3.6. Let s write out the sample space for this random experiment:

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1 STAT 5 3 Let s write out the sample space for this ranom experiment: S = {(, 2), (, 3), (, 4), (, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)} This sample space assumes the orering of the balls selecte oes not matter There are ( ) 5 N = = 0 2 outcomes (sample points) in S In this problem, we assume the outcomes are equally likely (a) Define Y = largest of the two sample numbers We can calculate the value of Y for each outcome ω S: Outcome ω S Y (ω) = y Probability P ({ω}) (, 2) Y ((, 2)) = 2 0 (, 3) Y ((, 3)) = 3 0 (, 4) Y ((, 4)) = 4 0 (, 5) Y ((, 5)) = 5 0 (2, 3) Y ((2, 3)) = 3 0 (2, 4) Y ((2, 4)) = 4 0 (2, 5) Y ((2, 5)) = 5 0 (3, 4) Y ((3, 4)) = 4 0 (3, 5) Y ((3, 5)) = 5 0 (4, 5) Y ((4, 5)) = 5 0 The pmf of Y is given by p Y (y) = P (Y = y) = P ({all ω S : Y (ω) = y}) = which can be epicte in the following table: P ({ω}), ω S Y (ω)=y y p Y (y) /0 2/0 3/0 4/0 It is easy to see that this is a vali pmf (b) Define Y = sum of the two sample numbers We can calculate the value of Y for each outcome ω S just like we i in part (a); see next page PAGE

2 STAT 5 The pmf of Y is given by Outcome ω S Y (ω) = y Probability P ({ω}) (, 2) Y ((, 2)) = 3 0 (, 3) Y ((, 3)) = 4 0 (, 4) Y ((, 4)) = 5 0 (, 5) Y ((, 5)) = 0 (2, 3) Y ((2, 3)) = 5 0 (2, 4) Y ((2, 4)) = 0 (2, 5) Y ((2, 5)) = 7 0 (3, 4) Y ((3, 4)) = 7 0 (3, 5) Y ((3, 5)) = 8 0 (4, 5) Y ((4, 5)) = 9 0 p Y (y) = P (Y = y) = P ({all ω S : Y (ω) = y}) = which can be epicte in the following table: P ({ω}), ω S Y (ω)=y y p Y (y) /0 /0 2/0 2/0 2/0 /0 /0 It is easy to see that this is a vali pmf 30 Define the ranom variable Y = number of ays between a pair of rentals Clearly, Y is a nonnegative ranom variable with support R = {y : y =, 2, 3, 4, } The one ay in five phrase means the probability of a rental on any ay is 02 (an hence the probability of there being no rental on any ay is 08) Inepenence means we multiply the each ay probabilities together Here are the probabilities associate with each value of y: A general formula for the pmf of Y is y p Y (y) = P (Y = y) 02 2 (08)02 3 (08)(08)02 4 (08)(08)(08)02 5 (08)(08)(08)(08)02 p Y (y) = P (Y = y) = (08) y (02), for y =, 2, 3, This is an example of a geometric probability istribution It is easy to check this is a vali pmf PAGE 2

3 STAT 5 32 The values of E(Y ), E(/Y ), an E(Y 2 ) are E(Y ) = ( ) E Y = E(Y 2 ) = yp Y (y) = (04) + 2(03) + 3(02) + 4(0) = 2 y= y= y p Y (y) = (04) + 2 (03) + 3 (02) + (0) (y 2 )p Y (y) y= = ( 2 )(04) + (2 2 )(03) + (3 2 )(02) + (4 2 )(0) = 4 Another way to calculate E(Y 2 ) is to first calculate the secon moment: Then observe that E(Y 2 ) = y 2 p Y (y) = 2 (04) (03) (02) (0) = 5 y= E(Y 2 ) = E(Y 2 ) = 5 = 4 Finally, we can get V (Y ) by using the variance computing formula V (Y ) = E(Y 2 ) [E(Y )] 2 = = 32 First we can calculate E(R 2 ), where R is the istance (in blocks) the fire engine can cover The pmf of R, p R (r), is given in the problem E(R 2 ) = 2 r=2 r 2 p R (r) = 2 2 (005)+22 2 (020)+23 2 (030)+24 2 (025)+25 2 (05)+2 2 (005) = 549 With C = 8, the expecte number of resiential homes that can be serve is or about 3,800 homes E(N) = E(8πR 2 ) = 8πE(R 2 ) = 8π(549) , 329 There is a typo in the problem; it shoul rea E(Y ) = P (Y k); k= ie, the inex of summation shoul be k, not i Because the support is R = {, 2, 3,, }, we can write out what the probabilities P (Y k) are; see next page PAGE 3

