Section 3.7 Suppose the number of occurrences in a “unit” interval follows a Poisson distribution with mean. Recall that for w > 0, P(interval length to obtain the first occurrence  w) = 1 – P(interval length to obtain the first occurrence > w) = 1 – P(no occurrences in interval of length w) =. Consider a random variable W = interval length to obtain the first occurrence with distribution function G(w) = P(W  w) = 1 – e –H(w) for w > 0, where H(w) = (phone calls)(one hour) (10 phone calls per hour) 1 – e – w 0 w (t) dt.

If the function (t) is a constant, say, then H(w) =, so that the p.d.f. of W is g(w) = which is the p.d.f. for an distribution. A non-constant function (t) implies that the mean of the Poisson process changes over an interval. For instance, the number of occurrences in a “unit” interval may steadily increase (or decrease). We observe that in general, H / (w) =, so that the p.d.f. of W is g(w) = When W represents length of life (for an item or living organism), (w) is called the failure rate or force of mortality, and we find that (w) = If H(w) =for  > 0 and  > 0, we say W has a Weibull distribution. exponential() 1 — w e – w if w > 0, (phone calls per hour)(increase as time goes on) (w) H / (w)e –H(w) = (w)e –H(w) if w > 0. g(w) —— = e –H(w) g(w) ————. 1 – G(w) w—  w—   Consider (t) = 10 versus (t) = 5t / 2 when W = hours until first phone call

For W = hours until first phone call, compare P(W > 1/10) when (t) = 10 and (t) = 5t / 2. When (t) = 10, the p.d.f. of W is P(W > 1/10) = When (t) = 5t / 2, the p.d.f. of W is P(W > 1/10) = 10e –10w if w > 0. (5w / 2)e –(5/4)w if w > 0. 2 We can say that W has a Weibull distribution with  = and  = 11/10 2 2/  5 e –1 = e –1/80 =

1. H(w) =in order to demonstrate that W has a Weibull distribution. A random variable W has p.d.f. g(w) = H / (w)e –H(w) if w > 0, where H(w) = 243w 5/3. Find the values of  and  so that we may write w—  w—    =  = 5/31/27

2. (a) Find the distribution function for W. Suppose the random variable W represents length of life and has p.d.f. g(w) = H / (w)e –H(w) if w > 0, with an exponential force of mortality (w) =for a > 0 and b > 0. g(w) ———— = ae bw 1 – G(w) 0 w (t) dt = H(w) = 0 w ae bt dt = ae bw – a ——— b The distribution function for W is G(w) = if w  0 if 0 < w 0 1 – exp a – ae bw ——— b

(b)Find the p.d.f. for W (known as Gompertz law). if w > 0.The p.d.f. for W is g(w) = ae bw exp a – ae bw ——— b

Let X be a random variable with c.d.f. F(x). In general, for any value v, P(X = v) = P(X  v)  P(X < v) = P(X  v)  Lim P(X  x) = x   vx   v F(v)  Lim F(x). x   vx   v Example. Let X be a random variable with c.d.f.F(x) = 0 if x < 5 1/4 if 5  x < 6 (x  5) / 4 if 6  x < 9 1 if 9  x 5 1/4 0 1/2 3/ Suppose v is a value less than 5, say 4.8. P(X = v) = P(X = 4.8) = F(4.8)  Lim F(x) = x    0 = 0

Example. Let X be a random variable with c.d.f.F(x) = 0 if x < 5 1/4 if 5  x < 6 (x  5) / 4 if 6  x < 9 1 if 9  x 5 1/4 0 1/2 3/ Suppose v is 5. P(X = v) = P(X = 5) = F(5)  Lim F(x) = x   5 1/4  0 = 1/4 Suppose v is a value greater than 5 but less than 6, say 5.8. P(X = v) = P(X = 5.8) = F(5.8)  Lim F(x) = x   5.8 1/4  1/4 = 0

Example. Let X be a random variable with c.d.f.F(x) = 0 if x < 5 1/4 if 5  x < 6 (x  5) / 4 if 6  x < 9 1 if 9  x 5 1/4 0 1/2 3/ Suppose v is a value such that 6  v < 9, say 7.5. P(X = v) = P(X = 7.5) = F(7.5)  Lim F(x) = x   / 8  (7.5  5) / 4 = 0 Suppose v is a value greater than or equal to 9, say 9.8. P(X = v) = P(X = 9.8) = F(9.8)  Lim F(x) = x    1 = 0

