7 Notes for Math 34, Part
Chapter Conditional epectations, variances, etc.. Conditional probability Given two events, the conditional probability of A given B is defined by P[A B] = P[A B]. P[B] P[A B] is the probability of A assuming that B have happened. To find a conditional probability, we restrict the sample space to B. Given a r.v. X and A, B R, P[X A X B] = P[X A B]. P[X B] Problem.. (# 7, November ) The loss due to a fire in a commercial building is modeled by a random variable X with density function.5( ) for < < f() otherwise Given that a fire loss eceeds 8, what is the probability that it eceeds 6? Answer: 9 Now, and Solution: We have to find P[X > 6) = 6 P[X > 6 X > 8] =.5( ) d = P[X > 8] = 8 ( ) 6 d = So, P(X > 6 X > 8) = 4 =. 9 73 P[X > 6] P[X > 8]. ( ) d = ( ) 4 ( ) 4 8 = 4. 6 = 4 4
74 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. Problem.. (# 39, May ) An insurance company insures a large number of homes. The insured value, X, of a randomly selected home is assumed to follow a distribution with density function 3 4 for > f() = otherwise Given that a randomly selected home is insured for at least.5, what is the probability that it is insured for less than? Answer:.578 Now, and So, Solution: We have to find P[X X.5] = P[.5 X ] = P[.5 X] = P[X X.5] =.5.5 P[.5 X ]. P[X.5] 3 4 d = ( 4 ) 3 4 d = ( 4 ).5 3 3.5 3 =.5.5 =.5 3 3 =.5 3. ( ) 3 3 = 55 4 64 =.578. Problem.3. (# 7, November ) A group insurance policy covers the medical claims of the employees of a small company. The value, V, of the claims made in one year is described by V =,Y where Y is a random variable with density function k( y) 4 for < y < f(y) = otherwise where k is a constant. What is the conditional probability that V eceeds 4,, given that V eceeds,? Answer:.3 Solution: To make f a density k = 5. We have to find Now, and So, P[V 4, V,] = P[,Y 4,,Y,] = P[Y.4 Y.] = P[Y.4] = P[Y.] =.4. P[Y.4] P[Y.]. 5( y) 4 dy = ( y) 5 5( y) 4 dy = ( y) 5.4. = (.6) 5 = (.9) 5 P[V 4, V,] = (.6)5 (.9) 5 = 3 43 =.37.
.. CONDITIONAL PROBABILITY 75 Problem.4. (# 3, November ) The number of injury claims per month is modeled by a random variable N with P [N = n] =, where n. Determine the probability (n+)(n+) of at least one claim during a particular month, given that there have been at most four claims during that month. Answer:. 5 Now, and So, Solution: We have to find P[N N 4] = P[ N 4]. P[N 4] P[ N 4] = P[N = ] + P[N = ] + P[N = 3] + P[N = 4] = + + + = 3 3 4 4 5 5 6 6 P[N 4] = P[N = ] + P[N = ] + P(N = ) + P[N = 3] + P[N = 4] = + + + + + = 5 3 3 4 4 5 5 6 6 P[N N 4] = 5. We also can define the conditional epectation and the conditional variance given that X B. For a discrete r.v. X, the conditional pmf of X given X B, is p X X B () = p() P[X B] if B otherwise Hence, the conditional epectation is defined by E[X X B] = p X X B () = p() P(X B) = B B B p() B p(). For a continuous r.v. X, the conditional pmf of X given X B, is f X X B () = f() P(X B) if B otherwise The conditional epectation of X given X B is f() E[X X B] = P[X B] d = B B B f() d. f() d E[X X B] is the average value of X assuming that B happened. We restrict the sample space to B. To find the conditional epectation given X B, we do the usual epectation in B and we divide over P [X B]. Problem.5. (# 9, Sample Test) The distribution of loss due to fire damage to a warehouse is:
