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Course: AST 204, Spring 2008
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VII: LECTURE THE EARLY UNIVERSE, FLUCTUATIONS, AND THE DEVELOPMENT OF STRUCTURE AST 204 25 Feb 2008 I. The Planck Era and the Notion of Time and Space First thing today, well discuss the fact that theres an epoch, associated with a density, a temperature, a mass, and a timescale earlier than which it makes no sense to even discuss the notions of time and space. This is caleld the Planck time. The energy of a...

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VII: LECTURE THE EARLY UNIVERSE, FLUCTUATIONS, AND THE DEVELOPMENT OF STRUCTURE AST 204 25 Feb 2008 I. The Planck Era and the Notion of Time and Space First thing today, well discuss the fact that theres an epoch, associated with a density, a temperature, a mass, and a timescale earlier than which it makes no sense to even discuss the notions of time and space. This is caleld the Planck time. The energy of a photon or other highly relativistic ( >> 1) particle is just E = h = hc/. The Schwarzschild radius associated with a black hole of mass m is rs = 2Gm/c2 in ordinary units, or, since the energy E associated with a mass m is mc2 , rs = 2GE/c4 . Thus an energetic particle so energetic that its own self-gravity makes a black hole in which its wavelength just ts has a wavelength which satises = rs = 2GE 2Gh = , c4 c3 so this critical wavelength, the Planck length, is lP = = 2Gh = 5 1033 cm c3 The energy, the Planck energy, is EP = hc = hc5 = 4 1016 erg = 2 1019 GeV 2G And, of course, an associated temperature, the Planck temperature TP = EP /k = 3 1032 K. The associated mass, the Planck mass, is mP = E/c2 = 4 105 g There is a density associated with this notion, namely one Planck mass per Planck length cubed, or 3 P = mP /lP = 3 1092 g/cm3 and a time which might be either of two expressionsthe inverse of the frequency associated with the wavelength lP or the expansion time of the universe associated with P . These two quantities are the same within a factor of two; the Planck time is usually called lp /c = 2 1043 sec. The fact that the two notions agree is profound. Can you gure out why they agree? 1 We do not know whether the universe was ever this dense, hot, or rapidly expanding, but we do know that it does not make sense to discuss conditions even more extreme than this. In order to study a more extreme condition (smaller length scales, say), one would have to resolve ner detail, which would involve particles of even shorter wavelength. But Planck energy particles already cannot propagate because they are within their own event horizon. So the concepts of space and time as a classical arena for events breaks down here, or probably actually well beforeit makes no sense to talk about the spacetime continuum on scales shorter than the Planck length and time; the universe is a fundamentally quantum gravitational entity at this epoch. The notion of even what quantum gravity is is rapidly evolving at the present time, and the whole eld of fundamental physics is currently doing a grand experiment of an unprecedented kind today. There is much hope that superstring theory will somehow produce the Holy Grail of the Theory of Everything, but the motivation now is mathematical beauty rather than any real connection with physics. We will see (hopefully within your lifetimes, but I suspect not within mine) whether this crusade will succeed. Perhaps the ination is associated with this epoch, not with the GUT epoch at all (see Aureliens lectures next week), and with the birthplace of this and possibly/probably an innity of other universes in a seething timeless quantum gravitational froth, some tiny uctuation in which spawned the whole thing. V. A Summary Table, From the Beginning Till Now We present below a table which summarizes the physical conditions in the universe from the Planck time to the present, which well use in the next few lectures. Cosmological Conditions as Functions of Time and Redshift (sec) R( ) 1+z T (K ) kT (eV ) T g/cm3 m g/cm3 R 0 uh (cm) rh (cm) mh M description 4(17) 8(12) 4(11) 10 3(7) 2(11) 2(34) 1(34) 2(36) 1(43) 2/3 1 1500 7000 5(9) 8(12) 1(15) 1(28) 2(45) 5(62) ?? 2.7 4000 2(4) 1(10) 1(13) 1(15) 1(28) ?? 1(28) 2(32) 2.3(4) 0.3 1.9 1M 1G 100G 1(15)G ?? 1(15)G 2(19)G 9(30) 1.2(20) 2.5(18) 5(5) 1.6(18) 4(26) 3(77) 3(77) 3(77) 3(92) 3(30) 1(20) 1(18) 0.2 5(8) 1(15) 1(54) (106) 2(158) ?? 4(28) 1.1(27) 1.7(26) 4(21) 1.4(17) 1.2(5) 1.5(3) 3.5(20) 8(37) ?? 