Example. Let X be a random variable with c.d.f.F(x) = 0 if x < 5 1/4 if 5  x < 6 (x  5) / 4 if 6  x < 9 1 if 9  x 5 1/4 0 1/2 3/ Consequently, we see that if v  5, then P(X = v) = 0 but P(X = 5) = 1 / 41 / 4 We also see that P(6 < X  9) = P(X  9)  P(X  6) = F(9)  F(6) =1  1 / 4 = 3 / 4 We see then that the space of X is {5}  {x | 6 < x < 9}

We see then that the space of X is {5}  {x | 6 < x < 9} What kind of random variable X could have this space? A dart is thrown at a circular board of radius 9 inches. If the dart lands in the area between 3 and 9 inches from the center, a score of 6 is assigned; if the dart lands within 3 inches of the center, then a score of 9  x is assigned where x is the distance from the center. Blue Area Score = 9  x where x = inches from the center. Yellow Area Score = 6 The random variable X is not a discrete type random variable and is not a continuous type random variable; X is called a random variable of the mixed type.

Let F(x) be the distribution function for a random variable X, then (1) F(x) must be an function, (2) as x  – , F(x) , and (3) as x  + , F(x) . If there exists a discrete set of points at which F(x) is discontinuous with F / (x) = 0 at all other points, then X must be a type random variable whose space is the set of points of discontinuity. (The probability of observing a given value in the space of X is equal to the “length of the jump in F(x)” at the given value.) increasing0 1 discrete If F(x) is continuous everywhere, then X is a type random variable whose space is the set of points where F / (x) > 0 (i.e., the set of points where F(x) is strictly increasing). continuous If there exists a discrete set of points at which F(x) is discontinuous and also one or more intervals where F / (x) > 0, then X is called a random variable of the mixed type. Such a random variable is neither of the discrete type or the continuous type and has no p.m.f. or p.d.f. The space of X consists of the union of the set of points of discontinuity and the intervals where F / (x) > 0.

3. (a) For each distribution function, (1) sketch a graph, and (2) either find the corresponding p.m.f. or p.d.f., or state why no p.m.f. or p.d.f. can be found. 1 X 1 has distribution functionF 1 (x) = 1 – ——— if –  < x < . (1 + e x ) 0 1/2 1 X 1 is a continuous type random variable with p.d.f. f 1 (x) = F 1 (x) = e x ———if –  < x < . (1 + e x ) 2

(b) X 2 has distribution function F 2 (x) = 0if x < – 3 (x ) / 35if – 3  x < 2 1if 2  x 0– X 2 is a continuous type random variable with p.d.f. f 2 (x) = F 2 (x) = 3x 2 —if – 3  x < 2. 35

(c)X 3 has distribution functionF 3 (x) = 0if x < – 2 1/6if – 2  x < 0 1/3if 0  x < 2 2/3if 2  x < 4 11/12if 4  x < 6 1if 6  x 0– /6 1/3 2/3 4 11/12 6 X 3 is a discrete type random variable. The space of X is The p.m.f. of X is f 3 (x) = {– 2, 0, 2, 4, 6}. 1/6 if x = – 2 1/6 if x = 0 1/3 if x = 2 1/4 if x = 4 1/12 if x = 6

(d) X 4 has distribution function F 4 (x) = 0if x < – 2 (x + 2) / 12if – 2  x < 0 (2 + 3  x ) / 12if 0  x < 4 11 / 12if 4  x < 5 1if 5  x 0– /6 2/3 4 11/12 5 X 4 has a distribution of the mixed type. X 4 has no p.m.f. or p.d.f. The space of X 4 is {x | – 2 < x < 4}  {4,5}.

(e)X 5 has distribution function F 5 (x) = 0if x < – 2 (x + 2) / 12if – 2  x < 0 (5 + 3  x ) / 12if 0  x < 4 11 / 12if 4  x < 5 1if 5  x 0– /6 5/ /12 5 X 5 has a distribution of the mixed type. X 5 has no p.m.f. or p.d.f. The space of X 5 is {x | – 2 < x < 0}  {0}  {x | 0 < x < 4}  {5}.

