2. This problem is very easy if you remember the exponential cdf; i.e., 0, y 0 F Y (y) = = ln(0.5) = ln = ln 2 = φ 0.5 = β ln 2.

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1 FALL 8 FINAL EXAM SOLUTIONS. Use the basic counting rule. There are n = number of was to choose 5 white balls from 7 = n = number of was to choose ellow ball from 5 = ( ) 7 5 ( ) 5. there are n n = ( )( ) = 34 5 = possible was to select 6 numbers in the lotter. There is onl wa to select all numbers correctl. Assuming all possible was are equall likel, the probabilit of winning is This problem is ver eas if ou remember the exponential cdf; i.e., {, e /β, >. To find the median φ.5, we set F Y (φ.5 ) = P (Y φ.5 ) =.5 and solve for φ.5. We have.5 set = F Y (φ.5 ) = e φ.5/β = e φ.5/β =.5 = φ.5 β ( ) = ln(.5) = ln = ln = φ.5 = β ln. Note: If ou did not remember the exponential cdf, then ou can integrate the pdf. We have.5 set = F Y (φ.5 ) = P (Y φ.5 ) = = β φ.5 Then solve for φ.5 in the same wa as above. β e /β d ( βe /β) φ.5 = (e /β) φ.5 = e φ.5/β. 3. Write m Y (t) = t = ( t ). This makes it easier to take derivatives. The first derivative of m Y (t) is d dt m Y (t) = ( )( t ) ( t) = t( t ). E(Y ) = d dt m Y (t) = ()( ) =. t= PAGE

2 FALL 8 FINAL EXAM SOLUTIONS The second derivative of m Y (t) is d dt m Y (t) = d dt [t( t ) ] = ( t ) + (t)( )( t ) 3 ( t). There is no need simplif this, because all we do is evaluate it at t =. We have E(Y ) = d dt m Y (t) = ( ) + ()( )( ) 3 ( )() =. t= The variance of Y is V (Y ) = E(Y ) [E(Y )] = =. 4. The support R = {(, ) : >, >, + < } is shown at the top of the next page (left). The joint pdf f Y,Y (, ) is a three-dimensional function which takes the value 4 over this region and is otherwise equal to zero. Note that the boundar of the support is + = = =, a linear function of with intercept and slope. We can find P (Y <.5) b finding the volume under f Y,Y (, ) over the set B = {(, ) : < <.5, >, + < }, which is shown at the top of the next page (right). The double integral limits come from this picture. We have P (Y <.5) = = =.5 = =.5 =.5 =.5 = 4 [ ( 4 d d ) ] d = ( ) d = ( + 3 ) d ( = ).5 = ( = 8 + ) = Note: You could also do this problem b finding the marginal pdf of Y and then calculating P (Y <.5) as an area under this pdf. For < <, the marginal pdf of Y is [ ( ) ] f Y ( ) = 4 d = 4 = ( ). = = f Y ( ) = { ( ), < <, otherwise. PAGE

3 FALL 8 FINAL EXAM SOLUTIONS.5 You would then calculate P (Y <.5) as just like in the previous solution..5 = ( ) d = Let Y denote the number of people arriving for treatment per hour. In the problem, we assume Y Poisson(λ = 5). (a) The probabilit at most two people arrive for treatment is P (Y ) = P (Y = ) + P (Y = ) + P (Y = ) = 5 e 5! + 5 e 5! + 5 e 5! = e 5 ( ).5. (b) The time until the second person arrives, T, follows a gamma distribution with α = and β = λ = 5. The variance of T is σ = ( ) ( ) = σ = = Let Y denote the number of events that occur (out of n) so that Y b(n, p). We want to calculate P (Y = Y ). Using the definition of conditional probabilit, we have P (Y = Y ) = P (Y = and Y ). P (Y ) PAGE 3

4 FALL 8 FINAL EXAM SOLUTIONS The denominator is P (Y ) = P (Y = ) = ( ) n p ( p) n = ( p) n. When =, the numerator is because {Y = and Y } =, the null set. When =,,..., n, then {Y = } {Y }, so {Y = and Y } = {Y = }. the numerator is ( ) n P (Y = and Y ) = P (Y = ) = p ( p) n. P (Y = Y ) = ( n, = ) p ( p) n ( p) n, =,,..., n. 7. The cdf P (Y ) is defined for all R. we consider three cases: Case : When, Case : When < <, f Y (t)dt = dt =. Case 3: When, f Y (t)dt = dt + = (t + t 3 )dt 6 = ( ) t 6 + t4 4 t 6 ( + t )dt = 6 ( ) = 4 ( + ). f Y (t)dt = dt + t 6 ( + t )dt }{{} =, 4 ( + ), < <,. + dt =. 8. The marginal pmfs are 3 p Y ( ) and PAGE 4

5 FALL 8 FINAL EXAM SOLUTIONS 3 p Y ( ) The marginal means are E(Y ) = (.5) + (.45) + 3(.3) =.5 and We now calculate E(Y ) = (.3) + (.4) + 3(.3) =. E(Y Y ) = ()(.) + ()(.5) + 3()(.5) + ()(.) + ()(.) + 3()(.) + (3)(.5) + (3)(.) + 3(3)(.5) = 4.5. Using the covariance computing formula, we have Cov(Y, Y ) = E(Y Y ) E(Y )E(Y ) = 4.5.5() = Note that ( ay ) = a Y Y + a a. Now take the expectation of this: [ ( Q(a) = E ay ) ] = a E(Y ) E(Y ) + a a. To minimize Q(a), we take its derivative, set it equal to zero, and solve for a. We have d da Q(a) = d [a E(Y ) E(Y ) + a ] da = ae(y ) a 3. Setting this equal to zero, we have ae(y ) a 3 set = = ae(y ) = [ ] a 3 = a 4 E(Y /4 ) = = a = E(Y. ) From the variance computing formula, we have E(Y ) = V (Y ) + [E(Y )] = σ + µ. ( ) /4 a = σ + µ. To verif this solution minimizes Q(a), note that d da Q(a) = E(Y ) + 6 a 4 >. the solution a = (σ + µ ) /4 is a minimizer b the second derivative test. PAGE 5

