CS 798: Homework Assignment 2 (Probability)

Transcription

1 0 Sample space Assgned: September 30, 2009 In the IEEE 802 protocol, the congeston wndow (CW) parameter s used as follows: ntally, a termnal wats for a random tme perod (called backoff) chosen n the range [, 2 CW ] before sendng a packet If an acknowledgement for the packet s not receved n tme, then CW s doubled, and the process s repeated, untl CW reaches the value CWMAX The ntal value of CW s CWMIN What s the sample space for (a) the value of CW? ( the value of the the backoff? Soluton: The sample space for CW s the dscrete set {CWMIN, 2* CWMIN, 4* CWMIN, 2 n *K*CWMIN}, where K s chosen so that 2 n *K*CWMIN < CWMAX The sample space for backoff, gven CW s a subset of the real lne defned by [0, CW] 20 Interpretatons of probablty Consder the statement: gven the condtons rght now, the probablty of a snowstorm tomorrow mornng s 25% How would you nterpret ths statement from the perspectve of an objectve, frequentst, and subjectve nterpretaton of probablty (assumng these are possble)? Soluton: An objectve nterpretaton would be that we have a complete weather model that has an ntrsc source of randomness Gven ths model and the current weather condtons, the model predcts that the probablty of a snowstorm s 25% A frequentst approach would be to look at all pror days where today s weather condtons also held, and look at the number of such days where there was a snowstorm the next mornng We would see that 25% of the tme, gven the current weather, there was as snowstorm A subjectve nterpretaton would be that an expert, who knew all the varables, would take 4: odds (or bettter) on a bet that t would snow tomorrow 30 Condtonal probablty Consder a devce that samples packets on a lnk (a) Suppose that measurements show that 20% of packets are UDP, and that 0% of all packets are UDP packets wth a packet sze of 00 byteswhat s the condtonal probablty that a UDP packet has sze 00 bytes? ( Suppose 50% of packets were UDP, and 50% of UDP packets were 00 bytes long What fracton of all packets are 00 byte UDP packets? Soluton: (a) We have P(UDP) = 02, and P(UDP AND 00) = 0 So, P(00 UDP) = 0/02 = 05 ( Here, P(UDP) = 05 and P(00 UDP) = 05 So, P(00 AND UDP) = 05*05 = Condtonal probablty agan Contnung wth Ex 3: How does the knowledge of the protocol type change the sample space of possble packet lengths? In other words, what s the sample space before and after you know the protocol type of a packet? Soluton: Before you know the protocol type of a packet, the sample space s all possble packet lengths of all possble protocol types After you know the protocol type, the sample space only nclude packet lengths for that protocol Page of 5

2 50 Bayes rule For Exercse 3(a), what addtonal nformaton do you need to compute P(UDP 00)? Settng that value to x, express P(UDP 00) n terms of x Soluton: P(UDP 00) = (P(00 UDP)P(UDP))/P(00) We need P(00) = x Then, P(UDP 00) = 05*02/x = 0/x 60 Cumulatve dstrbuton functon (a) Suppose dscrete random varable D take values {, 2, 3,,,} wth probablty /2 What s ts CDF? ( Suppose contnuous random varable C s unform n the range [x, ] Whats s ts CDF? Soluton: (a) F D () = --- = -2 - ( f C(x) = x x, so F C (x) = = x dx x x 70 Expectatons Compute the expectatons of the D and C n Exercse 6 Soluton: (a) E[D] = --- ( By geometry, E[C] = ( +x )/2 (you can also derve ths analytcally) 80 Varance 2 j j = Prove that V[aX] = a 2 V[X] 2 j j = x x Soluton: V[aX] = E[a 2 X 2 ] - (E[aX]) 2 = a 2 (E[X 2 ] - (E[X]) 2 ) = a 2 V[X] 90 Bernoull dstrbuton A hotel has 20 guest rooms Assumng outgong calls are ndependent and that a guest room makes 0 mnutes worth of outgong calls durng the busest hour of the day, what s the probablty that 5 calls are smultaneously actve durng the busest hour? What s the probablty of 5 smultaneous calls? Soluton: Consder the event E defned as Room X s makng an outgong call durng the busy hour Clearly, P(E) =p = /6 The probablty of 5 smultaneous calls s and of 5 smultaneous calls s = = 4*0 8 Page 2 of 5

