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#12 22S8 (Random Variables – 43) (Russo)

By Judy Reynolds,2014-12-09 16:14
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#12 22S8 (Random Variables – 43) (Russo)

     #12 22S:8 (Random Variables 4.3) (Russo)

(1) Let X be the result of one spin of the spinner shown. 22

    What is X ? Answer: We don't know until we do the experiment. X is called a random variable. We can list

    the possible values of X and the corresponding probabilities.

Possibility 1 2 5 draw the probability histogram ?

    Probability 1/2 1/3 1/6

    pdf of X: f(x) = 1/(x+1) x = 1, 2, 5 ___________________________________ ? 1 2 5

    probability density function

    E(X) = µ = sum of possibilities × probabilities = 1(1/2) + 2(1/3) + 5(1/6) = 2

     2E(X) =

     2Var(X) = E[ (X - µ)] =

SD(X) = sqrt Var(X) =

     22Important computational formula: Var(X) = E(X) - µ =

     2(2) Suppose the random varaible Y has pdf f(y) = y/55 y = 1, 2, 3, 4, 5

    E(Y) = ___________________________________

     1 2 3 4 5

Var(Y) =

P(Y > 2) = P(Y is odd valued) =

     2P(Y + 3 < 12) = P(Y = 1 or Y = 2) =

P( Y < 4 | Y > 1 ) =

#12a

     1/2(3) Suppose the random variable Q has pdf f(q) = C(q+1) q = 0, 3, 8, 24

Find the value of C.

     __________________________________

     0 3 8 24

     ?

    E(Q) = probability

     histogram

Var(Q) =

P(5Q + 3 is even valued) =

(4) Suppose T = 3X + 5 where X is the random variable from problem (1).

What are the possible values of T ?

Find the pdf of T.

    E(T) = How are E(T) and E(X) related ?

    Var(T) = How are Var(T) and Var(X) related ?

     How are SD(T) and SD(X) related ?

IMPORTANT RULES

     2E(aX + b) = aE(X) + b Var(aX + b) = a Var(X)

E(X + Y) = E(X) + E(Y) Var(X + Y) = Var(X) + Var(Y)

     ?

     if X and Y are independent

    #12b (application of the rules for expectation & variance)

Suppose X, Y & Z are independent random variables (representing the value of a

    single share of stock in three different companies one year from today), and that

E(X) = 12 E(Y) = 8 E(Z) = 10 Var(X) = 5 Var(Y) = 3 Var(Z) = 2

E(8X) = E(5X + 3Y - 2Z -1) =

     2E(X) =

     22 E(5X) = E(5X- 8) =

     22E(5X - 2Y + 3Z + 4) =

Var(5Y) = Var(4Y + 8) = Var(-Z) =

Suppose X, Y & Z are independent.

Var(X + Y) = Var(X + Y + 2) =

Var(3X + 4Y) = Var(3X + 4Y + 2Z -18) =

    Suppose X & Y are independent random variables with common mean µ

    2and common variance σ. Find the mean and variance of (X+Y)/2.

     E[(X+Y)/2] = E[?X + ?Y] = ?E(X) + ?E(Y) = ? µ + ? µ = µ

    2 22Var[(X+Y)/2] = Var[?X + ?Y] = ?Var(X) + ?Var(Y) = ? σ+ ? σ = σ/2

#12c

X = the number on the card drawn from the box 2 2 4 12 2E(X) = 5 E(X) = 42 Var(X) = 17

Y = the number on the card drawn from the box 3 9 2) = 45 Var(Y) = 9 E(Y) = 6 E(Y

    Consider the random variable W = 3X + 5. To observe this random variable, we can observe X, triple it, & add 5. Or, we can select from 11 11 17 41

We can calculate directly...

    E(W) = 11(1/2) + 17(1/4) + 41(1/4) = 20, or write... E(W) = E(3X + 5) = 3E(X) + 5 = 20

We can calculate directly... 22222E(W) = 11(1/2) + 17(1/4) + 41(1/4) = 553, & Var(W) = 553 20 = 153

     2or write... Var(W) = Var(3X + 5) = 3Var(X) = 9(17) = 153

    Consider the random variable T = X + Y. The easy way to find E(T) & Var(T)...