6 FALL 8 FINAL EXAM SOLUTIONS. Treating each candidate as a distinct object, there are N = 9! = 3688 possible permutations. We can think of one outcome (i.e., ordering of the candidates) in the underling sample space S as having the following structure: Define the event ( ). A = {candidates within each part grouped together}. The number of outcomes in A can be found using the basic counting rule with n = number of was to permute R, D, and I ordering = 3! n = number of was to permute R candidates = 4! n 3 = number of was to permute D candidates = 3! n 4 = number of was to permute I candidates =! The number of outcomes in A is n a = 3! 4! 3!! = 78. P (A) = n a N = Note: You could also treat candidates within each part as indistinguishable; e.g., ou cannot tell one R candidate from another, etc. Then there would be ( ) 9 N = = outcomes in S. The number of outcomes in A is n a = n = number of was to permute R, D, and I ordering = 3! = 6. The steps to calculate n, n 3, and n 4 in the other solution are now vacuous because ou cannot tell candidates in the same part from each other.. For 3, the pdf of Y is f Y () = d d d d P (A) = n a N = [ e ( 3)/5] = ( ) e ( 3)/5 = 5 5 e ( 3)/5. f Y () = 5 e ( 3)/5, 3, otherwise. PAGE 6

7 FALL 8 FINAL EXAM SOLUTIONS The mean of Y is In the last integral, let E(Y ) = R f Y ()d = 3 u = 3 = du = d. 5 e ( 3)/5 d. Note that with this substitution, the limits change as well; i.e., E(Y ) = 3 : 3 = u :. 5 e ( 3)/5 d = (u + 3) 5 e u/5 du. Note that 5 e u/5 is the exponential pdf with mean β = 5 and we are integrating over (, ). the last integral equals E(U + 3), where U exponential(5). We have E(U + 3) = E(U) + 3 = = 35 minutes.. The support R = {(, ) : >, < < } is shown below. The joint pdf f Y,Y (, ) is a three-dimensional function which takes the value 6 / 3 over this region and is otherwise equal to zero. The support is rectangular and does not involve a constraint between and. Note that we can write f Y,Y (, ) = 6 3 = 6 3 = g( )h( ), where g( ) = 6 3 and h( ) = are both are nonnegative functions. Because f Y,Y (, ) can be written in this wa, we know Y and Y must be independent. PAGE 7

8 FALL 8 FINAL EXAM SOLUTIONS Note: You could also derive the marginal pdfs f Y ( ) and f Y ( ) and show that For >, the marginal pdf of Y is f Y ( ) = f Y,Y (, ) = f Y ( )f Y ( ). = 6 3 For < <, the marginal pdf of Y is f Y ( ) = = 6 3 f Y ( ) = d = 6 3 [ ( ) ] = 8 = 3. [ ( d = 6 ) ] ( = 6 + ) =. = 8 8, > 3, otherwise and {, < <, otherwise. Note that 6 3 = f Y,Y (, ) = f Y ( )f Y ( ) = Define the two events A = {resident smoked} D = {lung cancer death}. The first bullet sas P (A) =.. The second bullet sas P (D) =.6. The third bullet sas P (D A) = P (D A). We want P (A D). Use Baes Rule: P (A D) = We get P (D A) b using LOTP; i.e., P (D A)P (A) P (D) = P (D A)(.)..6.6 = P (D) = P (D A)P (A) + P (D A)P (A) = P (D A)(.) +.P (D A)(.) = P (D A)[. +.(.8)] = P (D A) =.6.8. P (A D) = = We have the hierarchical model: Y Y = exponential( ) Y U(, 5). PAGE 8

9 FALL 8 FINAL EXAM SOLUTIONS To find V (Y ), we will use Adam s Rule; i.e., V (Y ) = E[V (Y Y )] + V [E(Y Y )]. The conditional distribution of Y Y = exponential( ). remembering what we know about the mean and variance of the exponential distribution, and E(Y Y = ) = = E(Y Y ) = Y V (Y Y = ) = ( ) = 4 = V (Y Y ) = 4Y. V (Y ) = E(4Y ) + V (Y ) = 4E(Y ) + 4V (Y ), where Y U(, 5). The mean of Y U(, 5) is E(Y ) = 5, the midpoint of and 5. The second moment of Y is Finall, 5 E(Y ) = 5 d = 5 ( ) 3 5 = V (Y ) = E(Y ) [E(Y )] = 5 ( ) 5 3 = 5. V (Y ) = 4E(Y ) + 4V (Y ) = 4 ( ) ( ) 5 = The first moment is the mean E(Y ). Using the definition of mathematical expectation, we have ( ) E(Y ) = p Y () = = ( ). ln ln = = = }{{} =, see below We recognize = ( ) as an infinite geometric sum with common ratio r = /. ( ) ( ) ( ) = = = =. = = E(Y ) = ln. PAGE 9