4(28) 7(23) 2.4(22) (11) 1.8(4) 1.2 1.5(25) 1.5(25) 1.5(25) 5(33) 3.8(23) (18) 3(16) 130 5(12) 3(18) 7(54) 0.1 1(51) ?? present combination equal m&r nuclear reac quark-gluon baryogenesis end ination mid-ination beg. ination Planck era 1/2 eH ?? The numbers in parentheses are powers of 10, so that, for example, 1(43) = 1 1043 . The columns in the table are mostly self-explanatory, but briey are as follows: is the 2 cosmic time at the epoch in question, Rlaw the relation between the scale factor and . 1+ z the redshift factor, T the temperature in K , kT the typical thermal energy in eVlater MeV and GeV; T the total mass-energy density in g cm3 . The column headed m is the density of rest mass Actually, this is a swindle for early times, because what it is is simply the present rest mass density multiplied by (1 + z )3 , and as such is just a tracer for the mass today. Clearly early on at energies where protons and neutrons do not exist, we cannot easily calculate the rest mass density, and during ination, if the identities of protons and neutrons were carried along, the densities would become ridiculous. These entries are noted by the presence of a (*). Note that this simple bookkeeping calculation assumes that the baryons, were they conserved, continue to carry one proton mass, which is probably wrongin fact, they may carry no mass at all, because the mechanism which gives particles mass may do so only at relatively low energies. But this calculation illustrates graphically that baryon number cannot be conserved in ination. R0 uh is the comoving size of the particle horizon computed as if there were no ination referred to the present universe. At and prior to the end of the ination era, we just follow an incoming light ray which is at that horizon radius at the end of ination, which illustrates how immense the eective horizon becomes during ination. Note that the physical size of the horizon remains at the event horizon during this time, but the redshift factors become so large that the comoving distances become huge. About 55 efolds are required for the inated particle horizon to encompass the entire observable universe today, and after about 80, roughly the supposed number, the horizon is 7 orders of magnitude larger than the present observable universe. Thus any point in the universe at the end of ination had been able to receive information and send information to a volume which is now immensely larger than the present observable universe. The adjacent column, the physical size of the horizon at the relevant period, is just the comoving one divided by (1 + z ). The column mh is the rest mass, again referred to the present universe, within the horizon just the present rest mass density within the comoving radius R0 uh . We have talked about most of the phenomena identied in the table: The present, recombination, the equality of radiation and mass density, the era of nuclear reactions. Above a temperature corresponding to something like 1 GeV (though it is somewhat uncertain) baryons and mesons no longer exist and are replaced by a plasma of free quarks and gluons; at somewhat higher energies, above about 100 Gev, the electroweak symmetry is restored, and at something like these energies the processes which lead to baryogenesis probably freeze out and create the baryon number we see today. Before this, at least according to our present hazy understanding, not very much interesting happens over a very large range in temperature and expansion until the energies of the GUT era are reached. There are only the electroweak and strong forces, only the fundamental leptons, quarks, any decaying heavy remnants of the GUT era, the gauge 3 bosons responsible for the forces, including photons, Ws, Zs, and gluons. Then we hit the GUT era and ination, prior to which there may (or may not) have been a more-or-less ordinary expansion era from the Planck time. The numbers in this table are mostly, especially in the columns pertaining to the earliest times, very uncertain. The assumptions made in calculating the entries are that H0 is about 70km/s/Mpc, that 0 is unity but is made up currently of m = 0.3 and = 0.7; that baryogenesis is associated with the decay or interaction of weakly interacting particles of mass about 100 GeV, that ination is associated with the GUT energy scale at about 1015 GeV, and that the universe undergoes about 80 efolds of ination. III. Small Perturbations: The Linear Regime We will see that uctuations which arose in the ination era reappear across the particle horizon after ination ends; uctuations which enter the horizon before the end of radiation dominance at z 2500 do not grow appreciably until radiation dominance ends, though uctuations in the baryonic uid are coupled to the radiation and participate in acoustic wave motion both before the era of matter