λ or k=1 k λk = λeλ k=1 k 2 λk 1 = e λ + λe λ or k=1 k 2 λk = eλ λ(1 + λ) So E(X 2 ) = e λ k=1 k 2 λk = λ(1 + λ) and Var(X) = λ(1 + λ) λ 2 = λ We ll see later that both these can be derived easily from the moment generating function of the Poisson. Page 10
12 Poisson additivity Result: If X Pois(λ 1 ) and Y Pois(λ 2 ) are independent, then X + Y Pois(λ 1 + λ 2 ) Proof: To get the PMF of X + Y, we use the law of total probability: P(X + Y = k) = = = = = k P(X + Y = k X = j)p(x = j) j=0 k P(Y = k j)p(x = j) j=0 k j=0 e λ 2 λ k j 2 (k j)! e (λ 1+λ 2 ) k j=0 e λ 1 λ j 1 j! ( ) k λ j 1 j λk j 2 e (λ 1+λ 2 ) (λ 1 + λ 2 ) k (using the binomial theorem) Page 11
13 The Poisson as a limiting distribution We can show that a binomial distribution with large n and small p can be approximated by a Poisson, which is computationally much easier. Useful result: e v = lim n (1 + v n )n We can rewrite the binomial density for non-zero values as f(x; n, p) = x (n i+1) i=1 x! p x (1 p) n x. If np = λ, we can subsitute for p by ( λn ) to get lim n f(x; n, p) = lim n = e λ λ x x! = lim n x (n i + 1) i=1 x! x (n i + 1) i=1 n x [ n (n 1) = lim n..... n n ( λ ) x ( 1 λ ) n x n n λ x x! ( 1 λ ) n ( 1 λ ) x n n (n x + 1) n λ x ( 1 λ ) n ( 1 λ ) x ] x! n n (using the above result and the property that the limit of a product is the product of the limits) Page 12
14 A Poisson example We have a 300 page novel with 1, 500 letters on each page. Typing errors are as likely to occur for one letter as for another, and the probability of such an error is given by p = The total number of letters n = (300) (1500) = 450, 000 Using λ = np, the poisson distribution function gives us the probability of the number of errors being less than or equal to 10 as: P(x 10) 10 x=0 e 4.5 (4.5) x x! =.9933 Rules of Thumb: close to binomial probabilities when n 20 and p.05, excellent when n 100 and np 10. Page 13
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