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h formula in Theorem 3.15, asserting that H ∗(X × Y ; R) ≈ H ∗(X; R) ⊗ H ∗(Y ; R) under certain conditions, can now be extended to non-CW spaces since if X and Y are CW approximations to spaces Z and W , respectively, then X × Y is a CW approximation to Z × W . Here we are giving X × Y the CW topology rather than the p... |
by inclusion. From the long exact sequence of the pair (C±X, X) we see that this pair is n connected if X is (n − 1) connected. The preceding theorem then says that the middle map is an isomorphism for i + 1 < 2n and surjective for ⊔⊓ i + 1 = 2n . In Corollary 4.25. πn(S n) ≈ Z , generated by the identity map, for all ... |
all πn ’s, then a theorem of Serre, proved in [SSAT], says that the homotopy groups of such a space are finitely generated iff the homology groups are finitely generated. In this example, πn(S 1 ∨ S n) is finitely generated as a Z[π1] module, but there are finite CW complexes where even this fails. This happens in fact for... |
he inclusion 368 Chapter 4 Homotopy Theory X ֓ Y induces isomorphisms on all homotopy groups, and therefore this inclusion ⊔⊓ is a homotopy equivalence. Example 4.34: Uniqueness of Moore Spaces. Let us show that the homotopy type of a CW complex Moore space M(G, n) is uniquely determined by G and n if n > 1 , so M(G, n... |
. If we choose one such lift α , then all e n α under the deck transformations γ ∈ π1(X) . the other lifts are the images γ The special case proved in the preceding paragraph shows that the relative πn for the e e n α . By universal cover is the free abelian group with basis corresponding to the cells γ the relative ve... |
subgroup of unit complex numbers. For this bundle, the long exact sequence of homotopy groups takes the form ··· -→ πi(S 1) -→ πi(S 3) -→ πi(S 2) -→ πi−1(S 1) -→ πi−1(S 3) -→ ··· In particular, the exact sequence gives an isomorphism π2(S 2) ≈ π1(S 1) since the two adjacent terms π2(S 3) and π1(S 3) are zero by cellula... |
ually, the calculation that gi and hi are inverses needs only the following more elementary facts about octonions z, w , where the conjugate z of z = (a1, a2) is defined by the expected formula z = (a1, −a2) : (1) r z = zr for all r ∈ R and z ∈ O , where R ⊂ O as the pairs (r , 0) . (2) |z|2 = zz = zz , hence z−1 = z/|z... |
ified with n frames in Rn , and so p−1(U) is identified with U × Vn(Rn) . This argument works for k = ∞ as well as for finite k . In the case n = 1 the total spaces V1 are spheres, which are highly connected, and the same is true in general: Vn(Rk) is (k − n − 1) connected. Vn(Ck) is (2k − 2n) connected. Vn(Hk) is (4k − 4... |
omorphism will be studied in some detail in [VBKT]. ∗ include classes ηn ∈ 2π s 2n for n ≥ 4 , βn ∈ pπ s 2(p3−1)n−2p2−2p+1 for p ≥ 7 . The element βn appears in the diagram for p = 5 as the dot in the upper part of the diagram labeled by the number n . These βn ’s generate the strips along the upward diagonal, except w... |
on of all the Xn ’s, a CW complex X . Show πn(X) ≈ Gn for all n . 22. Show that Hn+1(K(G, n); Z) = 0 if n > 1 . [Build a K(G, n) from a Moore space M(G, n) by attaching cells of dimension > n + 1 .] 23. Extend the Hurewicz theorem by showing that if X is an (n − 1) connected CW complex, then the Hurewicz homomorphism h... |
e the following assertions. (1) The functors hn(X) = hX, K(G, n)i define a reduced cohomology theory on the category of basepointed CW complexes. (2) If a reduced cohomology theory h∗ defined on CW complexes has coefficient groups hn(S 0) which are zero for n ≠ 0 , then there are natural isomorphisms hn(X) ≈ H n(X; h0(S 0)... |
alences. The second and third of them have homotopy inverses In the lower row of the evident inclusion maps, indicated by the upward arrows. the diagram the maps are the obvious ones, except for the map X/A→SA which is the composition of a homotopy inverse of the quotient map X ∪ CA→X/A followed by the maps X ∪ CA→(X ∪... |
