Problem Set 9 Due December, 7

Transcription

1 EE226: Random Proesses in Sysems Leurer: Jean C. Walrand Problem Se 9 Due Deember, 7 Fall 6 GSI: Assane Gueye his problem se essenially reviews Convergene and Renewal proesses. No all exerises are o be urned in. Only hose wih he sign are due on hursday, Deember 7 h a he beginning of he lass. Alhough he remaining exerises are no graded, you are enouraged o go hrough hem. We will disuss some of he exerises during disussion seions. Please feel free o poin ou errors and noions ha need o be larified. Exerise 9.. Assume ha X n onverges in probabiliy o X and f( ) is a oninuous bounded funion. Show ha f(x n ) onverges in probabiliy o f(x) We wan o show ha ɛ > P r{ω : f(x n (ω)) f(x(ω)) > ɛ} X n onverges in probabiliy o X implies ha ɛ > P r{ω : X n (ω) X(ω) > ɛ} Sine f is oninuous, we also have ha for all ω and ɛ, δ > here is an n(w) > suh ha X n (ω) X(ω) < δ f(x n (ω)) f(x(ω)) < ɛ or f(x n (ω)) f(x(ω)) ɛ X n (ω) X(ω) δ hus, leing N = max{n(w)}, we have for n > N {ω : f(x n (ω)) f(x(ω)) ɛ } {ω : X n (ω) X(ω) δ } and P {ω : f(x n (ω)) f(x(ω)) ɛ } P {ω : X n (ω) X(ω) δ } whih ends he proof. Exerise 9.2. Bonus Assume ha X n onverges in disribuion o some random variable X. Show ha we an find (Y n, Y ), where Y n is a sequene of random variables suh ha Y n has same disribuion as X n, Y has he same disribuion as X, and Y n onverges almos surely (a.s) o Y. 9-

2 EE226 Problem Se 9 Due Deember, 7 Fall 6 Le Ω = (, ) be our even spae and for ω uniformly piked in Ω le Y n (w) = sup{x : F Xn (x) < ω}. Y n has disribuion F Xn. We wan o show ha Y n (ω) Y (ω) for all bu a ounable number of ω s, where Y d X. o show ha, le us define for all ω, a ω = sup{x : F X (x) < ω}, b ω = inf{x : F X (x) > ω}, and Ω o = {x : (a ω, b ω ) = } where (a ω, b ω ) is an open inerval wih he given end poin. In words, Ω o onains all ω Ω suh ha lim inf F (ω) = lim sup F (ω). Now le Y (ω) = F X (ω), ω Ω o. Ω Ω o is ounable sine he inervals (a ω, b ω ) are disjoin (beause of he use of sup and inf and eah nonempy inerval onains a differen raional number; reall ha he se of raional numbers is ounable). Now for any ω Ω o sup F X n (ω) F (ω) inf F X n (ω) aking he limi n we have Y n (ω) = F X n (ω) Y (ω), ω Ω o. Exerise 9.3. In he noes we have shown ha if X n onverges in probabiliy o X, hen i onverges in disribuion o X. Show ha, onversely, if X n onverges in disribuion o a onsan C, hen i onverges in probabiliy o C. Sine X n onverges in disribuion o C, we have Now le us ompue F Xn (x) δ(x > C), x P r[ X n C > ɛ] = P r[ X n C ɛ] hus X n onverges in probabiliy o C. Exerise 9.4. Problem 2.2 of he ourse noes. = P r[c ɛ < X n < C + ɛ] = (F Xn (C + ɛ) F Xn (C ɛ)) n ( ) Noe ha sine ɛ n, for all ɛ > here exiss n (k, m) suh ha ɛ n < ɛ and for all k, m n P r[ X k X m > ɛ] P r[ X k X m > ɛ n ] 2 n 9-2

3 EE226 Problem Se 9 Due Deember, 7 Fall 6 Le Z n (k, m) = X k X m for all k, m n. hen P r[z n (k, m) > ɛ] = P r[z n (k, m) > ɛ] + n < n n (k,m) n n (k,m) P r[z n (k, m) > ɛ] + n>n (k,m) n>n (k,m) P r[z n (k, m) > ɛ] hus Borel-Canelli Lemma implies ha P r[z n (k, m) > ɛ, i.o] =. his ells ha for all ɛ >, and for all ω Ω, here exiss n(ɛ, ω) = max{n (k, m)} > suh ha So X k (w) X m (w) < ɛ, k, m n(ɛ, ω) sup X k (w) X m (w) < ɛ, k, m n(ɛ, ω) m,k Hene X n is Cauhy a.s. Sine he se of real number is omplee, X n onverges almos surely o a limi in R. Exerise 9.5. Problem 2.9 of he ourse noes. For any given sae i, le N i () be he proess ha oun he number of arrivals when he CMC X() is in sae i. N i () is a Poisson proess wih rae λ i. he proporion of ime ha X() is in sae i is given by 2 n So We also have ha π i = lim N i () N() = i [Xs=i]ds λ i π i N i () Hene N() lim = lim = i N i () i lim N i () (9.) = i λ i π i 9-3