4 STAT 5 P (Y ) = P (Y = ) + P (Y = 2) + P (Y = 3) + P (Y = 4) + P (Y = 5) + P (Y = ) + P (Y 2) = P (Y = 2) + P (Y = 3) + P (Y = 4) + P (Y = 5) + P (Y = ) + P (Y 3) = P (Y = 3) + P (Y = 4) + P (Y = 5) + P (Y = ) + P (Y 4) = P (Y = 4) + P (Y = 5) + P (Y = ) + P (Y 5) = P (Y = 5) + P (Y = ) + P (Y ) = P (Y = ) + an the pattern continues Now, a own There is P (Y = ) term There are 2 P (Y = 2) terms There are 3 P (Y = 3) terms There are 4 P (Y = 4) terms There are 5 P (Y = 5) terms There are P (Y = ) terms, an so on P (Y k) = P (Y ) + P (Y 2) + P (Y 3) + P (Y 4) + P (Y 5) + P (Y ) + k= = P (Y = ) + 2P (Y = 2) + 3P (Y = 3) + 4P (Y = 4) + 5P (Y = 5) + P (Y = ) + = kp (Y = k), k= which is the efinition of E(Y ) when Y has support R = {, 2, 3,, } 333 Suppose Y is a iscrete ranom variable with pmf p Y (y) an support R (a) Using the efinition of mathematical expectation, E(aY + b) = (ay + b)p Y (y) = ayp Y (y) + bp Y (y) = a yp Y (y) + b p Y (y) However, p Y (y) = an yp Y (y) = E(Y ) This shows E(aY + b) = ae(y ) + b (b) Define X = ay + b We want to show V (X) = a 2 V (Y ) From part (a), we know E(X) = E(aY + b) = ae(y ) + b Also, an E(X 2 ) = E[(aY + b) 2 ] = E(a 2 Y 2 + 2abY + b 2 ) = a 2 E(Y 2 ) + 2abE(Y ) + b 2 [E(X)] 2 = [ae(y ) + b] 2 = a 2 [E(Y )] 2 + 2abE(Y ) + b 2 from the variance computing formula, we have V (X) = E(X 2 ) [E(X)] 2 = a 2 E(Y 2 ) a 2 [E(Y )] 2 = a 2 {E(Y 2 ) [E(Y )] 2 } = a 2 V (Y ) Note that in this part, we i not assume that Y was iscrete; in fact, this result hols for any ranom variable with finite variance PAGE 4

5 STAT (a) The first erivative of m Y (t) = [(/3)e t + (2/3)] 5 is t ( 3 et ) 5 = 5 E(Y ) = t m Y (t) ( 3 et ) 4 3 et = 5 3 et = 5 3 e0 ( 3 e0 + 2 ) 4 = ( 3 et + 2 ) 4 3 The secon erivative of m Y (t) = [(/3)e t + (2/3)] 5 is 2 ( t 2 3 et + 2 ) [ 5 = ( 5 3 t 3 et 3 et + 2 ) ] 4 3 E(Y 2 ) = 2 t 2 m Y (t) = 5 3 e0 From the variance computing formula, = 5 3 et ( 3 et ) ( 3 et 4 3 et + 2 ) et ( 3 e0 + 2 ) ( 3 3 e0 4 3 e0 + 2 ) e0 = = 35 9 V (Y ) = E(Y 2 ) [E(Y )] 2 = 35 9 ( 5 3 ) 2 = 0 9 (b) We calculate E(Y ) an V (Y ) with this mgf in the notes; see Example 33 (c) The first erivative of m Y (t) = exp[2(e t )] is t exp[2(et )] = 2e t exp[2(e t )] E(Y ) = t m Y (t) = 2e 0 exp[2(e 0 )] = 2 The secon erivative of m Y (t) = exp[2(e t )] is 2 t 2 exp[2(et )] = t 2et exp[2(e t )] = 2e t exp[2(e t )] + 2e t 2e t exp[2(e t )] E(Y 2 ) = 2 t 2 m Y (t) = 2e 0 exp[2(e 0 )] + 2e 0 2e 0 exp[2(e 0 )] = From the variance computing formula, V (Y ) = E(Y 2 ) [E(Y )] 2 = 2 2 = 2 Note: The mgf of Y in part (a) correspons to a b(n = 5, p = /3) istribution The mgf of Y in part (b) correspons to a geometric(p = /2) istribution The mgf of Y in part (c) correspons to a Poisson(λ = 2) istribution You ll see this later PAGE 5