3 00 Geometrc dstrbuton Consder a lnk that has a packet loss rate of 0% Suppose that every packet transmsson has to be acknowledged Compute the expected number of data transmssons for a successful packet+ack transfer Soluton: Packet and ack transmssons are geometrcally dstrbuted wth parameter p=0 So the expected number of packet transmssons s /p = 0 and the expected number of ack transmssons s also 0 These are ndependent events, so the expected number of data transmssons for successful packet+ack transfer = 0+0 = 20 0 Posson dstrbuton Consder a bnomally dstrbuted random varable X wth parameters n=0, p=0 (a) Compute the value of P(X=8) usng both the bnomal dstrbuton and the Posson approxmaton ( Repeat for n=00, p=0 Soluton: (a) Usng the bnomal dstrbuton, the value s 0 8 ( 0 )( 09 2 ) = 36*0-6 For the Posson 8 approxmaton, λ=, so the value s PX ( = 8) = e = 89*0-6 ( Usng the bnomal dstrbuton, the 8! value s 00 8 ( 0 )( ) = 4 For the Posson approxmaton, λ= 0, so the value s 8 PX ( = 8) = e = 2 It s clear that as n ncreases, the approxmaton greatly mproves 8! 20 Gaussan dstrbuton Prove that f X s Gaussan wth parameters ( μ, σ 2 ), then the random varable Y=aX + b, where a and b are constants, s also Gaussan, wth parameters( aμ + b, ( aσ) 2 ) Soluton: Consder the cumulatve dstrbuton of Y = F Y (y) = PY ( y) PaX ( + b y) P ( y X ( y = = = F a X f a > 0 a Then, f Y (y) = F Y (y) = F X ( y --f ( a a x y 2σ 2 2a a 2 σ 2 2a 2 σ 2 = = = = Comparng wth the standard defnton of a Gaussan, we see that the parameters of Y are ( aμ + b, ( aσ) 2 ) A smlar calculaton holds f a < 0 30 Exponental dstrbuton y b μ 2 a ( y b aμ) ( y ( b+ aμ) ) Suppose that customers arrve to a bank wth an exponentally dstrbuted nter-arrval tme wth mean 5 mnutes A customer walks nto the bank at 3pm What s the probablty that the next customer arrves no sooner than 3:5? Soluton: We have / λ =5 We need to compute -F(5) = ( e λx ) = e 5 = e 3 = 485 % Page 3 of 5

4 40 Exponental dstrbuton It s late August and you are watchng the Persed meteor shower You are told that that the tme between meteors s exponentally dstrbuted wth a mean of 200 seconds At 0:05 pm, you see a meteor, after whch you head to the ktchen for a bowl of cecream, returnng outsde at 0:08pm How long do you expect to wat to see the next meteor? Soluton: Because the exponental dstrbuton s memoryless, the expected watng tme s the same, e 200 seconds, no matter how long your break for cecream Isn t that nce? 50 Power law Consder a power-law dstrbuton wth x mn = and n the followng table: α = 2 and an exponental dstrbuton wth λ = 2 Fll x f power_law (x) f exponental (x) Soluton: x f power_law (x) f exponental (x) * * * * * *0-87 It should now be obvous why a power-law dstrbuton s called heavy-taled! 60 Markov s nequalty Consder a random varable X that exponentally dstrbuted wth parameter λ = 2 What s the probablty that X > 0 usng (a) the exponental dstrbuton ( Markov s nequalty Soluton: (a) We need -F(0) = e -20 = 206*0-9 ( The mean of ths dstrbuton s /2 So, 05 PX ( 0) = 005 It s clear that the bound s very loose 0 70 Jont probablty dstrbuton Consder the followng probablty mass functon defned jontly over the random varables, X, Y, and Z: Page 4 of 5

5 P(000) = 005; P(00) = 005; P(00) = 0; P(0)=03;P(00) = 005; P(0) = 005; P(0) = 0; P()=03 (a) Wrte down p X, p Y,p z,p XY,p XZ,p YZ ( Are X and Y, X and Z, or Y and Z ndependent? What s the probablty that X=0 gven that Z=? Soluton: (a) p X = {05, 05}; p Y = {02, 08}; p Z = {03, 07}; p XY = {0, 04, 0, 04}; p XZ = {05, 035, 05, 035}; p YZ = {0, 0, 02, 06} ( X and Y are ndependent because p XY = p X p Y X and Z are ndependent because p XZ = p X p Z (c) P(X=0 Z=) = P(X=0 AND Z=)/P(Z=) = 035/07 = 05 Page 5 of 5