76 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. Amount of Loss Probability.9 5.6,.3,.8 5,.,. Given that a loss is greater than zero, calculate the epected amount of the loss. Answer:, 9 Solution: We need to find E[X X > ]. We condition on X >. Since P(X > ) = P(X = ) =., we get the following conditional distribution 5 5 p X X> () = P[X=] P[X>].6.3.8.. Finally, E[X X > ] = > p X X>() = (5)(.6) + ()(.3) + ()(.8) + (5)(.) + ()(.) = 9.. Conditional probability mass functions and conditional densities For two r.v. s we can have conditional probabilities in the usual way. Given two r.v.s X and Y and A, B R, P[X A, Y B] P[X A Y B] =. P[Y B] Eample.. The joint probability density functions of X and Y is given by Find P[Y > X /]. f(, y) = for < <, < y <. Solution: By definition of conditional probability, P[Y > X /] = P[Y >, X /] P[X /] = / dy d /8 / = dy d /4 =. Given two discrete r.v. s X and Y, the conditional mass function of X given Y = y is defined by p X Y ( y) = P[X = Y = y] = p X,Y (, y), if p Y (y) >. p Y (y)
.. CONDITIONAL PROBABILITY MASS FUNCTIONS AND CONDITIONAL DENSITIES 77 Similarly, the conditional mass function of Y given X = is defined by p Y X (y ) = P[Y = y X = ] = p X,Y (, y), if p X () >. p X () From these formulas, it follows that p X,Y (, y) = p X ()p Y X (y ) = p Y (y)p X Y ( y). If p Y (y) =, p X Y ( y) is not defined. Since p Y (y) =, Y = y cannot happen. We can find probabilities and epectations assuming that something happens, if the probability of that something is positive. Given two jointly continuous r.v. s X and Y, the conditional density of X given Y = y is defined by f X Y ( y) = f X,Y (, y), if f Y (y) >. f Y (y) If f Y (y) =, f X Y ( y) is not defined. Similarly, the conditional mass function of Y given X = is defined by f Y X (y ) = f X,Y (, y), if f X () >. f X () From these formulas, it follows that f X,Y (, y) = f X ()f Y X (y ) = f Y (y)f X Y ( y). Eample.. Let X and be Y discrete random variables with joint probability function (+)(y+) for =,, ; y =,, 54 p(, y) = else Find the conditional pmf of Y given X =. Solution: We have that p X () = P(X = ) = p(, ) + p(, ) + p(, ) = 54 + 3 54 + 4 54 = 8 54 = 3. The conditional conditional pmf of Y given X = is p Y X ( ) = p X,Y (, ) p X () p Y X ( ) = p X,Y (, ) p X () p Y X ( ) = p X,Y (, ) p X () = 4/54 8/54 = 4 8 = 6/54 8/54 = 6 8 = 8/54 8/54 = 8 8 Eample.3. Let X and Y have the pdf 6(y ) if y f(, y) = else Find the marginal pfd s, find the conditional pdf s f Y X (y ) and f X Y ( y).