domination and after until recombination. (We will follow common practice and call the time and redshift of the switchover from radiation to matter dominance eq and zeq , though the epoch has no good name that one does not stumble over.) We believe, however, that most of the matter is dark matter which does not couple to the radiation at all, and so uctuations in it can begin to grow immediately after eq . The initial icker-noise power spectrum is modied on large scales, though it is essentially preserved in the dark matter on smaller scales. The table above shows that the horizon at the turnover is on comoving scales of about 2 1026 cm, about 70 Mpc, and that the horizon encloses a total mass or order 3 1016 solar masses. The largest bound objects we know in the universe are the great clusters, with masses of a few times 1015 solar masses; galaxies have masses three orders of magnitude smaller. So we are interested in small enough regions that we can proceed with purely Newtonian physics. We believe that the uctuations grow to become the clusters of galaxies and galaxies we see in the universe today, and in this lecture we will see how they grow and how they can be characterized. Since perturbations in the matter cannot grow before matter dominance, we can assume as the initial conditions, which we take at the beginning of matter dominance, the that power spectrum is whatever emerged from ination, and that the Hubble constant is uniform, since the radiation, which up until the initial time provided essentially all the energy density, had time to smooth itself out on the length scales we are considering. Consider for the sake of argument a tophat perturbation in which the matter density is dierent from the critical density in the universe by some factor (1 + + ) in some spherical region of initial radius ri , and is the average density in the universe outside this region. Now if m is dierent from unity, whether or not the total is dierent from unity, the mean density is dierent from the critical density as well. Let (1 + e ) be the ratio of the mean matter density to the critical density c (e stands for external); note that e is probably negative. 4 Now when we were developing the description for the expanding universe, we wrote a Newtonian equation which was the direct analog of the Friedman equation for the radius of a shell in an expanding universe, just balancing the kinetic and potential energies: 8Gr2 = 2 (1) 3 We begin, we argue, with the same value of H = r/r for all the shells in our model, inside the perturbation and out. Let us think about the behavior of a perturbation with the density just such as to make = 0. It is clear that in this case = c and r = C 2/3 , because this is just part of a critical-density universe. If the external density is less than c , it will expand faster and leave the little tophat alone. r2 Now consider a density in the tophat greater than the critical density. Then is negative, and the material in the tophat will reach some maximum radius at some point in the future, turn around, and collapse again. How does the density excess behave early in the expansion? We can perturb Equation 1 to follow the small dierence in behavior from the critical case, which we can solve exactly, but it is messy and we do not need to; we know the answer already. The cosmological relations you derived in your problem sets say that at early times 1 0 1 m (1 + z )1 m (1 + z ) and furthermore that any contribution to the density in the present universe in the form of a cosmological constant is completely negligible early in the universe. The perturbation is a little piece of a higher-density universeit does not know about the exterior, remember, and c = = + 1 m = 1 c c The shell in question is not expanding, to be sure, exactly the way the universe is, but when the perturbation is still small, it is certainly expanding approximately the way the universe is, so we can neglect for a while the dierence in the factor (1+z) inside and out; so we get, for no work, + (z ) (1 + z )1 R Of course, the same proportionality applies to e , so whether we refer the density perturbation to the critical density or the mean density, the contrast R. This is a very important result. Think again about the energetics. Let us write the energy equation (1) in the following way for the initial conditions: 8Gc r2 8G( c )r2 = 2. 3 3 Here we have just added and subtracted the term 8Gc r2 /3. But the rst two terms just cancel (this is the denition of the critical density), and leave us with (dividing by 2): r2 4G( c )r2 = . 