2004-2005 2005.5.11

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When a large collection of numbers is assembled, as in a census, we are usually interested not in the individual numbers, but rather in certain descriptive quantities such as the average or the median. In general, the same is true for the probability distribution of a numerically-valued random variable. In this and in the next section, we shall discuss two such descriptive quantities: the expected value and the variance.

()Expected Value or Expectation or Mean or Average Number

Expected Value of Discrete Random Variables

, , , ., , , . 7 , .

, . , ?

12 , 4 5 , 3 6 , 5 7 , .12 ,

:4/12 , 3/ 12 5/12567 .

12 . X (: ) , 5, 6, 7. X

1 X: P(X=Xk)=pk , k=1,2,,.

Example 2 Let an experiment consist of tossing a fair coin three times.HHHHHTHTHTHHTTHTHTHTTTTTTTTTTH,THT,HTTHHT,HTH,THHHHH1/83/83/81/8

Example 1 Let an experiment consist of tossing a fair coin three times. HHH,HHT,HTH,THH,TTH,THT,HTT,TTTLet X denote the number of heads which appear. Then the possible values of X are 0, 1, 2 and 3. The corresponding probabilities are 1/8, 3/8, 3/8, and 1/8. Thus, the expected value of X equals

(n, p)E(X)=npProof:B(3,1/2)E(X)=3/2

Exercise In Las Vegas, a roulette wheel has 38 slots numbered 0, 00, 1, 2, . . . , 36. The 0 and 00 slots are green, and half of the remaining 36 slots are red and half are black. A croupier spins the wheel and throws an ivory ball. If you bet 1 dollar on red, you win 1 dollar if the ball stops in a red slot, and otherwise you lose a dollar. We wish to calculate the expected value of your winnings, if you bet 1 dollar on red.

Let X be the random variable which denotes your winnings in a 1 dollar bet on red in Las Vegas roulette. Then the distribution of X is given by

, 1 , 2 , , 6 , 7/2 ? X, EX ? 100 1007/2 = 350

3 n. .: X,P(X=k)= 1/n , k=1,2,,n

Xf (x),x0

[Xi, Xi+1) xixi+1, [xi, xi+1)xi.. r.v

.2 X f (x),,X,.

X~U(a,b),X( a,b),

, X N(0,1),E(X)=0

1.68.

The time between arrivals is an exponentially distributed random variable X with density functionFind the expected value of the time between arrivals.

1. CE(C)=C; 4. XY E(XY)=E(X)E(Y); 2. kE(kX)=kE(X); 3. E(X1+X2) = E(X1)+E(X2);Xi:E(XY)=E(X)E(Y)X,Y

6 9,,0.8.3? ii 9= 1+ 2+ 3+ 9E=E(1+ 2+ 3+ +9) =E1+ E2+ E3++E9 =9 E1

S: success F: Failure P(=1) =P(S)=0.8P(=2)= P(FS)=0.20.8P(=3) =P(FFS+FFF) =P(FFS)+P(FFF) = 0.20.20.8+ 0.20.20.2=0.2 0.2

A coin is tossed twice. Xi = 1 if the ith toss is heads and 0 otherwise. We know that X1 and X2 are independent. They each have expected value 1/2.Thus E(X1 X2) = E(X1)E(X2) = (1/2)(1/2) = 1/4.

A,BP(AB)=P(A)P(B)A,B .

We flip a coin and let X have the value 1 if the coin comes up heads and 0 if the coin comes up tails. Then, we roll a die and let Y denote the face that comes up.It reasonable to assume that X and Y are independent.

What does X + Y mean, and what is its distribution? This question is easily answered in this case, by considering the joint random variable Z = (X, Y ), whose outcomes are ordered pairs of the form (x, y)

r.v .

(X,Y)r.v

(X,Y)r.v

1 X~B(n,p)Xn .X .

npXnp. X~B(n,p), X= X1+X2++Xn= npi=1,2n P(Xi =1)= p,P(Xi =0)= 1-pXn .

10000 . : 120 , , 10000 . 0.006, ().

X 10000 , X n =10000; p = 0.006 ,

X , X , (120-X)

10000 . : 120 , , 10000 . 0.006, .X 10000 , X ~ B(10000; 0.006), EX = 10000*0.006 = 60. , 60 . , 120-60 60 .

1. XXXg(X).

? g(X)X. g(X)E[g(X)]. g(X) .

g(X)XE[g(X)].

E(X): XY=g(X) X,P(X= xk)=pk ; X,Xf(x).

: E[g(X)], g(X)X. .

g(X): k .

0.2510502

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, X, X~U(2000,4000): . 1,311

s2000

X

EX. , EY.