78 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. Solution: The marginal density of X is f X () = The conditional density of Y given X = is f Y X (y ) = f X,Y (, y) f X () 6(y ) dy = 3(y ) = 3( ) for < < = 6(y ) (y ) = for y. 3( ) ( ) The marginal density of Y is y y f Y (y) = 6(y ) d = (6y 3 ) = 6y 3y = 3y for < y < The conditional density of X given Y = y is f X Y ( y) = f X,Y (, y) f Y (y) = 6(y ) 3y = (y ) y for y. Figure.: 3, May Problem.6. (# 3, May ) An insurance policy is written to cover a loss X where X has density function 3 8 f() = for otherwise The time (in hours) to process a claim of size, where, is uniformly distributed on the interval from to. Calculate the probability that a randomly chosen claim on this policy is processed in three hours or more. Answer:.7
.. CONDITIONAL PROBABILITY MASS FUNCTIONS AND CONDITIONAL DENSITIES 79 Solution: Let Y be the time to process a claim. The conditional distribution of Y given X = is f Y X (y ) = if < y <. The joint density of X and y is if < < y < <. Hence, f X,Y (, y) = f Y X (y )f X () = 3 8 = 3 8 P(Y 3) = 4 3 = 4 3 3 6 y/ 3 y/ 8 dy = 4 3 d dy ( 3 3y 4 64 ) dy = 64 =.78. Figure.: 34, November Problem.7. (# 34, November ) An auto insurance policy will pay for damage to both the policyholder s car and the other driver s car in the event that the policyholder is responsible for an accident. The size of the payment for damage to the policyholder s car, X, has a marginal density function of for < <. Given X =, the size of the payment for damage to the other driver s car, Y, has conditional density of for < y < +. If the policyholder is responsible for an accident, what is the probability that the payment for damage to the other driver s car will be greater than.5? Answer: 7 8. Solution: The marginal density of X is f X () = for. The conditional density of X given Y is f Y X (y ) = if < < and < y < +. So, the joint density of X and Y is f X,Y (, y) = f X ()f Y X (y ) = if < < and < y < +
8 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. So, [ P Y ] = P[Y / ] = / dy d = / ( ) d = 7 8..3 Conditional epectations Given two discrete r.v. s X and Y and let A R, the conditional probability that X A given Y = y is defined by P[X A Y = y] = A p X Y ( y). Similarly, the conditional probability that Y A given X = is defined by P[Y A X = ] = y A yp Y X (y ). Given two jointly continuous r.v. s X and Y, and let A R, the conditional probability that X A given Y = y is defined by P[X A Y = y] = f X Y ( y) d. Similarly, the conditional probability that Y A given X = is defined by P[Y A X = ] = yf Y X (y ) dy. Problem.8. (#, May ) A company offers a basic life insurance policy to its employees, as well as a supplemental life insurance policy. To purchase the supplemental policy, an employee must first purchase the basic policy. Let X denote the proportion of employees who purchase the basic policy, and Y the proportion of employees who purchase the supplemental policy. Let X and Y have the joint density function f(, y) = ( + y) on the region where the density is positive. Given that % of the employees buy the basic policy, what is the probability that fewer than 5% buy the supplemental policy? Answer:.47 Solution: The joint density of X and Y is f X,Y (, y) = ( + y), for y. The marginal density of X is f X () = So, the conditional density of Y given X = is For =., we have f Y X (y ) = f X,Y (, y) f X () f Y X (y.) =. + y.3 = A y A ( + y) dy = 3. + y 3 for y. = 3 + y for. y. 3
.3. CONDITIONAL EXPECTATIONS 8 So, the conditional probability we are looking for is.5 ( P(Y.5 X =.) = 3 + ) 3 y dy = 5 =.466. Problem.9. (# 5, November ) Once a fire is reported to a fire insurance company, the company makes an initial estimate, X, of the amount it will pay to the claimant for the fire loss. When the claim is finally settled, the company pays an amount, Y, to the claimant. The company has determined that X and Y have the joint density function f(, y) = ( ) y ( )/( ), >, y >. Given that the initial claim estimated by the company is, determine the probability that the final settlement amount is between and 3. Answer: 8 9 Solution: We need to find P[ Y 3 X = ]. The marginal density of X is f X () = f X,Y (, y) dy = = for >. 3 = y ( )/( ) ( ) + The conditional density of Y given X = is For =, we have f Y X (y ) = f X,Y (, y) f X () = ( ) y ( )/( ) dy ( )y ( )/( ) for y >. f Y X (y ) = y 3 for y >. The conditional probability we are looking for is P[ Y 3 X = ] = 3 f Y X (y ) dy = 3 y 3 dy = 8 9. Given two discrete r.v. s X and Y, the conditional epectation of X given Y = y is defined by E[X Y = y] = p X Y ( y). The conditional epectation of Y given X = is defined by E[Y X = ] = y yp Y X (y ). Given two jointly continuous r.v. s X and Y, the conditional epectation of X given Y = y is defined by E[X Y = y] = f X Y ( y) d. The conditional epectation of Y given X = is defined by E[Y X = ] = yf Y X (y ) dy.