4 EE226 Problem Se 9 Due Deember, 7 Fall 6 where o ge equaion 9. we use he fa ha X() is posiive reurren and does no explode, hus N() mus be finie (hus i is bounded and we an use he dominaed onvergene heorem o jusify swapping he lim and he sum). Exerise 9.6. Reall ha in he sudy of renewal proess we defined he iner-arrival imes = i+ i, i o be iid wih disribuion F (). Le f() = ( + ) 2 o be he orresponding pdf. Find E[τ] where τ is he ime unil he nex jump for he saionary proess, and λ he rae of jumps. his exerise misses he poin i was supposed o make. Answer he he nex quesion: Find a disribuion for whih < λ < and E[τ] =. Hin A simple example is f() = ( + ) 3 Find he onsan and see nex exerise for a more ineresing disussion. Exerise 9.7. In he derivaion of E[τ] in lass we wroe E[τ] = = λ 2 = λ 2 λ( F ())d λ( F ())d 2 [ 2 ( F ()) ] + λ 2 2 f()d (9.2) We laimed ha he firs erm in he RHS of equaion 9.2 vanishes beause he mean of = 2 (iner-arrival ime) should be finie. his argumen is no quie orre; show he orre argumen ha is: 2 ( F ()) as if and only if E[ 2 ] <. hanks o all he suden for being auious/pessimisi in his exerise. 2 ( F ()) as if E[ 2 ] <. One ounerexample is: F () =, <, and F () = 9-4 ( + ) 2 ln( + ),

5 EE226 Problem Se 9 Due Deember, 7 Fall 6 I is easy o verify ha 2 ( F ()) as if E[ 2 ] =. If however E[ 2 ] < we have 2 ( F ()) = 2 f(s)ds Bu sine s 2 f(s)ds < = lim 2 f(s)ds s 2 f(s)ds s 2 f(s)ds = bse < s Exerise 9.8. Bonus Considering again he renewal proess seing given in lass, show ha if he iner-arrival imes are iid uniform in [, ], hen ɛ-oupling ours in finie ime. o show his we will sar wo proesses, one (N s ()) wih he saionary disribuion as iniial disribuion, and he oher (N x ()) wih some oher iniial disribuion, hen we will show ha a some pariular imes (e.g. jus afer jumps of N x ()), here is a posiive probabiliy ha he 2 proesses jump wihin and inerval of lengh ɛ. Le he jump imes of he proess N x () denoed, 2,... and onsider some pariular ime i. Le a (, ) be he ime spen by proess N s () sine he mos reen jump and le s ompue he probabiliy ha here will be a jump in he nex ɛ seond ( i, i + ɛ). Given ha he proess N s () did no jump in he firs a seond, he ime arrival of he nex jump is uniformly disribued in (a, ). hus he probabiliy ha his jump happens in he nex ɛ seond is P r[τ ( i, i + ɛ)] = P r[τ (a, ) τ > a] = ɛ a > ɛ Hene he probabiliy ha he wo proesses do no jump wihin ɛ seond is less han ɛ. Sine we have a renewal proess and here is an infinie number of jumps, he probabiliy ha he 2 proesses do no ouple in finie ime is P r[no ouple] = ( ɛ) = hus wih probabiliy, here is ɛ-oupling in finie ime. Exerise 9.9. In lass we have shown ha for a posiive reurren oninuous-ime Markov hain wih rae marix Q, and invarian disribuion π, we have [x =i]d as j 9-5 [j=i] π j = π i

6 EE226 Problem Se 9 Due Deember, 7 Fall 6 Le f : X R be a bounded funion from he sae spae X o he real line R. Show ha f(x )d as f(j)π j j Noe ha we have f(x )d = = j f(j) [ x = j]d j f(j) [ x = j]d (9.3) We know ha [x =i]d as π i aking he limi as in equaion 9.3, we have for any finie sae Markov hain wih number of saes N N f(j) N [ x = j]d as f(j)π j j= beause f( ) is a bounded funion, so we an safely hange he order of he limi and he summaion using he Dominaed Convergene heorem. For a general posiive reurren Markov hain, we will use some kind of runaion argumen. he inuiion is ha for large enough, here is a finie number of visied saes, and all saes j ha have no been visied mus have very small π j. Le N be he se of saes visied by ime we have: j= f(j) j= [ x = j]d = j N f(j) [ x = j]d and f(j)π j = f(j)π j j= j N j N f(j)π j + We jus need o show ha he seond erm in he RHS goes o zero as. Now noie ha: 9-6

7 EE226 Problem Se 9 Due Deember, 7 Fall 6 - N is an inreasing se 2- he saes in N are suh ha f(j)π j M π j j N j N where we have used he fas ha f( ) is bounded and ha he saes in N and smaller π j s Combining his wih he previous remark (finie sae), we ge he resul. have smaller Exerise 9.. In Prof. X s group, John Lazy, a very daydreaming nework manager has se up a priner wihou queue. Any reques ha finds he priner busy (i.e. already prining) is jus los. Assume ha requess arrive a he priner aording o a Poisson proess wih rae λ, and he amoun of ime needed o prin a reques is a random variable having disribuion G wih mean µ G and independen for eah reques. (a) Wha is he rae a whih requess are aeped (i.e. requess ge prined)? (b) Wha is he proporion of saisfied requess? Compue i for λ = 2 requess per seond and µ G = 2 seonds. (a) Beause of he memoryless propery of he Poisson proess, he mean ime beween enering requess is µ = λ + µ G (mean ime i akes for a reques o arrive plus mean servie ime). Hene he rae a whih requess are aeped is µ = λ + λµ G (b) Requess arrive a rae λ and are aeped wih rae /µ. So he fraion of aeped reques is given by f = /µ λ = λ + λµ G λ = + λµ G For λ = µ G = 2 we have f = /5 meaning ha ou of 5 requess is aeped. GOOD LUCK! 9-7