Interesting Probability Problems

Transcription

1 Interesting Probability Problems Jonathan Mostovoy University of Toronto August 9, 6 Contents Chapter Questions a) b) Chapter Questions 3 a) b) Chapter 3 Questions 4 3a) b) c) d) e) f) Chapter 4 Questions 9 4a) b) c) Chapter 5 Questions 5a) b) Non-Textbook Problems 3 6a) A b) B c) C d) D

2 Chapter Questions a).8.7 A deck of 5 cards contains four aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 3 cards, what is the probability that all four aces will be received by the same player? Answer: Since agents i =,..., 4 the probability that agent i will receive 4 aces is ) 3 4. Since 4 agents, it = 4 ) 3 4 scenarios where one agent receives all 4 aces. Therefore, and since there are ) 5 4 ways of arranging the four aces in a deck of 5 cards, the probability that all four aces will be received by the same player is: P r4 aces recieved by same agent) = 4 ) ) ,3,3,9 ) = 4 ) = b) ,3,3,3 Let A,..., A n be n arbitrary events. Show that the probability that exactly one of these n events will occur is: P ra i ) P ra i A j ) + 3 P ra i A j A k ) + ) n+ np ra A A n ) i= i<j i<j<k Proof. = i= Let Pn) = n P ra i \A, A,..., A i, A i+,..., A n )), which we shall prove: i= P ra i ) P ra i A j ) + 3 i<j i<j<k P ra i A j A k ) + ) n+ np ra A A n ) Our plan is to form induction on the function Pn). We begin at n =. We define an alternate definition of the set A and B as A = x : x A\B or A B} and B = y : y B\A or A B}. Both sets for which all x or y in A or B are trivially disjoint to one other since A\B) A B) =. Thus, P ra) = P ra\b)+p ra B). From here, we find P ra)+p rb) = P ra\b)+p rb\a)+ P ra B). Now since all three sets are disjoint, we can replace our additions to unions to yield: P ra) + P rb) P ra B) = P) = P ra\b) P rb\a) We now have our inductive hypothesis assuming validity of our Theorem for Pn), thus for k = n+: Pn + ) = n+ i= P ra i \A, A,..., A i, A i+,..., A n, A n+ )) But by our inductive hypothesis, we know what Pn) is, thus all we need to do is add on the remove all intersections of A n+ and add back it s own probability in a sense, take away the overlaps that A n+ could have on everything else) terms in our Pn) formula, i.e.: Pn + ) = P ra i ) + P ra n+ ) P ra i A j ) P ra i A n+ ) +... i= i<j i=

3 + ) n+ np ra A A n ) + ) n+ np ra A 3 A n n + ) ) n+ P ra A n+ ) n+ n+ = P ra i ) P ra i A j )+3 i= i<j n+ i<j<k P ra i A j A k ) + ) n+ n+)p ra A A n+ ) Chapter Questions a)..4 A machine produces defective parts with three different probabilities depending on its state of repair. If the machine is in good working order, it produces defective parts with probability.. If it is wearing down, it produces defective parts with probability.. If it needs maintenance, it produces defective parts with probability.3. The probability that the machine is in good working order is.8, the probability that it is wearing down is., and the probability that it needs maintenance is.. Compute the probability that a randomly selected part will be defective. Answer: We first define the notation: g, w, b, d as good working order, wearing down, needs maintenance bad) and production of defective parts respectively. We summarize the information given as: P rd g) =., P rd w) =., P rd b) =.3, P rg) =.8, P rw) =. = P rb). We now recall the Law of Total Probability, which states: given events B,..., B k form a partition of the space S and P rb j ) > for j =,..., k. Then, for every event A in S: P ra) = k P rb j )P ra B j ) j= Since g, w & b must be independent and sum to, our answer is a direct application of the Law of Total Probability: Prd) = i P ri)p rd i), j = g, w, b) =.8).) +.).) +.).3) =. b).. Suppose that A,..., A k form a sequence of k independent events. Let B,..., B k be another sequence of k events such that for each value of j, j =,..., k), either B j = A j or B j = A c j. Prove that B,..., B k are also independent events. Hint: Use an induction argument based on the number of events B j for which B j = A c j. Proof. We let n, n k denote the number of B j s where the relation B j = A c j holds, which = k n cases where B j = A j. For n =, our desired relation trivially holds because we assume A,..., A k is independent. For higher order cases, we recall the identity of P ra B) = P ra) P ra B c ) and the fact that if A,..., A l is independent, then so too will A,..., A v where v l. 3