8 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. Eample.4. Let X and be Y discrete random variables with joint probability function (+)(y+) for =,, ; y =,, 54 p(, y) = else Find E[Y X = ]. Solution: By a previous eample, the conditional pmf of Y given X = is p Y X ( ) = 4 p Y X ( ) = 6, p 8 Y X( ) = 8. So, 8 E[Y X = ] = () 4 8 + () 6 8 + () 8 8 = 8. Eample.5. Let X and Y have the pdf 6(y ) if y f(, y) = else Find E[X Y = y] and E[Y X = ]. So, So, Solution: By a previous eample, f Y X (y ) = 6(y ) (y ) = for y. 3( ) ( ) E[Y X = ] = y (y ) ( ) dy = = 3( ) ( ) 3 + 3( ) The conditional density of X given Y = y is f X Y ( y) = ( y 3 3( ) 3 ( ) ) y ( ) = 3+3 = + 3( ) 3 (y ) y for y. E[X Y = y] = ( ) y y (y ) d = y 3 = y. y y 3y 3 Problem.. (# 37, Sample test) An insurance contract reimburses a family s automobile accident losses up to a maimum of two accidents per year. The joint probability distribution for the number of accidents of a three person family (X, Y, Z) is p(, y, z) = k( + y + z), where =,, y =,,, z =,, and, y, z are the number of accidents incurred by X, Y, and Z, respectively. Determine the epected number of unreimbursed accidents losses given that is not involved in any accidents. Answer: 7 9. Solution: First, we find P[X = ] = p(,, ) + p(,, ) + p(,, ) + p(,, ) + p(,, ) +p(,, ) + p(,, ) + p(,, ) + p(,, ) = k + k + k + 3k + 4k + 4k + 5k + 6k = 7k 8,
.4. CONDITIONAL VARIANCES 83 (y, z) (,) (,) (,) (,) (,) (,) (,) (,) (,) P[Y = y, Z = z X = ] 7 7 7 The number of unreimbursed accident losses is ma(x + Y + Z, ). Assuming that X =, the number of unreimbursed accident losses is ma(y + Z, ). So, E[ma(Y + Z, ) X = ] = P[Y =, Z = X = ] + P(Y =, Z = X = ) + P[Y =, Z = X = ] = () 4 + () 4 + () 4 = 7 7 7 7 7.4 Conditional variances Given two discrete r.v. s X and Y, the conditional variance of X given Y = y is defined by 7 3 7 4 7 4 7 5 7 6 7 Var(X Y = y) = E[(X E[X Y = y]) Y = y] = ( E[X Y = y]) p X Y ( y). It can be shown that Var(X Y = y) = E[X Y = y] (E[X Y = y]) = ( p X Y ( y) p X Y ( y)). Similarly, the conditional variance of Y given X = is defined by Var(Y X = ) = E[(Y E[Y X = ]) X = ] = y (y E[Y X = ]) p Y X (y ). It can be shown that Var(Y X = ) = E[Y X = ] (E[Y X = ]) = y ( y p Y X (y ) yp Y X (y )). Given two jointly continuous r.v. s X and Y, the conditional variance of X given Y = y is defined by Var(X Y = y) = E[(X E[X Y = y]) Y = y] = Usually, it is easier to do Var(X Y = y) = E[X Y = y] (E[X Y = y]) = ( f X Y ( y) d f X Y ( y) d). Similarly, the conditional variance of Y given X = is defined by Var(Y X = ) = E[(Y E[Y X = ]) X = ] = Usually, it is easier to do Var(Y X = ) = E[Y X = ] (E[Y X = ]) = ( y f Y X (y ) dy y f Y X (y ) dy). ( E[X Y = y]) f X Y ( y) d. (y E[Y X = ]) f Y X (y ) d. y