    E(T) = E(X + Y) = E(X) + E(Y) = 11 Var(X + Y) = Var(X) + Var(Y) = 26

Alternatively, we can find the pdf of T

    X 2 2 4 4 12 12

     3 9 3 9 Y 3 9

    probability ? ? 1/8 1/8 1/8 1/8

    T= X + Y 5 11 7 13 15 21

    E(T) = 5(?) + 11(?) + 7(1/8) + 13(1/8) + 15(1/8) + 21(1/8) = 11

     2222222E(T) = 5(?) + 11(?) + 7(1/8) + 13(1/8) + 15(1/8) + 21(1/8) = 147

so that

     2Var(T) = 147 11 = 26

    2Q: Var(X + W) = Var(X + 3X + 5) = Var(4X + 5) = 4Var(X) = 272

    but this is not the same as Var(X) + Var(W) = 17 + 153 = 170 HOW COME?

#12d expected life & expected residual life

    Suppose that an organism lives a whole number of years (from 1 to 10) with probabilities as indicated in the following table:

Lifetime 1 2 3 4 5 6 7 8 9 10

    probability .02 .04 .08 .10 .12 .18 .24 .15 .05 .02

Find the expected lifetime of the organism

    E(lifetime) = sum of possibilities x corresponding probabilities

     = 1(.02) + 2(.04) + 3(.08) + …. + 10(.02) = 5.95

Now suppose that the organism has lived to the age of 5.7

    Find his/her expected residual life (expected additional life beyond 5.7)

The answer is NOT expected lifetime current age = 5.95 5.7 = .25

    By this faulty reasoning, an organism that has lived to age 6.2 would have a negative expected residual life = 5.95 6.2 = -.25

    Soln: we need to find the conditional probabilities that the organism lives to age 6,7,8,9,10 given that it has lived to age 5.7 For example….

P(L6|L5.7)

P(L6andL5.7)P(L6).18.18.28P(L5.7)P(L5.7).18.24.15.05.02.64

    Similarly, divide .24, .15, .05 & .02 by P(L > 5.7) to get the other conditional probabilities (note, these probabilities add up to .999 because of round-off error)

Lifetime 6 7 8 9 10

    Cond. .281 .375 .234 .078 .031

    probability

Conditional expected lifetime = E(L | L > 5.7)

     = 6(.281) + 7(.375) + 8(.234) + 9(.078) + 10(.031) = 7.195

Expected residual life = 7.195 5.7 = 1.495 (not .25)

#13 (Joint distributions & Covariance) (Russo)

    THIS MATERIAL WILL NOT BE COVERERED ON THE FINAL The 16 equally likely outcomes when a fair coin is tossed 4 times

     stHHHH HHTT TTTH Let X = number of heads in 1 2 tosses HHHT HTHT TTHT

    HHTH HTTH THTT Let Y = total number of heads HTHH THHT HTTT

    THHH THTH TTTT

     TTHH Y

     0 1 2 3 4 The JOINT DISTRIBUTION of X and Y 0

     x 1

(1) Find the marginal pdfs of X & Y 2

(2) E(X) =

     E(Y) =

(3) Var(X) =

     Var(Y) =

    (4) P(X + 1 = Y) = (5) are X & Y independent?

Cov(X, Y) = E[ (X - E(X))(Y - E(Y)) ] = E(XY) - E(X)E(Y)

Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y)

ρ = Corr(X,Y) = Cov(X, Y)

    SD(X) SD(Y)

Suppose the random variables Q and W W

    have the joint pdf given in the table. 0 1 2

     1 .1 0 .2

    (5) E(Q) = E(W) = Q

     2 .2 .1 .1

    (6) Var(Q) = Var(W) =

     3 0 .3 0

(7) Var(Q+W) =

(8) P(Q > W) =

    #13a THIS MATERIAL WILL NOT BE COVERERED ON THE FINAL

     W

     0 1 2

    1 .1 0 .2 .3 marginal pdf of Q in red

     Q 2 .2 .1 .1 .4

     3 0 .3 0 .3 marginal pdf of W in blue

     .3 .4 .3

E(Q) = 1(.3) + 2(.4) + 3(.3) = 2 22222E(Q) = 1(.3) + 2(.4) + 3(.3) = 4.6 Var(Q) = 4.6 - 2 = .6