4 Thus, through induction, we assume P rb B k ) are independent, and n of these B j s satisfy B j = A c j. Therefore, for s = n +, we want to see if we can factor P rb B i, Bi c B i+ B k ) where it was B i that now became Bi c from n to n +. We see this relationship to be true from: P rb B i B i+ B k B c i ) = P rb B i B i+ B k ) P rb B i B i+ B k B i ) = P rb ) P rb i ) P rb i+ ) P rb k ) P rb B i B i+ B k B i ) = P rb ) P rb i ) P rb i+ ) P rb k ) P rb ) P rb i ) P rb i+ ) P rb k ) P rb i ) = P rb ) P rb k ) P rb i )) = P rb ) P rb i ) P rb c i ) P rb i+) P rb k ) 3 Chapter 3 Questions 3a) 3..3 An ice cream seller takes gallons of ice cream in her truck each day. Let X stand for the number of gallons that she sells. The probability is. that X =. If she doesn t sell all gallons, the distribution of X follows a continuous distribution with a p.d.f. of the form: cx for < x < fx) = otherwise where c is a constant that makes P rx < ) =.9. Find the constant c so that P rx < ) =.9 as described above. Answer: All we need to remember that the integral of our p.d.f over our sample space must be equal to the the total probability of occurrence, which in this case is equal to.9, but usually. Thus: cx =.9 = c =.9 = c = 9 =.45 3b) Prove that the quantile function F of a general random variable X has the following three properties that are analogous to properties of the c.d.f.:. F is a non-decreasing function of p for < p <.. Let x = lim p, p> F p) and x = lim p, p< F p). Then x equals the greatest lower bound on the set of numbers c such that P rx c) >, and x equals the least upper bound on the set of numbers d such that P rx d) >. 3. F is continuous from the left; that is F p) = F p ) for all < p <. Proofs: 4

5 . Proof. Assume p, q, ) and p q. Then, by definition of the quantile function and since F x) is non-decreasing, F is non-decreasing since p, q: ) ) F p) = minx R : F x) p} F q) = minx R : F x) q}. Proof. x : Let z > z > z 3 >... be a decreasing sequence of numbers such that lim n z n =. = x = lim p, p> F p) = minx R : F x) z n } = minx R : F x) c} n= for some c R where no other x R is less than c and F x) >, which is equivalent to saying c is the least upper bound of all possible numbers where F c) > the condition P rx c) > x : Let y < y < y 3 <... be an increasing sequence of numbers such that lim n y n = = x = lim p, p< F p) = minx R : F x) y n } = minx R : F x) d} n= for some d R where no other x R is greater less than d and F x) <, which is equivalent to saying d is the least upper bound of all possible numbers where F d) < the condition P rx d) > 3. Proof. Let y < y < y 3 <... be an increasing sequence of numbers such that lim n y n = p. We immediately see that: F p) = minx R : F x) p} = minx R : F x) y n } n= = F p) = lim n F y n ) = F p ) 3c) Suppose that X and Y have a continuous joint distribution for which the joint p.d.f. is defined as follows: cy for x and y fx, y) = otherwise Determine:. the value of the constant c. P rx + Y > ) 3. P ry < /) 5

6 4. P rx ) 5. P rx = 3Y ) Answers: cy dxdy = = P rx + Y > ) = P ry < ) = P rx ) = cy dxdy = x cy dy = c 3 = = c = 3 3y dxdy = 3 8 3y dxdy = 8 3y dxdy = 5. P rx = 3Y ) = since with an n-dimensional continuous probability space, the probability that an n )-dimensional event occurs is effectively ; i.e., since X = 3Y is a line in a -dimensional continuous probability space, the probability this happens must be. 3d) Suppose that the joint p.d.f. of X and Y is as follows: 4xy for x, y and x + y fx, y) = otherwise Are X and Y independent? Answer: We recall Theorem 3.5.5, which states: when dealing with a joint p.d.f. fx, y), the random variables, X and Y, will be independent fx, y) = h x)h y) where h i z) is a nonnegative function only dependent on z. For every point within the defined triangle, we can define two functions which work, e.g; kx for x [, ] h x) = otherwise where k R, and h y) = 4 k y for x [, ] otherwise However, if we choose a point within the the unit square outside of our triangle, we need fx, y) =, but h x) > and h y) > which thus leads to a contradiction which = X and Y are NOT independent. Thus, we can conclude the following generalization: If x i within fx, x,... ) whose domain is a function of at least other variable, x j where j i, then some dependency amongst the variables x, x,.... 6