84 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. Problem.. (# 4, May ) The stock prices of two companies at the end of any given year are modeled with random variables X and Y that follow a distribution with joint density function for < <, < y < + f(, y) = otherwise What is the conditional variance of Y given that X =? Answer: Solution: f X () = + d =, for < <. The conditional density of Y given X = is f Y X (y ) = for < y < +. This is a uniform distribution on the interval (, + ). So, its variance is. Problem.. (# 4, November ) A diagnostic test for the presence of a disease has two possible outcomes: for disease present and for disease not present. Let X denote the disease state of a patient, and let Y denote the outcome of the diagnostic test. The joint probability function of X and Y is given by: P [X =, Y = ] =.8, P [X =, Y = ] =.5, P [X =, Y = ] =.5, P [X =, Y = ] =.5, Calculate Var(Y X = ). Answer:. Solution; First, we find the conditional pmf of Y given X =. We have that P[X = ] =.75. So, P[Y = X = ] =.5.75 = 7 and P[Y = X = ] =.5.75 = 5 7. Now, E[Y X = ] = 7 + 7 = 7 E[Y X = ] = 7 + 7 = 7 and Var(Y X = ) = 7 ( 7) =.4. Problem.3. (#, May ) An actuary determines that the annual numbers of tornadoes in counties P and Q are jointly distributed as follows: Annual number of tornadoes in county P Annual number of tornadoes in county Q 3..6.5..3.5..3.5.5..
.5. FURTHER PROPERTIES OF THE CONDITIONAL EXPECTATIONS 85 Calculate the conditional variance of the annual number of tornadoes in county Q, given that there are no tornadoes in county P. Answer:.99 Solution: Since P[P = ] =.5, the conditional pmf of Q given P =, is q 3 P[Q = q P = ].48.4..8 So, E[Q P = ] = ().48 + ().4 + (). + (3).8 =.88 E[Q P = ] = ().48 + ().4 + (). + (3).8 =.76 Var(Q P = ) = E[Q P = ] (E[Q P = ]) = (.76).88 =.9856..5 Further properties of the conditional epectations Given two r.v. s X and Y and a function g : R R, we may define E[g(X, Y ) X = ]. We have the following properties of the conditional epectation: Theorem.. Let X and let Y be two r.v. s, let g, g : R R and let h : R R. Then, E[ag (X, Y ) + bg (X, Y ) X = ] = ae[g (X, Y ) X = ] + be[g (X, Y ) X = ] and E[ag (X, Y ) + bg (X, Y ) Y = y] = ae[g (X, Y ) Y = y] + be[g (X, Y ) Y = y]. E[h(X) X = ] = h() and E[h(Y ) Y = y] = h(y). If X and Y are independent, then E[h(Y ) X = ] = E[h(Y )] and E[h(X) Y = y] = E[h(X)]. Given two r.v.s X and Y, E[X Y = y] is a function of y. So, we find the epectation and the variance of E[X Y = y]. Theorem.. (Double epectation theorem for epectations) Given two r.v.s X and Y, E[E[X Y = y]] = E[X] and E[E[Y X = ]] = E[Y ]. Proof. We only to the continuous case, E[X Y = y] is a function on y. So, E[E[X Y = y]] = By definition E[X Y = y] = f X Y ( y) d. Hence, E[E[X Y = y]] = f X Y ( y) df Y (y) dy = E[X Y = y]f Y (y) dy. By definition of conditional density, f Y (y)f X Y ( y) = f X,Y (, y). So, E[E[X Y = y]] = f X,Y (, y) d dy = E[X]. f Y (y)f X Y ( y) d dy. Q.E.D.
