Final Exam: Probability Theory (ANSWERS)

Transcription

1 Final Exam: Probability Theory ANSWERS) IST Austria February :00-1:30) Instructions: i) This is a closed book exam ii) You have to justify your answers Unjustified results even if correct will not be accepted iii) The exam is an individual activity collaboration will not be tolerated iv) You can solve the problems in any order You do not have to solve all problems to receive the highest grade Partial credits can be given v) The total working time is 150 minutes vi) Write legibly Clearly cross out the parts that you deem wrong Question 1 A football team consists of 0 offensive and 0 defensive players The players are to be paired in groups of for the purpose of determining roommates If the pairing is done at random: a) What is the probability that there are no offensive-defensive roommate pairs? b) What is the probability that there are 10 offensive-defensive roommate pairs? The answer should be expressed in terms of factorials or binomial coefficients Solution 1 a) There are 40 39/ ways to select the first pair 38 37/ ways to select the second pair etc This way we see that there are 40!/ 0 ordered pairs The number of ways to split the players into unordered pairs is thus: 40! 0 0! There will not be offensive-defensive pairs if and only if these two subsets of players are paired among themselves It follows that there are 0! 10 10! such divisions Hence the probability of no offensive-defensive roommate pairs call it P 0 is given by P 0 0! 10 10! 40! 0 0! 0!)3 10! 40! b) To determine P k the probability that there are k offensive-defensive pairs we first note that there are 0 k ways of selecting k offensive players and k defensive players who are to be in the o-d pairs These 4k players can then be paired up into k)! possible o-d pairs This is so because the first offensive player can 1

2 be paired with any of the k defensive players the second offensive player with any of the remaining k 1 defensive players and so on) As the remaining 0 k offensive and defensive) players must be paired among themselves it follows that there are divisions which lead to k o-d pairs Hence Question 0 0 k)! k)! k 10 k 10 k)! P k 0 k)! 0 k)! k 10 k 10 k)! 40! 0 0! Suppose there are two manufacturers A and B of a smartphone battery The factory A manufactures the fraction p 0 1) of all the batteries used in the smartphones while the factory B manufactures the rest Assume that the lifetime of each battery is a continuous random variable T λ parametrised by λ > 0 with a density function: { f λ t) : λ 1 t λ ) when 0 t λ ; 0 otherwise The parameter λ depends on the manufacturer of the battery If the battery is manufactured by A then λ 1 while if it is manufactured by B then λ Suppose you buy a new smartphone that is randomly picked among all the produced phones a) Compute the expectation and variance of the lifetime of the battery of your new phone? b) Suppose the battery functions properly after time t > 0 What is the probability that your phone has a battery manufactured by A? Solution a) The lifetime of the battery in a new phone is a random variable T : T K where K is a random variable independent of T λ s with P{K 1} p and P{K } 1 p Since T 1 T and K are independent ET m ) p ET m 1 ) + 1 p) ET m ) m 1 Now ET m λ ) λ Using this in the previous formula yields: The variance is hence λ 0 t m 1 t λ ) dt ET ) p p) 3 3 p 3 ET ) p p) 6 1 λ m m + )m + 1) 4 3p 6 σ : ET ) ET 34 3p) p) 18 4 p p 18 b) As T 1 1 it remains to consider t [0 1] Since K is independent of T λ s PT λ t K k) PT λ t) Bayes formula yields: P "battery manufactured by A" "phone has lasted time t" ) PK 1 T t) PT K t K 1) PT t) p PT 1 t) p PT 1 t) + 1 p)pt t) 1)