3 5 We can rephrase this in the suggestive way Gm = r where m is just the excess mass in the perturbation interior to the shell at radius r compared to the mass that would be contained if the shell contained just the critical density. Clearly m = m + where m is the total mass inside the shell. Then m goes like 2/3 as well, and this can be thought of protably and interchangeably in two very dierent ways. The excess mass is growing because the density contrast is growing because the shell is expanding somewhat less rapidly than the external universe. This is correct and precise. The excess mass is growing because the density contrast is growing because matter is slowly moving inward with respect to a shell which is expanding at the same rate as the external universe is expanding. This is also correct, but expresses a rather dierent point of view. We have restricted ourselves here to spherical perturbations; clearly one does not need to restrict oneself to uniform ones. Everything we have said applies to a shell in a general spherical perturbation, with all the densities and density contrasts replaced with the mean density inside the shellwhat matters, after all, is just how much mass there is within the shell. Rather more surprising is that these results are not conned to spherical perturbations while the perturbations are very smallif one has some general density excess eld (u) = ((u) )/, then ( u) R This result is true as long as << 1 but is crudely correct as long as < 1i.e. , as long as the perturbation in the density is not much larger than the density itself. This approximation is the linear approximation and the realm in which it is valid is called the linear regime. IV. Spherical Perturbations in the Nonlinear Regime Now if we are going to make a galaxy or a cluster or any other kind of bound structure, we are clearly interested in perturbations with negative . We are interested in structures with enormous density contrast, and furthermore structures which are bound. Remember that most of the matter is dark matter, which does not radiate or otherwise interact with either itself or other matter except gravitationally, which means that it cannot lose energy. The galaxy is bound now and must have been bound for all of its existence if most of its mass cannot get rid of any energy. So consider a shell in a perturbation which has positive + and hence negative . Its energy, we have seen, is Gm/ri initially, and this is conserved. At some point in its history, it will reach some maximum radius, and then collapse. We can easily calculate how big it will get. At its maximum radius, its energy is all potential, and is Gm/rmax . So Gm/ri = Gm/rmax , 6 and we obtain the exquisitely simple result rmax 1 m = +. = ri m if a shell at the initial time has interior to it a 1 percent density excess, it will expand a factor of 100 before turning around. If we do a little work, we can also calculate how long it takes to reach maximum expansion; twice this is the collapse time, the time it takes the structure it is forming to reach high density, and, well form. We will solve the equation of motion, which we can write dr d 2 1 1 r rmax = 2Gm ; clearly, as we saw above, = Gm/rmax . Then we write 2Gm 3 rmax 1/2 dr/rmax d = rmax /r 1 dq = 1/q 1 qdq = , 1q where we have here let q = r/rmax . If we let q = sin2 (/2), (dont worry about where the /2 comes fromit is convenient later but is not necessary) (1 q ) is cos(/2) and dq = sin(/2) cos(/2)d, so 2Gm 3 rmax 1/2 d = sin2 (/2)d = 1 cos d. 2 Notice that the term (1 cos )/2 = sin2 (/2) is just r/rmax again. We can integrate this trivially, and get the parametric solution 1/2 3 rmax = 8Gm r 1 cos . = rmax 2 ( sin ) For the geometry acionados among you, this is the pair of equations which describe the development of a cycloid, the locus of a point on a circular hoop of unit radius as it rolls along the x-axis, starting at x = 0, y = 0, and ending at x = 2 , y = 0; is the angle 3 through which the hoop has rolled. In this identication, x is 8Gm/rmax ; y is 2r/rmax . 7 Look at the behavior of r/rmax . It is zero for = 0 and 1 for = , so = corresponds to maximum expansion. Then it is zero again for = 2 , which is fully collapsed. If we let 3 3 tc = 2 rmax /(8Gm) = rmax /(2Gm) then the expression for is = tc 2 ( sin); when is , we are at maximum expansion, and = tc /2; when is 2 , we have collapsed and = tc . The quantity tc is the collapse time. With a little manipulation you can also relate the collapse time to the initial conditions: tc = + 3/2 ( ) Hi where Hi is the Hubble constant at the initial conditions. You will be asked to derive this in the next problem set. We have done a lot of manipulation; where has it gotten us? We see that a positive perturbation is like a little piece of a closed