7 3e) Let the conditional p.d.f. of X given Y be g x y) = 3x y for < x < y and otherwise. Let the 3 marginal p.d.f. of Y be f y), where f y) = for y but is otherwise unspecified. Let Z = X Y. Prove that Z and Y are independent and find the marginal p.d.f. of Z. Proof. By the definition of conditional probability, the joint p.d.f. of X, Y ) is 3x f y) fx, y) = y if x > 3 otherwise We recall that if Z = X Y, we can define the dummy second random variable W = Y so that the Jacobian of our inverse transformation of x = zw and y = w is: x x J = z w y y = w z Thus, z w gz, w) = fx, y)w = fzw, w)w = 3zw) f w)w w 3 = 3z f w) Where the bounds on our variables were previously established in the question. Thus, since gz, w) = f z)f w) we may conclude independence. Further, the marginal p.d.f. of Z will be: 3z if z, ) f z) = otherwise Since f w)dw = if f w) is a proper probability function. 3f) 3..6 Let X, X be two independent random variables each with p.d.f. f x) = e x for x > and f x) = for x. Let Z = X X and W = X X.. Find the joint p.d.f. of X and Z.. Prove that the conditional p.d.f. of X given Z = is: e x for x > h x ) = otherwise 3. Find the joint p.d.f. of X and W. 4. Prove that the conditional p.d.f. of X given W = is: 4x e x for x > h x ) = otherwise 7

8 5. Notice that Z = } = W = }, but the conditional distribution of X given Z = is not the same as the conditional distribution of X given W =. This discrepancy is known as the Borel paradox. In light of the discussion that begins on page 46 about how conditional p.d.f. s are not like conditioning on events of probability, show how Z very close to is not the same as W very close to. Hint: Draw a set of axes for x and x, and draw the two sets x, x ) : x x < ɛ} and x, x ) : x x < ɛ} and see how different they are.. Answer: First, let us define the Jacobian from the two equations X = V and X = V Z: x x J = v z = = x v x z gv, z) = fx, x ) J = e v e v z) = e v e z e x e z. Proof. By definition, h x ) = gx,) g ). Thus, we first find g z): = g z) = max,z) [ gx, z)dx = ] x= e z e x x =max,z) e h x ) = e x = e x and if x e = e z if z ez if z < 3. Answer: First, let us define the Jacobian from the two equations X = V and X = V W : = J = x v x v x w x w = w v w gv, w) = fx, x ) J = v w e v e v w ) = v w e v+ w ) x w e x+ w ) for v/x, w > 4. Proof. By definition, h x ) = gx,) g ). Thus, we first find g w): = g w) = gx, w)dx = h x ) = x w e x+ w ) w + w ) [ + x w + w ) e x+ w ) ] x= x = = 4x e x and if x w= = w + w ) 5. The difference between x, x ) : x x < ɛ} and x, x ) : x x below, which = h x w) h x z). < ɛ} can be seen 8