3 Now it remains to compute the probabilities Clearly PT λ t) 0 if t λ Otherwise λ PT λ t) 1 s λ t λ λ t) λ ) ds λ 1 λ + t λ ) [λ t) λ t ] In particular PT 1 t) 1 t and PT t) 1 t/ Plugging these numbers in 1) yields P "battery manufactured by A" "phone has lasted time t" ) { p 1 t p 1 t +1 p) 1 t/) when t [0 1] ; 0 when t [1 ] Question 3 Individuals of age 50 have a probability 10 5 of having a certain type of cancer A test is developed with a false positive rate of 10 3 ie it diagnoses cancer in a healthy individual once in 1000 times) Conversely it has a false negative rate of 10% ie it diagnoses 10% of individuals with cancer as healthy) If the cancer is detected early there is a 90% chance of survival after treatment whereas if untreated it is always fatal However the treatment itself is risky and kills 1% of healthy patients a) What fraction of positive diagnoses are correct? b) Would screening a large population using this test save lives? c) The incidence of this cancer increases with age; the test might be more effective when applied to an older age group Assuming that the other parameters stay the same at what incidence would the test be worthwhile? λ Solution 3 a) Using Bayes formula we obtain: P "cancer" "diagonosis" ) P "cancer" "diagnosis" ) P "diagnosis" ) ) b) The screening would not safe lives: The expected number of people with cancer saved per individual screened would be The number of people who would die because they were treated after mis-diagnosis is ) which is larger c) Substituting an incidence rate I we have I lives saved and 1 I) dying through mis-diagnosis Thus screening will save lives overall if I > However this does not account for the costs financial and human) of the test and treatment Question 4 A bacterium divides in two at a rate µ and dies at rate λ per unit time All bacteria are equivalent: that is they do not age Initially there is a single bacterium a) What is the expected number at time t? b) What is the probability that a single bacterium will found a growing population? c) Suppose µ > λ After a long time what is the expected number of bacteria conditional on survival of the population? HINT: Consider what happens in a small time interval δt so small that only at most one event can happen namely division with probability µδt or death with probability λδt HINT: You might find useful that 1 + a δt) t/δt e at as δt 0 3

4 Solution 4 a) Dividing time into small time units δt we end up with a branching process where each node has two children with probability µ δt one child with probability 1 µ+λ) δt and no child with probability λ δt The difference to the branching process in the lecture is that the population size at time kδt is the number of nodes in the k-th layer of the tree The expected number of children of any given node is µ δt+1 1 µ δt λ δt)+0 λ δt 1 + µ λ)δt To obtain the expected number at time t we let tδt time units pass and obtain the expectation 1 + µ λ)δt) t/δt e µ λ)t as δt 0 b) Let the probability of ultimate extinction starting from a single cell be Q In time δt there is a probability λ δt of death in which case extinction is certain and a probability µ δt of division in which case the chance that both offspring leave no descendants is Q Therefore conditioning on the first time step we obtain and so Q λ δt + 1 λ + µ) δt) Q + µδt Q 0 λ λ + µ)q + Q This is solved by Q 1 and Q λ/µ Therefore the probability of ultimate survival is 0 if µ < λ µ λ)/µ otherwise c) As the time t approaches infinite the fate of the population becomes certain expected size of the population is µ λ µ N t) For very large times t the where N t) is the expected size of the population at time t conditional on survival Combining this with the part a) we get N t) µ µ λ eµ λ)t Question 5 Consider an Erdös-Rényi random graph with n vertices that we label by the integers 1 n Suppose any given edge is drawn in the graph with probability p 0 1) We say that there exists a path of length l connecting two given vertices v w if there exist vertices v 1 v l 1 such that v v 1 v v l 1 w where v w means that there is an edge between v and w For each 1 l n define the random variable X l : "number of paths of length l connecting the vertices 1 and " a) Compute the expectation of X l for 1 l n b) What is the probability that X 0? c) What is the probability that X 3 0? d) Suppose n 198 and p 1/ Estimate PX < 84) using the principle of central limit theorem HINT: You may find it useful that if F x) : 1 π x e t / dt then F ) 0977 Solution 5 a) By definitions we have X l : i 1i l 1 1{1 v 1 v 1 v v l 1 } where the sum is over ordered but distinct vertices v i taken from the set {3 n} Taking expectation yields EX l ) P{1 v 1 v 1 v v l 1 } i 1i l 1 P{1 v 1 }P{v 1 v } P{v l 1 } i 1i l 1 n )n 3) n l) p l 4