universe, which reaches some maximum radius and then collapses. If we draw a spacetime diagram, the positive part of the perturbation separates out from the expanding universe and collapses to form a bound object. We might well ask what the interface looks likedo we leave a vacuum at the (admittedly unrealistic) sharp edge of the tophat? The answer is no. Remember that the maximum radius and collapse time depend on the average density inside a shell. A shell just outside the tophat feels the same average density as the last shell in the tophat, so it collapses in the same time. As we go away from the edge of the tophat the average density clearly tends to the mean density in the universe, and the collapse time gets longer and longer; if the mean density of the universe is less than the critical density, there is a last bound shell. Let us see how this goes: The average density inside a shell is av = 3m 3, 4ri but m is just the mass belonging to the mean density in the universe, 4 r3 /3, plus the excess mass in the tophat, which we can call M : 3 M = 4Ri c ( + e )/3 Here everything refers to conditions at the initial epoch, and capital R and M to the whole tophat, i.e. at its edge. Then clearly, outside the tophat, av = + 3M 3, 4ri 8 ri > Ri If we subtract the critical density and divide by it, we can write the much more intuitive expression: 3 Ri , ri > Ri , + (ri ) = e + ( + e ) 3 ri = + , ri < Ri . here + with no argument is the density excess with respect to the critical density inside the tophat. Thus we see that if e is negative, there will be a radius ri,last , ri,last = Ri e + e 1/3 which has energy zero. All shells inside of this eventually fall into the object being formed, and can represent an amount of mass large compared to the mass in the original tophat, especially if the original tophat is a large-amplitude perturbation. In the next lecture we will investigate the energetics of the object being formed, and see how the present properties of galaxies and clusters connect to the initial conditions. 9
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YASAR UNIVERSITY TESTING AND ASSESSMENT OFFICEANSWER KEY / SAMPLE FLATPAPER AUSE OF ENGLISH / VOCABULARYPART A1. C2. A3. B4. B5. C6. D7. B8. A9. D10. A11. C12. B13. C14. A15. D16. B17. C18. D19. B20. APART B1.2.3.4.INSPITE OF
Princess Sumaya University for Technology - ENG - 101
YASAR UNIVERSITY TESTING AND ASSESSMENT OFFICEANSWER KEY / SAMPLE FLATPAPER BUSE OF ENGLISH / VOCABULARYPART A1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.IN SPITE OF HER HEADACHE SHE STILL EATS CHOCOLATE.IF ONLY I HAD AN IPODTHE POLICE WHO H
Princess Sumaya University for Technology - ENG - 101
CLOZE TEST 1The biggest obstacle to learning is boredom. Try to become truly interested inthe subject as it is easier to remember something (1)_fascinates you. A good example for this is the encyclopaedic knowledge mostfootball fans have about their s
Princess Sumaya University for Technology - ENG - 101
Lineer Denklem SistemleriNTE3YazarYrd. Do.Dr. Nezahat ETNAmalarBu niteyi altktan sonra; Lineer Denklem ve Lineer Denklem Sistemleri kavramlarnrenecek, Lineer Denklem Sistemlerinin zmlerinin varln tartabilecek, Lineer Denklem Sistemlerinin zm ynt
Princess Sumaya University for Technology - ENG - 101
YASAR UNIVERSITY ENGLISH PREPARATORY SCHOOLSAMPLE FLATUSE OF ENGLISH/VOCABULARY(A)NAME: _CLASS: _Official UseONLYDATE: SAMPLETIME SUGGESTED: 90 min.PART A (20x1MARK=20)CIRCLE THE CORRECT ANSWER . (0) IS AN EXAMPLEGeorge: Hi Vicky, where (0) _A
Princess Sumaya University for Technology - ENG - 101
YASAR UNIVERSITY ENGLISH PREPARATORY SCHOOLSAMPLE FLATUSE OF ENGLISH/VOCABULARY(B)NAME: _CLASS: _Official UseONLYDATE: SAMPLETIME SUGGESTED: 90 min.PART A (15x2MARKS=30)USING THE WORD GIVEN, COMPLETE THE SECOND SENTENCE SO THAT IT HAS A SIMILAR
Princess Sumaya University for Technology - ENG - 101
Verbask someone outask aroundadd up to somethingback something upback someone upblow upblow something upbreak downbreak downbreak something downbreak inbreak into somethingbreak something inbreak inbreak upbreak upbreak outbreak out in s
Princess Sumaya University for Technology - ENG - 101
Allow for:Hesaba katmak, dikkate almak(You should allow for the possibility that we might not finish it on time.)Ask for:istemek, talep etmek(He asked me for a glass of water.)Back up:desteklemekdocuments.)(He backed up his claims with someBe
Princess Sumaya University for Technology - ENG - 101
SAMPLE TESTENGLISH PROFICIENCY EXAMINATIONANSWER KEYPART I : LANGUAGE USE ( 50 x 0.5 = 25 pts)1. C2. B3. A4. D5. B6. A7. D8. B9. C10. A11. D12. C13. D14. D15. B16. C17. B18. C19. C20. D21. A22. B23. C24. D25. B26. D27. C28.