86 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. Theorem.3. (Double epectation theorem for variances) Given two r.v.s X and Y, Var(X) = Var(E[X Y = y]) + E[Var(X Y = y)] and Var(Y ) = Var(E[Y X = ]) + E[Var(Y X = )]. Problem.4. (# 4, Sample Test) An automobile insurance company divides its policyholders into two groups: good drivers and bad drivers. For the good drivers, the amount of an average claim is 4, with a variance of 4,. For the bad drivers, the amount of an average claim is, with a variance of 5,. Sity percent of the policyholders are classified as good drivers. Calculate the variance of the amount of a claim for a policyholder. Answer:,4 Solution: Let X be the amount of a claim. Let if the driver is good Y = if the driver is bad We know that P(Y = ) =.6, P(Y = ) =.4, E[X Y = ] = 4, Var(X Y = ) = 4, E[X Y = ] =, Var(X Y = ) = 5. We use the formula of the double epectation theorem for the variance, Var(X) = E[Var(X Y )] + Var(E[X Y ]). So, Net, and E[Var(X Y )] = P[Y = ]Var(X Y = )] + P[Y = ]Var(X Y = ) = (.6)(4) + (.4)(5) = 4. E[E[X Y ]] = P[Y = ]E[X Y = ] + P[Y = ]E[X Y = ] = ()(.6) + (4)(.4) = 76, E[(E[X Y ]) ] = P[Y = ](E[X Y ]) + P[Y = ](E[X Y ]) = () (.6) + (4) (.4) = 384 Finally, Var(E[X Y ]) = E[(E[X Y ]) ] (E[E[X Y = ]]) = 384 (76) = 864. Var(X) = E[Var(X Y )] + Var(E[X Y ]) = 4 + 864 = 4. Problem.5. (#, May ) Let X and Y denote the values of two stocks at the end of a five year period. X is uniformly distributed on the interval (, ). Given X =, Y is uniformly distributed on the interval (, ). Determine Cov(X, Y ) according to this model. Answer: 6
.5. FURTHER PROPERTIES OF THE CONDITIONAL EXPECTATIONS 87 Solution : Since X is uniformly distributed on the interval (, ), then E[X] = 6. Y given X = has a uniform distribution on the interval (, ), E[Y X = ] =. So, [ ] X E[Y ] = E[E[Y X = ]] = E = 3 E[XY ] = E[E[XY X = ]] = E[XE[Y X = ]] = E[X X ] = 3 d = 7 = 4. Finally, Cov(X, Y ) = E[XY ] E[X]E[Y ] = 4 (6)(3) = 6. Solution : Since X is uniformly distributed on the interval (, ), f X () =, for. The conditional density of Y given X = is f Y X (y ) = for y (, ). So, the joint density is So, Finally, f X,Y ( y) = f X ()f Y X (y ) =, for y. E[X] = E[Y ] = E[XY ] = dy d = 6 y dy d = 3 y dy d = 4 Cov(X, Y ) = E[XY ] E[X]E[Y ] = 4 (6)(3) = 6. Problem.6. (#, May ) Two life insurance policies, each with a death benefit of, and a one-time premium of 5, are sold to a couple, one for each person. The policies will epire at the end of the tenth year. The probability that only the wife will survive at least ten years is.5, the probability that only the husband will survive at least ten years is., and the probability that both of them will survive at least ten years is.96. What is the epected ecess of premiums over claims, given that the husband survives at least ten years? Answer: 897 Solution: Let and let Y = Y = if the husband dies before years if the husband survives at least years if the wife dies before years if the wife survives at least years The ecess of premiums over claims is Y Y. We know that P [Y =, Y = ] =.5, P [Y =, Y = ] =., P [Y =, Y = ] =.96. The ecess of