9 4 Chapter 4 Questions 4a) Let X be a random variable with mean µ and variance σ, and let ψ t) denote the m.g.f. of X for < t <. Let c be a given positive constant, and let Y be a random variable for which the m.g.f. is: ψ t) = e cψt) ) for < t < Find expressions for the mean and the variance of Y in terms of the mean and the variance of X.. Answer: We first summarize: ψ ) =, ψ ) = µ and ψ ) = σ + µ. We compute: dψ t) dt = cψ )e cψ) ) = cµ t= d ψ t) dt = cψ )e cψ) ) + c ψ )) e cψ) ) = cσ + µ ) + c µ t= Therefore, Mean = cµ and Variance = cσ + µ ) + c µ cµ) = cσ + µ ). 4b) 4.7. Suppose that X and Y are random variables such that EY X) = ax + b. Assuming that CovX, Y ) exists and that < VarX) <, determine expressions for a and b in terms of EX), EY ), VarX), and CovX, Y ). Answer: We recall EEX X )) = EX ) = EEY X)) = EY ) = EaX +b) = aex)+b. Thus, we have our first equation: EY ) = aex) + b. Next, we apply EXfX, Y )) to both sides of the equation, yielding on the left: EXEY X)) = EEXY X)), and by what we had originally noted above, = EXY ), which = EXaX + b)) = EaX + bx) = aex ) + bex). Therefore, we have our second equation: EXY ) = aex ) + bex). So we must solve the following linear equations: 9

10 . EXY ) = aex ) + bex). EY ) = aex) + b Which yields the equation: 4c) EY ) aex) = EXY ) aex ) EX) = EX)EY ) EXY ) = aex)) ) EX ) = a = EXY ) EX)EY ) EX ) EX)) = CovX, Y ) VarX) = b = EY ) aex) = EY ) EX) CovX, Y ) VarX) Suppose that X,..., X n are random variables for which VarX i ) has the same value σ for i =,..., n and ρx i, X j ) has the same value ρ for every pair of values i and j such that i j. Prove that ρ n. Proof. Let us first note that if we want to find the variance of the sum of: Z = X + + X n, then: VarZ) = i= VarX i ) + i<j CovX i, X j ) In this scenario, we have VarX i ) = σ i and CovX i, X j ) = ρσ i j, which = ) n V arz) = nσ + ρσ = nσ + nn )ρσ We next recall V ary ) Y. Therefore, and in recalling σ, n > : nσ + nn )ρσ = nσ nn )ρσ = n )ρ = ρ n 5 Chapter 5 Questions 5a) In this exercise, we shall prove that the three assumptions underlying the Poisson process model do indeed imply that occurrences happen according to a Poisson process. What we need to show is that, for each t, the number of occurrences during a time interval of length t has the Poisson distribution with mean λt. Let X stand for the number of occurrences during a particular time interval of length t. Feel free to use the following extension of Eq ): For all real a, lim + au + ou)) u = e a u

11 . For each positive integer n divide the time interval into n disjoint subintervals of length t n each. For i =,..., n, let Y i = if exactly one arrival occurs in the i th subinterval, and let A i be the event that two or more occurrences occur during the i th subinterval. Let W n = n i= Y i. For each non-negative integer k, show that we can write P rx = k) = P rw n = k) + P rb), where B n i= A i.. Show that lim n P r n i= A i) =. Hint: Show that P r n i= Ac i ) = + ou)) u where u = n. 3. Show that lim n P rw n = k) = e λ λt) k k!. Hint: lim n n! n k n k)! =. 4. Show that X has the Poisson distribution with mean λt.. Proof. Trivially from chapter, we know X = k} = X = k} A) X = k} A c ) sets A, and the two sets, X = k} A) and X = k} A c ) are disjoint. If we let A = n i= A i, then X = k} n i= A i) c ) = W n = k}. We now trivially note X = k} n i= A i)) A = P rx = k) = P rw n = k) + P rb) where B = X = k} n i= A i)) A.. Proof. First, we note as is stated in the part, n disjoint subintervals of length t n = A,..., A n are independent and that P ra i ) = P ra j ) i, j. Therefore, P r n i=a c i) = n P ra c i) = [P ra c )] n = [ P ra )] n i= It was our assumption that P ra i ) = o n ) = ou) = lim P ra) = lim + n n n o n ))n = e = 3. Proof. We recall that if n Bernoulli R.V. s with the parameter p = λt n + ou), and if W n = n i= Y i, then: ) ) k n λt P rw n = k) = k n + ou) λt ) n k n ou) Next, we note lim n n k λt n Therefore, + ou))k = λt) k, and lim n n k λt n ou))n k = e λt. lim P rw n = k) = λt)k e λt n k! lim n n! n k n k)! = e λt λt) k k! 4. Proof. From part, we already saw P rx = k) = P rw n = k) + P rb), since P rx = k) fn) = P rx = k) = lim P rw n = k) + lim P rb) = e λt λt) k + lim n n k! + n n o n ))n = e λt λt) k k! By parts -3