natural. α S n W The situation for cohomology is quite similar, but there is one point in the ar- gument where a few more words are needed. For cohomology theories the cellular cochain groups are the direct product, rather than the direct sum, of copies of the coefficient group G = h0(point ) , with one copy per cell. T... |
+ n + 1) skeleton and we can obtain µ by extending the inclusion Sm ∧ S n = Sm+n ֓ Km+n . It is not hard to prove the basic properties of cup product using this definition, and in particular the commutativity property becomes somewhat more transparent from this viewpoint. For example, when R = Z , commutativity just co... |
t : Ep→Ep whose second coordinate is the restriction of the paths γ to the interval [t, 1] . Since the endpoints of the paths γ are unchanged, ht is fiber-preserving. We have h0 = 11, h1(Ep) ⊂ E , and ht(E) ⊂ E for all t . If we let i denote the inclusion E ֓ Ep , then ih1 ≃ 11 via ht and h1i ≃ 11 via ht || E , so i is ... |
tower for X happen to be principal, we have a diagram as at the right, where each Xn+1 is, up to weak homotopy equivalence, the homotopy fiber of the map kn : Xn→K(πn+1X, n + 2) . The map kn is equivalent to a class in H n+2 Xn; πn+1X called the nth k invariant of X . These classes specify how to construct X inductivel... |
obstruction theory, which produce essentially the same result in the end. In the more elementary approach one tries to construct the extension or lifting one cell of W at a time, proceeding inductively over skeleta of W . This approach has an appealing directness, but the technical details of working at the level of c... |
tion if f is a homotopy equivalence. 12. Show that for homotopic maps f , g : A→B the fibrations Ef →B and Eg→B are fiber homotopy equivalent. 13. Given a map f : A→B and a homotopy equivalence g : C→A , show that the fibrations Ef →B and Ef g→B are fiber homotopy equivalent. [One approach is to use Corollary 0.21 to reduc... |
is a normal covering space Since π1(X) is the product of n free groups on two generators, X is the covering e space of X corresponding to the kernel of the homomorphism π1(X)→Z sending each of the two generators of each free factor to 1 . Since X is a K(π , 1) , so is X . For example, when n = 1 , X is the union of tw... |
nvariant Section 4.B 427 In §2.2 we used homology to distinguish different homotopy classes of maps S n→S n via the notion of degree. We will show here that cup product can be used to do something similar for maps S 2n−1→S n . Originally this was done by Hopf using more geometric constructions, before the invention of c... |
X is uniquely determined by the cup product ring H ∗(X; Z) . In particular, this applies to any simply-connected closed 4 manifold. j of m 2 spheres S 2 Proof: By the previous proposition we may assume X is a complex Xϕ obtained from j by attaching a cell e4 via a map ϕ : S 3 wedge sum j ) is free with basis the Hopf m... |
; Z) ≈ (c) H ∗(Sp(n); Z) ≈ Z[x3, x5, ··· , x2n−1] . Z[x3, x7, ··· , x4n−1] . Λ Λ Z[x1, x3, ··· , x2n−1] , the exterior algebra on These are ring isomorphisms, and the proof will involve bundles where the iso- Λ morphism in the Leray–Hirsch theorem happens to be an isomorphism of rings. U(n), U(n−1) Proof: For (a), assu... |
---→ H i(E; R) initial portion of the Gysin sequence gives isomorphisms p∗ : H i(B; R) for i < n − 1 , and the more interesting part of the sequence begins p∗------------→ H n−1(E; R) -→ H 0(B; R) p∗------------→ H n(E; R) -→ ··· `e --------------→ H n(B; R) 0 -→ H n−1(B; R) In the case of a product bundle E = S n−1 × ... |
tive ring with identity, in which case a ‘generator’ is an element with a multiplicative inverse, so all elements of R are multiples of the generator. A Thom class with Z coefficients gives rise to a Thom class with any other coefficient ring R under the homomorphism H n(E, E′; Z)→H n(E, E′; R) induced by the homomorphism ... |