88 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. premium over claims is Y Y. The epected ecess of premiums over claims, given that the husband survives at least ten years is because E[ Y Y Y = ] = E[Y Y = ] = P[Y = Y = ] =..97 = 896.8, P[Y = Y = ] = P[Y =, Y = ] P[Y = ] =..97. Problem.7. (# 9, Sample Test) An insurance company designates % of its customers as high risk and 9% as low risk. The number of claims made by a customer in a calendar year is Poisson distributed with mean θ and is independent of the number of claims made by a customer in the previous calendar year. For high risk customers θ =.6, while for low risk customers θ =.. Calculate the epected number of claims made in calendar year 998 by a customer who made one claim in calendar year 997. Answer:.4 Solution: Let Y = if the customer is high risk if the customer is low risk Let X be the number of claims in 997 and let X be the number of claims in 998. We have to find E[X X = ] = E[X I(X = )]. P[X = ] Conditioning on Y, and So, E[X I(X = )] = (.)E[X I(X = ) Y = ] + (.9)E[X I(X = ) Y = ] = (.)e.6 (.6)(.6) + (.9)e. (.)(.) =.79 P[X = ] = (.)P[X = Y = ] + (.9)P(X = ) Y = ) = (.)e.6 (.6) + (.9)e. (.) =.44, E[X X = ] =.79.44 =.439. Eample.6. The number of claims received in a year by an insurance company is a Poisson random variable with λ = 5. The claim amounts are independent and uniformly distributed over [, 5]. If the company has $4, available to pay claims, what is the probability that it will have enough to pay all the claims that come in?.6 Miture of densities Given densities f,..., f n and real numbers α,..., α n with n j= α j =, then f() = n j= α jf j () is a density. f is a miture of the densities f,..., f n. Alternatively, a r.v. with the density f can obtained as follows. Let Y be a discrete r.v. with values,..., n
.6. MIXTURE OF DENSITIES 89 and P(Y = j) = p j for j n. Let X be a r.v. such that the distribution of X given Y = j is continuous with density f j. Then, for each A R, n n P[X A] = P[Y = j]p[x A Y = j] = p j f j () d = f() d. j= Problem.8. (# 7, May ) An auto insurance company insures an automobile worth 5, for one year under a policy with a, deductible. During the policy year there is a.4 chance of partial damage to the car and a. chance of a total loss of the car. If there is partial damage to the car, the amount X of damage (in thousands) follows a distribution with density function.53e / if < < 5, f() = otherwise. What is the epected claim payment? Answer:38 Solution: Let j= if there is no damage to the car Y = if there is partial damage to the car if there is total damage to the car Let X be the amount of damage of the car. Then, X given Y = is zero. X given Y = has the density above. X given Y = is 5,. The epected payment is By a change of variables, 5 E[ma(X, )] = E[E[ma(X, ) Y ]] = P[Y = ]P[X Y = ] + P[Y = ]P[ma(X, ) Y = ] +P[Y = ]P[ma(X, ) Y = ] = (.4) 5 ( ( )(.53)e / d + (.)(4) 5 =. e / d ) 5 e / d + 8 e / d = 7.5.5 4ue u d = ( 4)e u (u + ) 7.5.5 = 6e.5 34e 7.5 A A and 5 e / d = (e.5 e 7.5 ) The epected payment is. ( 6e.5 34e 7.5 (e.5 e 7.5 ) ) + 8 = 38.9.73. Problem.9. (# 33, May ) For Company A there is a 6% chance that no claim is made during the coming year. If one or more claims are made, the total claim amount is normally distributed with mean, and standard deviation,. For Company B there is a 7% chance that no claim is made during the coming year. If one or more claims are made, the total claim amount is normally distributed with mean 9, and standard deviation