4.An ordinary, fair, six-sided die is rolled, and each of the two spinners displayed on the right are spun, with the random variables Y 1 and Y 2 respectively being set equal to the numbers selected from Spinner #1 and Spinner #2. Note that Y 1 and Y 2 have respective U(  2,2) and U(3,5) distributions. The random variable X is defined as follows:  2  Spinner # Spinner #2 Y 1 if the die roll results in 1 2if the die roll results in 2 Y 2 if the die roll results in 3, 4, or 5 6if the die roll results in 6 X = (a) Find the distribution function of X.

The space of X is {x | – 2 < x < 2}  {2}  {x | 3 < x < 5}  {6} P(X  x) = F(x) = 0 if x < – 2 In order for – 2 < X < 2, the die roll must result in 1, so that X = Y 1. For – 2 < x < 2, P(X  x) = P({die roll results in 1}  {Y 1  x}) = P(die roll results in 1) P(Y 1  x) = (1 / 6)(2 + x) / 4 P(X  x) = F(x) = 2 + x ——— if – 2  x < 2 24 In order for X = 2, the die roll must result in 2, and P(X = 2) = We see then that P(X  x) = for 2  x < 3. 1 / 6. 1 / 31 / 3

F(x) = 0 if x < – x ——— if – 2  x < — if 2  x < 3 3 In order for 3 < X < 5, the die roll must result in 3, 4, or 5, so that X = Y 2. For 3 < x < 5, P(X  x) = P({X < 3}  {3  X  x}) = P(X < 3) + P(3  X  x) = 1 / 3 + P(3  X  x) = 1 / 3 + P({die roll results in 3, 4, or 5}  {Y 2  x}) = 1 / 3 + P(die roll results in 3, 4, or 5) P(Y 2  x) = 1 / 3 + (1/2) (x  3) / 2 3x  5 ——— if 3  x < 5 12

1 if 6  x F(x) = 0 if x < – x ——— if – 2  x < — if 2  x < 3 3 3x  5 ——— if 3  x < — if 5  x < 6 6 It is not possible that 5 < X < 6, and in order that X = 6, the die roll must result in 6, and P(X = 6) = We see then that P(X  x) = for 5  x < 6. 1 / 6. 5 / 65 / 6

(b)Graph the distribution function of X. 1 if 6  x F(x) = 0 if x < – x ——— if – 2  x < — if 2  x < 3 3 3x  5 ——— if 3  x < — if 5  x < 6 6 0– 224 1/3 5/ /6

When X is a random variable of the mixed type, the expected value of a function u(X) is calculated by applying the following rules: (1) (2) (3) Multiply each value u(x) corresponding to the discrete list of values for x at which F(x) is discontinuous by its probability (i.e., by the “length of the jump in F(x)” at x). Integrate u(x)F / (x) over each interval where F / (x) > 0. Sum the results obtained from (1) and (2).

(c)Find E(X). 1 if 6  x F(x) = 0 if x < – x ——— if – 2  x < — if 2  x < 3 3 3x  5 ——— if 3  x < — if 5  x < 6 6 E(X) = 1 (2) — (6) — – 2 1 x — dx x — dx = 4 1 — = 10 — 3

5. (a) For each random variable, graph the distribution function, and find the expected value. The random variable W = "the length of time in hours a brand W light bulb will burn" has distribution function F(w) = 0 if w < 0 1 – e –w/300 if 0  w 0 1 F(w)F(w) Note that W is a continuous type random variable with p.d.f. f(w) = F (w) = e –w/300 —— if 0 < w <  300(an exponential(300) p.d.f.). E(W) =300

(b)The random variable X = "the length of time in hours a brand X light bulb will burn" has distribution function G(x) = 0 if x < 0 1 – (9/10)e –x/300 if 0  x 0 1 G(x)G(x) 1/10 X has a distribution of the mixed type. X has no p.m.f. or p.d.f. E(X) = 1 (0) — + 10  0 9e –x/300 x ——— dx = e –x/300 —x —— dx =  9 — (300) =

(c)The random variable Y = "the length of time in hours a brand Y light bulb will burn" has distribution function H(y) = 0 if y < 0 1 – (9/10)e –y/300 if 0  y < if 120  y 0 1 H(y)H(y) 1/10 1 – 0.9e –2/5 120 Y has a distribution of the mixed type. Y has no p.m.f. or p.d.f.

E(Y) = 1 (0) — e –y/300 y ——— dy (120) (0.9e –2/5 ) = – 3( y) e –y/300 ———————— + 10 y = e –2/5 = – 3( ) e –2/ ———————— + —— e –2/5 = – 270 e –2/5