Princess Sumaya University for Technology - ENG - 101
Forge: demir ocaFlourish: savurmak, gelimek,ilerlemekClumsy: acemi, sakarInvade/InvasionCowardice: korkaklk/cowardNobleman: soylu, asilzadeAccomplishment: baarTransitory: geiciPermanent: kalcAspect: grn, durumFlow: akmak, tamak,dklmekCounter
Princess Sumaya University for Technology - ENG - 101
Uzayn Analitik GeometrisiNTE9YazarDo.Dr. Hseyin AZCANAmalarBu niteyi altktan sonra; Dzlemde gelitirilen analitik geometri modeline benzer ekilde boyutlu uzay iin de bir analitik geometri modeligelitirilecek, Bu modelin dorular tanmlanp salamas ge
Bowling Green - MATH - 3
Chemical Engineering Science 62 (2007) 4159 4176www.elsevier.com/locate/cesMass transfer around freely moving active particles in the dense phaseof a gas uidized bed of inert particlesFabrizio Scala Istituto di Ricerche sulla CombustioneCNR, P.le Tec
Bowling Green - MATH - 4
*FCC..FCC.FCC :.FCCFCC..[ ]abtin_e_a @yahoo.com1Fluidization*.[]..[ ]..FCCFCC&quot;...[ ]....[]1Fixed bedExpanded bed3Impact2TTs..&quot;.&quot;..Ts(T)T1Ts T.Push Back().T1[].T1():():.Push Back:():()()
A.T. Still University - ECON - 12314
tousCmosehacelaFIV?GineclogoEnlaFIVgeneramosunaovulacincontrolada,generalmentedevariosvulosydespusdeunosdasdeinduccin,cuandolosfolculosalcanzanundeterminadotamaoserealizaunapuncinecoguiadadelosfolculosconelfinderecuperarlosvulos.Posteriormenteen
Maryland - ECE - 660
Department of Electrical and Computer EngineeringUniversity of Maryland, College ParkFall 2011ENEE 660Homework Set 1 Solution1. Exercise 1.2Solution.U = cfw_x R5 |[x]1 = 3[x]2 , [x]3 = 7[x]4 Choose31 0 , b2 = b1 = 0000710 , b3 = Nuno
UCLA - ANTHRO - 8
9/29/10!Early Archaeology Recap! Archaeology out of antiquarian studies of1500s-1800s&quot; Breakthroughs in the 1800s&quot; Progressive view of culture&quot; Antiquity of Man&quot; By 1900 archaeology as a discipline&quot; In Europe-independent or with history&quot; In US wi
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 1Due Thursday September 81. (2 points) A salad bar has 3 choices of greens, 5 veggies, 4 fruits, 3 dairy items, and 4dressings. In how many ways can I make myself a salad with at least two of the g
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 2Due Thursday September 151. (2 points) Let E , F , and G be three events. Determine which of the following statements arecorrect and which are incorrect. Justify your answers.(a) (E EF ) F = E F.
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 3Due Thursday September 221. (2 points) Clint Westwood has collected six old guns. The probability of hitting thebulls-eye on a target when these guns are properly aimed and fired is 0.6, 0.5, 0.7,
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 4Due Thursday September 29[1] (2 points) Two fair dice are tossed 6 times, independently. What is the probabilitythat the 6th sum is not a repetition of a previous sum?[2] (2 points) Suppose that
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 5Due Thursday October 6[1] (2 points) For each of the following, determine the value(s) of k for which p(.) is aprobability frequency function. Note that in part (d), n is a positive integer.(a) p
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 7Due October 13[1] (2 points) Before having any children, a woman and her husband want to decide in advance how manychildren they will have. They want to have at least a 95% chance of having at lea
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 7Due Thursday October 201. (2 points) Let X be a standard Cauchy random variable. Show that E [|X | ] exists if 0 &lt; &lt; 1and does not exist if 1.2. (2 points) A farmer Phil Dert who has two boards o
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 8Due Thursday October 27, 20111. (2 points) Suppose X and Y have joint density1fX,Y (x, y ) = ey I(0,) (y )I(y,y) (x)2Compute P [X 1, Y 3].2. (2 points) Let X have an exponential distribution w
Johns Hopkins - STATISTICS - 420
550.420 Introduction to ProbabilityFall 2011Homework 9Due Thursday November 3, 20111. (2 points) Suppose X and Y have joint density1fX,Y (x, y ) = ey I(0,) (y )I(y,y) (x).2Compute P [X 1|Y = 3].2. (2 points) Suppose X and Y have joint density1f