Spring 2016 Network Science

Transcription

1 Spring 206 Network Science Sample Problems for Quiz I Problem [The Application of An one-imensional Poisson Process] Suppose that the number of typographical errors in a new text is Poisson istribute with mean λ Two proofreaers inepenently rea the text Suppose that each error is inepenently foun by proofreaer i with probability p i, i =, 2 Let X enote the number of errors that are foun by proofreaer but not by proofreaer 2 Let X 2 enote the number of errors that are foun by proofreaer 2 but not by proofreaer Let X enote the number of errors that are foun by both proofreaers Finally, let X enote the number of errors foun by neither proofreaer (a) Describe the joint probability istribution of X, X 2, X, X Sol: Each of a Poisson numbers of event is classifie as either being of type I (if foun by proofreaer, but not by proofreaer 2), type 2 (if foun by proofreaer 2 but not by proofreaer ) or type (if foun by both), or type (if foun by neither) So the {X i } are inepenent Poisson ranom variables with means E[X ] = λp ( p 2 ), E[X 2 ] = λ( p )p 2, E[X ] = λp p 2, E[X ] = λ( p )( p 2 ) (b) Show that Sol: So we can show E[X ] E[X ] = p 2 p 2 an E[X 2] E[X ] = p p E[X ] E[X ] E[X 2 ] E[X ] = λp ( p 2 ) λp p 2 = p 2 p 2, = λ( p )p 2 λp p 2 = p p (c) By using X i as an estimator of E[X i ], i =, 2,, present estimators of p, p 2, an λ Sol: Using that E[X 2] = E[X ] p = X 2 X, we get X 2 + X X p p X X 2 + X So p is estimate by the fraction of the errors foun by proofreaer 2 that are also X foun by proofreaer Similarly, we can estimate p 2 by X +X The total number of errors foun is X + X 2 + X that has mean [ ] E[X + X 2 + X ] = λ[ ( p )( p 2 )] λ (X 2 + X )(X + X ) Hence, we can estimate λ by λ X + X 2 + X (X 2 +X )(X +X )

2 () Give an estimator of X, the number of errors not foun by either proofreaer Sol: Since λ is the expecte total number of errors, we can use the estimator of λ to estimate E[X ] That is, E[X + X 2 + X + X ] = λ E[X ] = λ E[X + X 2 + X ], E[X ] = (X 2 + X )(X + X ) λ = (X + X 2 + X ) [(X 2 + X )(X + X ) ] Problem 2 [Poisson Ranom Network] (a) What are the two unrealistic rawbacks of the Erös-Renyi ranom graph moel for social networks? What moels can fix these two rawbacks? Briefly explain why the moels you think can fix the rawbacks Sol: The the Erös-Renyi network moel has two rawbacks one is short path, the other is low clustering Using configuration moel can mitigate the short path problem since it can make sure that each noe has certain number of egree with high probability The small worl network moel such as a ring can fix the low clustering problem since all noes are originally places on a ring or a limite space (b) What is the threshol function for starting to observe at least one triangle in the Erös- Renyi network moel? Sol: The mean number of triangles in the network can be approximate by ( ) n E[number of triangles] p (n) n p (n), for n Thus, the threshol function is t(n) = /n (c) What is the threshol function for observing at least one four-noe component in the Erös-Renyi network moel? Sol:The mean number of -noe components in the network can be approximate by ( ) n E[number of triangles] p (n) n p (n), for n Thus, the threshol function is t(n) = n / () Consier a growing network process such that a newborn noe uniformly an ranomly forms m > links connecting to all other noes by preferential attachment Also, in any given perio, m links are estroye an they are selecte uniformly at ranom out of all links that exist at the beginning of the perio Fin the egree istribution (ie CDF of the egree) using a continuous time mean-fiel approximation What is the corresponing probability ensity function P ()? Is the mean of the egree finite? Sol: The mean-fiel expression for the change in the egree of a noe i < t over time can be written as i (t) = m i(t) mt/2 m i(t) mt 2 = i(t) t,