E(W) = 0(.3) + 1(.4) + 2(.3) = 1 22222E(W) = 0(.3) + 1(.4) + 2(.3) = 1.6 Var(Q) = 1.6 - 1 = .6

     W

     0 1 2

    1 0 1 2 E(QW) = 0(.1) + 1(0) + 2(.2)

     Q 2 0 2 4 + 0(.2) + 2(.1) + 4(.1)

     3 0 3 6 + 0(0) + 3(.3) + 6(0) = 1.9

Cov(Q,W) = E(QW) E(Q)E(W) = 1.9 (2)(1) = -.1

    Var(Q + W) = Var(Q) + Var(W) + 2Cov(Q,W) = .6 + .6 +2 (-.1) = 1

    Var(Q W) = Var(Q) + Var(W) - 2Cov(Q,W) = .6 + .6 -2 (-.1) = 1.4

    We can also find Var(Q + W) from the distribution of Q + W...

Q + W 1 2 3 4

    probability .1 .2 .3 .4 check that E(Q + W) = 3, Var(Q + W) = 1

Q - W -1 0 1 2

    probability .2 .1 .2 .5 check that E(Q - W) = 1, Var(Q - W) = 1.4

#14 22S:8 (Continuous r.v.’s) (Russo)

(1) Suppose X has density function f(x) = x/2 0 < x < 2 ______________

     0 2

P(X < 1) =

P(X > 1.2) = P(X > 1.2) =

P(.8 < X < 1.5) =

P( X > 1.2 | X > 1) =

P( X > .6 | X < 1.2) =

    (2) Y has density function f(t) = t/8 0 < t < 2

     ____________________

     1/4 2 < t < 5 0 2 5

P(Y < 3) =

P(Y > 3.2) =

P(1.4 < Y < 3.6)

P(Y > 1 | Y < 4) =

    (3) Suppose U is a U(0,60) random variable representing the number of minutes after 12 PM that a train arrives at Kendall Station.

P(arrival after 12:22 PM) =

    P(arrival after12:10PM | arrival between 12:05 & 12:25PM) =

    #14a 22S:8 (Continuous r.v.’s) (Russo)

     st(10 Find E(X) & Var(X) for the 1 random variable on worksheet 14.

    232~?xx4?()EXxdx(??0263)??0

    242~?xx?22 E(X)xdx2(??028)??0

    2222;;Var(X)E(X)[E(X)]24/39

    2Suppose X is a random variable with pdf f(x)3xfor0x1

     Draw the pdf

    12P(X.8)3xdx?.8

    .823xdx?P(.5X.8).5P(X.5|X.8).82P(X.8)3xdx?0

     E(X)

    2E(X)

    22Var(X)E(X)[E(X)]

#14b 22S:8 (The Poisson random variable) (Russo)

Consider a random variable X with the following pdf:

    (ke( (recall k! = k(k-1)(k-2)…(3)(2)(1), 0! = 1) P(Xk)fork0,1,2,...k!

     is a positive constant (a parameter). It can be shown that E(X) = Var(X) = . ((

    The Poisson variable is used to model the occurrences of events over intervals of time.

    Example 1 Suppose that the number of customers entering a clothing store over a one hour period is Poisson distributed with mean = 12. Find the probability that exactly 10 customers enter the store between 1PM & 2PM.

    121012e (10).1048PX10!

    Find the probability that no customers enter the store between 3PM & 4PM.

    120e1212 P(X0)e.0000060!

    Example 2 The number of earthquakes occurring in Umbria during a 10 year period is Poisson distributed with mean = 4. Find the probability that exactly 2 quakes occur in Umbria during the next three years.

    34The number E over the next three years is Poisson() =Poisson(1.2) distributed, so 10

    1.221.2e (2).2168PE2!

    Example 3 In (1), find the probability that exactly 12 customers enter the store between 1PM and 2:30PM.

    The number W entering the store over a 1.5 hour period has a Poisson(18) distribution, so

    181218e (12).0368PW12!

Example 4 In (2), find the probability that at least one quake occurs over a two year

    .80.8eperiod 1.55060!

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