6 STAT You shoul see that the mgf correspons to this pmf: m Y (t) = et + 2 e2t + 3 e3t y 2 3 p Y (y) / 2/ 3/ To see why, just apply the efinition of the mgf; ie, m Y (t) = E(e ty ) with this istribution this is your answer to part (c) To fin E(Y ) in part (a), we coul quickly calculate E(Y ) = 3 yp Y (y) = (/) + 2(2/) + 3(3/) = 4 y= Using the mgf shoul give us the same answer The first erivative of m Y (t) is ( t et + 2 e2t + 3 ) e3t = et + 4 e2t + 9 e3t E(Y ) = t m Y (t) = e0 + 4 e2(0) + 9 e3(0) = 4 To fin V (Y ) in part (b), we coul calculate irectly V (Y ) = 3 (y µ) 2 p Y (y) = y= = 3 y= ( y 4 ) 2 p Y (y) ( 4 ) 2 ( ) ( ) 2 ( ) ( ) 2 ( ) 3 = 5 9 using the pmf of Y above If we wante to the use the mgf, we coul first fin E(Y 2 ) The secon erivative of m Y (t) is 2 ( t 2 et + 2 e2t + 3 ) e3t = ( t et + 4 e2t + 9 ) e3t = et + 8 e2t + 27 e3t E(Y 2 ) = 2 t 2 m Y (t) = e0 + 8 e2(0) + 27 e3(0) = From the variance computing formula, V (Y ) = E(Y 2 ) [E(Y )] 2 = ( ) 4 2 = 5 9 PAGE

7 STAT We have a ranom variable Y with mgf m Y (t) Now, we consier the linear function The mgf of W is equal to W = ay + b m W (t) = E(e tw ) = E[e t(ay +b) ] = E(e aty +bt ) = E(e bt e aty ) = e bt E(e aty ) The last step is true because e bt is a constant so it factors outsie the expectation (all previous steps are just algebra) Now, what is E(e aty )? Let s = at Then, E(e aty ) = E(e sy ) = m Y (s) = m Y (at); ie, E(e aty ) is the mgf of Y, evaluate at at as claime m W (t) = e bt m Y (at), 32 Suppose Y is a ranom variable with mgf m Y (t) Consier the function r Y (t) = ln m Y (t) This is calle the cumulant generating function; it is use in higher level probability courses In this problem, we are aske to show t r Y (t) = E(Y ) 2 t 2 r Y (t) = V (Y ) Note that t r Y (t) = t ln m Y (t) = m Y (t) m Y (t), where m Y (t) = (/t)m Y (t) is the erivative of m Y (t) t r Y (t) = m Y (0) = E(Y ) m Y (0) Note that m Y (0) = E(e 0Y ) = E() = To show the secon part, take another erivative: 2 t 2 r Y (t) = 2 t 2 ln m Y (t) = [ m ] Y (t) t m Y (t) = m Y (t)m Y (t) m Y (t)m Y (t) [m Y (t)] 2, by using the quotient rule for erivatives Now, evaluate at t = 0, an you shoul see this reuces to E(Y 2 ) [E(Y )] 2 = V (Y ) because m Y (0) = E(Y 2 ), m Y (0) = E(Y ), an m Y (0) = PAGE 7