3 which has the following solution i (t) = m ( ) t i Thus, ( m ) F t () = P () = m So the istribution of the number of egrees follows a power law istribution = P () is finite So Problem (25%) [SIS Moel] Consier the SIS moel of isease iffusion Let ρ (t) enote the fraction of noes of egree infecte at time t in a society an θ(t) enote the probability that a given meeting is with an infecte iniviual at time t If P () is the egree istribution in the society, the probability that a meeting of noe i is with a egree noe is P ()/ Let v enote the transmission rate of infection an δ enote the recovery rate of an infecte iniviual (a) Assume the probability that a susceptible agent with egree becomes infecte in a perio [t, t+ɛ] is ɛvθ(t) α where α 0, an the fraction that recover to become susceptible again is ρ (t)δ/ Using the continuous-time mean fiel approximation, write own the ifferential equation of ρ (t) with respect to t (ie, ρ (t) =?) an solve this ifferential equation in the steay state (ie, fin ρ ( )) What is θ in the steay state? Sol: The ifferential equation of ρ (t) wrt t is given by ρ (t) In the steay state, we have ρ (t) ρ ( ) = vθα+ = ( ρ (t))vθ α ρ (t)δ = 0 an thus δ + vθ α+ an θ = P ()vθ α+2 (δ + vθ α+ ) (b) Let λ = v Fin λ if the egree istribution is regular (ie all noes have a fixe egree δ ) Sol: If the egree istribution is regular, we have θ = α+2 λθ ( + λθ α+ ) θ = (α+) > 0 λ = λ α+ ( θ) (c) Let H(θ) = P ()vθ α+2 (δ + vθ α+ ) = P ()λθ α+2 ( + λθ α+ ) What is the constrain on λ that results in a positive steay state infection rate θ? Sol: We know H (0) = λ α+2 >

4 is the conition to have a positive fixe point of H(θ) Hence, Problem [Game Theory] λ > α+2 (a) In the static form game below, (i) what is the values of x that make (M, L) be a pure strategy Nash Equilibrium? (ii) what values of y o Player have a strictly ominate strategy where the strategy is ominate by a pure strategy? (iii) if x = 5 an y =, list all pure strategy Nash equilibria Sol: (i) x (ii) y < 0 (iii) There is only one NE which is (M,L) Player 2 Player Left Centre Right Top (, 0) (2, -2) (y, -) Mile (, ) (, 0) (-, 0) Bottom (2, x) (5, 2) (0, ) (b) Consier the following game of Cournot Competition, which moels two firms proucing the same homogeneous goo an seeking to maximize their profits A utility function u i for each player i given by its total revenue minus its total cost, ie, u i (s, s 2 ) = s i p(s i, s i ) 2 s i, i =, 2, where s i is the amount of goo that firm i prouces, p(s i, s i ) = max{0, s i s i } is the price function for the total amount of goo an s i /2 is the cost for firm i For firm i, the best-response corresponence B i (s i ) is given by B i (s i ) = arg max s i 0 { s i p(s i, s i ) 2 s i } Fin B i (s i ) What is the Nash equilibrium for these two firms? Sol: B i (s i ) can be solve as follows { 5 B i (s i ) = s 2 i, s i 5 2, for i =, 2 0, otherwise Thus we know the NE shoul be on the cross point of the two lines B (s 2 ) an B 2 (s ) an it is ( 5 6, 5 6 ) Problem 5 [Nash Equilibrium] (a) Define what is meant by a strictly ominate strategy Sol: A strategy is strictly ominate if it is inferior to another for EVERY strategy of the other player

5 (b) In the game above, etermine if either player has a strictly ominate strategy Be sure to state precisely how one or more strategies ominate another Sol: First, we notice that Player 2 oes not have a strictly ominate strategy Although Player 2 oes not have a strictly an purely ominate strategy, Player s action B is ominate by C or a mixture of A an C For example, suppose Player puts probability p an p on A an C, respectively If player 2 plays Q, we require 0p + 8( p) > 7 or p < /8 If player 2 plays R, we require 8p + 6( p) > 6 This hols for all p > 0 Thus, for 0 < p < /8, the mixture of A an C ominates B (c) Fin all Nash Equilibria of the game above Sol: Given that players never play strictly ominate strategies, we can remove action B from consieration So we see that there are two pure strategy NE at (C,Q) an (A,R) 5