12 5b) Review the derivation of the Black-Scholes formula 5.6.8). For this exercise, assume that our stock price at time u in the future is S e µu+wu, where W u has the gamma distribution with parameters αu and β with β >. Let r be the risk-free interest rate.. Prove that e ru ES u ) = S µ = r α log β β ).. Assume that µ = r α log β β ). Let R be minus the c.d.f. of the gamma distribution with parameters αu and. Prove that the risk-neutral price for the option to buy one share of the stock for the price q at time u is S Rc[β ]) qe ru Rcβ), where: ) ) q β c = log + αu log ru β S 3. Find the price for the option being considered when u =, q = S, r =.6, α =, and β =.. Proof. We recall ψt) for the Gamma distribution is = β β t )α. Therefore, Thus, S = e ru ES u ) ES u ) = ES e µu+wu = S e µu Ee Wu ) = S e µu β β )α S = e ru S e µu β β )αu β β log) = ru+µu+αu log ) µ = r α log β β ). Proof. We recall the value of an option at time u will be hs u ), where hs) = s q if s > q and otherwise. Therefore, hs u ) > ) ) q β W > log + αu log ru = c β S The risk-neutral price of the option is the present value of EhS u )), which equals: [ ] β e ru EhS ) ) = e ru αu S e µu+wu q Γαu) wαu e βw dw c We split the integrand into two parts at the q. The second integral is then just a constant times the integral of a normal p.d.f., namely, The first integral is: qe ru c β αu Γαu) wαu e βw dw = qe ru Rcβ) e ru S c e µu+wu βαu Γαu) wαu e βw dw = S [Rcβ ))]

13 Combining these two integrals yields: S Rc[β ]) qe ru Rcβ) where ) ) q β c = log + αu log ru S β 3. Since q = S it = q c = log S ) + αu log From here we substitute c into ) β ru = log β S S ) ) + log S Rc[β ]) qe ru Rcβ) S [ R )) e.6 R )) ] 6 Non-Textbook Problems 6a) A S For an event B with P B) >, define QA) = P A B) for any event A. Show that Q satisfies Axioms -3 of probability and conclude Q is a probability. Let us first recall the first 3 Axioms of Probability:.) For every event A, P ra).) P rs) = 3.) If A, A,... is a countably infinite sequence of disjoint events, then P r i= A i) = i= P ra i). Proof. Since P r is a probability function, we know that sets X, P rx). Furthermore, we recall the formal definition of condition probability: P ra B) = P ra B) P rb). From here we note x A B), x B = P ra B) P rb) = P ra B) P rb). Thus, we can conclude the first Axioms hold. For the 3rd Axiom, We find if A, A,... is a countably infinite sequence of disjoint events, then i j A i B) A j B) = A i A j ) B = B) = = A i B) and A j B) are independent = P r i= A i B) = i= P ra i B) steps made due to the distribution law and definition of disjointness respectively). Let us call these findings Rule Z. Next, P r i= QA i)) = P r i= A i B)) =... = P r i= A i)) B) P rb) = P ra B) A B)... ) P rb) by distribution law 3

14 = i= = P ra i B) by Rule Z P rb) P ra i B) and thus completes our proof i= 6b) B Assume that X, X,... are an i.i.d. random variables having E X j p ) < for < p. Let µ = EX j ).. Show that X n = X+ +Xn n µ in L p as n.. Show that X n µ almost surely.. First, we recognize the function, fx j ) = X j p must be convex for p, ]. Therefore, we can establish the lower bound from Jensen s Inequality: E X n p ) EX n ) p and by the Theorems of 4., plimex n )) = µ. Therefore, from Jensen s and Sec. 4., X n c s.t. c µ as n. We now establish the upper bound by E X n p ) n i= E X i p ) = also by 4. findings) X n k, k µ Therefore, combining both our upper and lower bounds, we get X n µ as n. By our previous findings, we know each of these upper and lower bounds work for plim s, but they also do work for a.s. covergance, therefore, P lim n X n = µ) = as n. 6c) C Two random variables X and Y have bivariate normal distribution if the joint density is: ) pdf X,Y x, y) = πσ x σ y ρ e ρ x µx ) σx ) ρ x µx σx ) y µy y µy )+ σy σy ). Compute marginal probability density function of X.. Show conditional distribution of Y given X = x is Nν, τ ) and find ν and τ. 3. Show that X and Y are independent if and only if CovX, Y ) =.. Proof. Let QX, Y ) = ρ ) QX, Y ) as: QX, Y ) = = [ x µx ρ ) σ x x µ x σ x [ y µ y σ y ) ρ x µx σ x ) y µy σ y ) + y µy σ y ) ). We can re-write ) ) x µ x ) ρ + ρ ) ] x µ x ) σ x σ x ) ) ] y αx) +, where αx) = µ y + ρ σ y x µ x ) σ y ρ σ x 4