9 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC.,. Assume that the total claim amounts of the two companies are independent. What is the probability that, in the coming year, Company B s total claim amount will eceed Company A s total claim amount? Answer:.3 Solution: Company B s total claim amount will eceed Company A s total claim amount if either Company A has claim and B does or both companies have claims and the claim for B is bigger than the claim for A. We do (.6)(.3) + (.4)(.3)P[X < Y ] =.8 = (.)P [ =.8 + (.)P N(, ) () +() ] [ N(, ) = (.8) + (.)Φ(.35) = (.8) + (.)(.63) =.68. ] () +() Problem.. (# 6, November ) You are given the following information about N, the annual number of claims for a randomly selected insured P [N = ] =, P [N = ] = 3, P [N > ] = 6 Let S denote the total annual claim amount for an insured. When N =, S is eponentially distributed with mean 5. When N >, S is eponentially distributed with mean 8. Determine P [4 < S < 8]. Answer:. Solution: We have that P[4 < S < 8] = P[N = ]P[4 < S < 8 N = ] + P[N > ]P[4 < S < 8 N > ] 5 e /5 d + 6 8 e /8 d = 3 (e 4/5 e 8/5 ) + 6 (e 4/8 e 8/8 ) =.. = 3 8 4 8 4 Problem.. (#, Sample Test) An insurance policy covers the two employees of ABC Company. The policy will reimburse ABC for no more than one loss per employee in a year. It reimburses the full amount of the loss up to an annual company-wide maimum of 8. The probability of an employee incurring a loss in a year is 4%. The probability that an employee incurs a loss is independent of the other employee s losses. The amount of each loss is uniformly distributed on [, 5]. Given that one of the employees has incurred a loss in ecess of, determine the probability that losses will eceed reimbursements. Answer: 5 Solution: Let X be the amount of the loss of employee. Let if the employee has a loss Y = if the employee does not have a loss If Y =, then X =. If Y =, then X has the density f X ( ) = 4, for 5.
.7. PROBLEMS 9 Figure.3:, Sample Test Let X be the amount of the loss of employee. Let if the employee has a loss Y = if the employee does not have a loss If Y =, then X =. If Y =, then X has the density Then, f X ( ) = 4, for 5. P[X + Y 8 X ] = P[X +X 8] P[X ], and So, P[X + X 8] = ( ) 4 5 5 3 8 d 4 d = 5 3 d 3 () = ( 3) 5 = (5 3) = () () 5 3 P[X ] = 4 5 d 4 = 4 P[X + X 8 X ] = 5 3 3 = 3. 4 = 5..7 Problems. A player throws a fair die and simultaneously flips a fair coin. If the coin lands heads, then she wins twice the value that appears on the die, and if tails, then she wins one half of the value that appears on the die. Determine her epected winnings.. Let X be a random variable with uniform distribution on the interval [, ]. Find P X 5 X 3}.
9 CHAPTER. CONDITIONAL EXPECTATIONS, VARIANCES, ETC. 3. Let X and Y have the pdf f(, y) = 6(y ) if y else Find the marginal pfd s, the conditional pdf s and E[X Y = y] and E[Y X = ]. 4. Let X and be Y discrete random variables with joint probability function (+)(y+) for =,, ; y =,, 54 p(, y) = else What is E[Y X = ]? 5. Let X and Y be continuous random variables with joint density fucntion y for and y f(, y) = else What is P ( X Y X)? 6. Let X and Y be discrete random variables with joint probability function + y for =, and y =, 9 p(, y) = else Calculate E [ ] X Y 7. Let X and Y have the pdf f(, y) = 6(y ) if y else Find the marginal pfd s, f Y X (y ), E[X Y = y] and V(Y X = ). Check that E[E[Y X = ]] = E[Y ] and that V(Y ) = E[V(Y X)] + V(E[Y X]). 8. Let X and Y have a jointly continuous distribution with λe λ(y ) if y f Y X (y ) = else and f X () = Find the correlation between X and Y. λe λ if else
.7. PROBLEMS 93 9. Let Y and Y be two jointly continuous random variables with joint density function have density function 3y if y, y, y + y, f(y, y ) = else. Find P (Y 3/4 Y /).. Let f(, y) = 6, < < y <, and zero otherwise. Find f X Y ( y).. Let X and Y be two random variables satisfying that Var(X) = and Var(Y ) = Var(X Y ) = 4. Find the covariance of the random variables X and Y.