− f (y1, ··· , yℓ) = b′ ` e for some b′ ∈ H ∗(B; R) . Since b′ has lower dimension than b , we may assume by induction that b′ is a polynomial in y1, ··· , yℓ, e . Hence b = f (y1, ··· , yℓ) + b′ ` e is also a polynomial in y1, ··· , yℓ, e . Thus the natural map R[y1, ··· , yℓ, e]→H ∗(B; R) is surjective. To see that ... |
st as we did in the case of the functors hn(X) = hX, Kni . In the converse direction, if we have natural long exact sequences of pairs, then by applying these to pairs of the form (CX, X) we get natural isomorphisms hn(X) ≈ hn+1( maps of pairs (CX, X) uniquely determine the coboundary maps for all pairs (X, A) X) . Not... |
to connected CW complexes. Each functor hn satisfies the homotopy, wedge, and Mayer–Vietoris axioms, as we noted earlier, so the preceding theorem gives CW complexes Kn with hn(X) = hX, Kni . ral isomorphisms hn(X) ≈ hn+1( Kn→ ural bijection hX, Kni ≈ h of this bijection gives, for any map f : X→Kn , a commutative diag... |
plexes Xα of X . The homology theory defined by a spectrum satisfies this axiom, and the converse is proved in [Adams 1971]. Ω Spectra have become the preferred language for describing many stable phenom- ena in algebraic topology. The increased flexibility of spectra is not without its price, however, since a number of c... |
0← A ֓ X1 where the pair (X1, A) has the homotopy extension property. Another example is a sequence of inclusions X0 ֓ X1 ֓ ··· for which the pairs (Xn, Xn−1) satisfy the homotopy extension property, by the argument involving mapping telescopes in the proof of Lemma 2.34. However, without some conditions on X→ X is a h... |
ces to maps generalize the definition of ∇X from diagrams of spaces to complexes of spaces. As special cases of the constructions X and X we have direct limits and inverse limits for diagrams X0→X1→ ··· and ··· →X1→X0 , respectively. Since inverse limit is related to product and direct limit to coproduct, it is common p... |
aken to be wedges of spheres, and so Xn is obtained from Xn−1 by attaching an n cell for each Z summand of Hn(Y ) . The attaching maps may be taken to be cellular, making X into a CW complex whose cellular chain complex has trivial boundary maps. Similarly, finite cyclic summands of Hn(Y ) can be realized by wedge summa... |
sition 4I.3. K(Zm × Zn, 1) ≃ K(Zm, 1) ∨ Σ Σ Σ 3. Extending Proposition 4I.3, show that the (2k + 1) skeleton of the suspension of a high-dimensional lens space with fundamental group of order pn is homotopy equivalent to the wedge sum of the (2k + 1) skeleta of the spaces Xi , if these Xi ’s are chosen to have the mini... |
induces the suspension map πi(X)→πi+1( ΩΣ J(X) , we can identify the relative groups πi( n connected then the pair (J(X), X) is (2n + 1) connected since we can replace X by a complex with n skeleton a point, and then the (2n + 1) skeleton of J(X) is X) . Since this inclusion factors through X, X) with πi(J(X), X) . ΩΣ... |
then we get an extension of f over Dn × {0} ∪ ∂+Dn × I , with the constant homotopy on ∂+Dn × I and (Dn × {0}, ∂−Dn × {0}) mapping to (Y , C) . Condition (iii) then gives an extension over Dn × I , whose restriction to Dn × {1} shows that f is zero in πn−1(X, A) , so the kernel of πn−1(X, A)→πn−1(Y , C) is trivial. To ... |
The Dold–Thom Theorem Section 4.K 481 where b is an arbitrary point in B . The upper map in the diagram is an isomorphism by the five-lemma since the hypotheses imply that F1 induces isomorphisms πi(E)→πi(E′) and πi(p−1(b))→πi p−1(F 1(b)) for all i . The hypotheses also imply that the lower map and the right-hand map ar... |
ift to SP (X) would have the form xtαt , αt ∈ SP (A) , ending at x1α1 = aα1 , a point of SP (A) which is a multiple of a , so in particular there would be no lift ending at the basepoint of SP (X) . What we will show is that the projection SP (X)→SP (X/A) has instead the weaker structure of a quasifibration, which is st... |