15 Thus, since f X x) = f X,Y x, y)dy, we now have: We next recognize that: πσy ρ e f X x) = e x µx σx ) e y αx) σy ρ ) dy πσ x σ y ρ y αx) σy ρ ) dy the p.d.f. of the Nαx), σ y ρ ) distribution Therefore, Thus, ξx, y) = f X x) = ζx)ξx, y) = ζx) = πσy ρ e y αx) σy ρ ) dy = πσx e x µx σx ) Nµ x, σ x). Proof. We recall: f Y X y x) = f X,Y x,y) f X x). Thus, from part, we can immediately substitute in f X x) = ζx) and f X,Y x, y): f Y X y x) = [πσ xσ y ρ ] e = πσy ρ e [ x µx σx [ πσ x ] e x µx σx ) y αx) σy ) + y αx) σy ρ ) ] ρ ) Nαx), σ y ρ ) Thus, f Y X y x) does = Nν, τ), where ν = αx) and τ = σ y ρ. 3. Proof. We recall Corollary 3.5., which states two variables are independent f X,Y x, y) = f X x)f Y y). Using this information, and by the symmetry of x and y in the Bivariate Normal Distribution, we know f Y y) Nµ y, σ y). Therefore, f Y y)f X x) = e πσ x σ y [ x µx σx ) + y µy ) ] However, since ρ [, ], we have a problem with independence. If we let: gx, y, ρ) = ρ ) logπσ x σ y ) M gx, y, ρ), ρ) = e ρ gx,y,ρ) log ρ ), and x µ x σ x σy ) ρ x µ x σ x ) y µ y ) + y µ y ) ) σ y σ y Thus one can see that: M gx, y, ρ), ρ) = f X,Y x, y). Furthermore, by construction, it is impossible for M gx, y, ρ), ρ) = f X x)f Y y) unless ρ = since otherwise the ρ and log ρ ) terms will be and respectively, thereby shifting the density too much for the now non-zero ρ x µx σ x ) y µy σ y ) to bring back. Also, since ρ = CovX, Y ) = = X and Y are independent CovX, Y ) = 5

16 6d) D Suppose Y i ind. exponentialµ i ) where µ i = βx i where β > and x i >.. Find β maximizing: n pdf Yi y i ). Show that the ˆβ found in ) converges to β in probability. 3. Show that Var ˆβ). i=. Proof. We know f Yi y i ) = βx i e βx y i i Therefore, by standard MLE practices, we have L = n i= [ e βx y i n i = logl ) = log βx i β n i= n And Maximizing: n logβ) + log logl ) β = n β + n β n i= x i= i x i ) β y i x i = = ˆβ = n ) ] e n y i β i= x i i= y i x i = y i x i= i x iµ i n i=. First, we know E ˆβ) = E n y i n i= x i ) = n E n y i i= x i ) = yi n E x i ) = n nβ) = β. By theorems from 4., and by Chebyshev s Inequality, and noting EY i ) = /µ i = EY i /x i ) = = βxi x i = β. 3. Since there is only one parameter within an exponential distribution, and σ = µ for exponential = ˆβ = ˆσ. Since we already showed that plim ˆβ) = β = plim ˆβ ) = β. Next, we apply Chebyshev s Inequality, which results in the fact that Var ˆβ) as n else our previous findings would not hold under such assumptions. 6