5 b) Let v v mean that the there is no edge connecting the vertices v and v Using independence of the edges we compute: P { X 0 } P n { } ) 1 v or v v3 n ) 1 P{1 v}p{v } v3 1 p ) n c) Analogously P { X 3 0 } n P n v3 v3 n { } ) 1 v or v w or w w3 : w v n ) 1 P{1 v}p{v w}p{w } w3 : w v 1 p 3 ) n )n 3) d) The random variables Z v : 1{1 v } appearing in the sum X Z v v3 are iid Hence the central limit theorem says that their sum X is close to a Gaussian random variable G with the same expectation and variance as X provided n is large enough that n is large enough is assumed here) To this end we compute: EX ) vw3 E 1{1 v }1{1 w } ) P{1 v}p{v } + v3 The variance of X is hence vw3 : v w n )p + n )n 3)p 4 σ : EX ) EX P{1 v}p{v }P{1 w}p{w } [ n )p + n )n 3)p 4] [ n )p ] n )1 p )p Plugging in n 198 and p 1/ yields nice round numbers: 1 EX ) 198 ) 98 σ 198 )1 1/)1/) 7 Now we use use the central limit theorem to approximate X by a standard Gaussian random variable G: P { X < 84 } { } X P < P{G < } 1 P{G } Note that here the symmetry of the normal distribution has been used 5

6 Question 6 A sample of k genes is taken from a large population of constant effective size N e genes There is a constant rate of mutation µ and every mutation is distinct from all the others the "infinite sites" model) a) What is the expected number of mutations that will be seen in the sample for k 3 10? b) What is the variance of the number of mutations for k 3 10? c) How does the standard deviation divided by the mean number of mutations change as k becomes large? HINT: You may find the following formulas useful: k j1 Here γ 057 a is numerical constant Solution 6 1 j γ + ln k and 1 j π 6 a) With j lineages present coalescence in the next generation happens with probability jj 1) N e The coalescence time of j lineages down to j 1 is thus given by an exponential distribution with mean N e jj 1) This time during which there are j lineages contributes j N e j j 1) to the expected length of the genealogy Summing from k down to gives the expected length as k j N e j 1 The expected number of mutations is obtained by multiplying with µ For k 3 10 this is N e µ N e µ 15 N e µ 83; for large k we have N e µ γ + ln k) b) Similarly the variance of length contributed by the time during which there are j lineages is the variance of the above exponential distribution Ne jj 1) and we obtain for the number of mutations k Ne µ j 1 For k 3 10 this is N e µ N e µ 15 N e µ 154; for large k this converges to j N e µ π 6 Note: The intervals between successive coalescence events are independent and hence we can just sum the variance c) The standard deviation divided by the mean is for k 3 10 and tends to for large k π 6γ + ln k) j1 6

7 Question 7 Challenge) Consider n arbitrary vectors v 1 v n R n of unit length v i n vi 1 i 1 n i1 Show that there exists n numbers ɛ i { 1 +1} such that ɛ1 v 1 + ɛ v + + ɛ n v n n ) HINT: Randomise the problem ie consider random coefficients ɛ i and compute the expectation of an appropriate quantity Solution 7 Let ɛ i s to be iid random variables with Pɛ i ±1) 1/ and set X : ɛ 1 v 1 + ɛ v + + ɛ n v n Recall that v v v Hence using the linearity of v w wrt the arguments v and w we get X ɛ i ɛ j v i v j ij1 Since ɛ i s are independent Eɛ i ɛ j ) δ ij taking expectation yields: EX) v i v j Eɛ i ɛ j ) ij1 v i n Since X 0 and EX) is just a convex combination of the term on the left hand side of ) evaluated over all possible choice of the tuples ɛ 1 ɛ n ) { 1 +1} n there must exists at least one choice of ɛ 1 ɛ n) such that X n Actually there must also be another one such that X n holds) Taking square roots completes the argument i1 7