ations are actually constructed, but only on their basic properties. This is similar to the situation for ordinary homology and cohomology, where the axioms generally suffice for most applications. The construction of Steenrod squares and powers and the verification of their basic properties, or axioms, is rather interest... |
en p = 3 , there is no chance that all the even-dimensional cohomology will be connected by the P i ’s. In fact, we showed in Proposition 4I.3 that K(Z3, 1) in dimensions congruent to 2 and 3 mod 4 , while X2 has the remaining cohomology. Thus the best one Σ could hope would be that all the odd powers of α are connecte... |
S 2p→S 3 such that in the mapping cone Cf = S 3 ∪f e2p+1 , the first Steenrod power P 1 : H 3(Cf ; Zp)→H 2p+1(Cf ; Zp) is nonzero, hence f is nonzero in π s 2p−3 . The construction starts with the fact that a generator of H 2(K(Zp, 1); Zp) has nontrivial p th power, so the operation P 1 : H 2(K(Zp, 1); Zp)→H 2p(K(Zp, 1)... |
4 . Proof: Let η : S n+1→S n be a suspension of the Hopf map, with mapping cone Cη obtained from S n by attaching a cell en+2 via η . If we assume the composition η)η is nullhomotopic, then we can define a map f : S n+3→Cη in the following ( way. Decompose S n+3 as the union of two cones CS n+2 . On one of these cones Σ... |
t to the same point in K(Z2, 2n) . This means that when we compose with the diagonal map S ∞ × X→S ∞ × X × X , (s, x)֏ (s, x, x) , there is an induced quotient map RP∞ × X→K(Z2, 2n) extending α2 : X→K(Z2, 2n) . This extended map represents a class in H 2n(RP∞ × X; Z2) . By the K¨unneth formula and the fact that H ∗(RP∞... |
comes the one-point compactification of R2 . The map T : S 1 ∧ S 1→S 1 ∧ S 1 then corresponds to reflecting R2 across the line x = y , so after a rotation of coordinates this becomes reflection of S 2 across the equator. Hence 1S 1 is obtained from the shell I × S 2 by identifying its inner and outer boundary spheres via ... |
lot cleaner. Lemma 4L.14. θi = 0 unless i = 2k(p − 1) or 2k(p − 1) + 1 . Proof: The group of automorphisms of Zp is the multiplicative group Z∗ p of nonzero elements of Zp . Since p is prime, Zp is a field and Z∗ p is cyclic of order p − 1 . Let r p . Define a map ϕ : S ∞× X ∧p→S ∞ × X ∧p permuting the factors Xj be a g... |
ying the calculation at the end H ∗(T (E)) , the of Example 3.11 in each fiber. Under the isomorphism H ∗(E, E − Lp) ≈ j (τj ) corresponds to ±λ(α) since both classes restrict to ±α⊗p in each class fiber S p ⊂ T (E) and λ(α) is uniquely determined by its restriction to fibers. j (Lp) , and the cup product j (τj ) ∈ H ∗ E,... |
i+j m(n−2j)−1 k m(n−2j) k ω2m(pn−2i) ⊗ ω2m(n−2j+2k)−1 ⊗ P i−kβP j(ι) ω2m(pn−2i)−1 ⊗ ω2m(n−2j+2k) ⊗ βP i−kP j(ι) (−1)i+j m(n−2j)−1 k (−1)i+j m(n−2j)−1 k ω2m(pn−2i)−1 ⊗ ω2m(n−2j+2k) ⊗ P i−kβP j(ι) ω2m(pn−2i)−1 ⊗ ω2m(n−2j+2k)−1 ⊗ βP i−kβP j(ι) Letting ℓ = mn+j −k , so that n−2j +2k = pn−2ℓ , the first of these five summatio... |
Condition (iii) is satisfied since it is a nontrivial condition only for the 2 cell. Φ Proof: We have already taken care of the ‘only if’ implication. For the converse, suppose inductively that X n−1 , the union of all cells of dimension less than n , is a α ’s as characteristic maps. The induction can start CW complex... |
W as a compact neighborhood of Ki × Li in Xi+1 × Yi+1 . To do this, we first choose for each x ∈ Ki compact neighborhoods Kx of x in Xi+1 and Lx of Li in Yi+1 such that Kx × Lx ⊂ W , using the compactness of Li . By compactness of Ki , a finite number of the Kx ’s cover Ki . Let Ki+1 be the union of these Kx ’s and let L... |
mplex X is obtained from a subcomplex A by attaching a cell ek via a map f : S k−1→A , and suppose that we have an embedding A ֓ Rm . Then we can embed X in Rk × Rm × R as the union of Dk × {0}× {0} , {0}× A× {1} , and all line segments joining points (x, 0, 0) and (0, f (x), 1) for x ∈ S k−1 . ⊔⊓ Spaces Dominated by C... |
same reasoning shows that continuity of C × Y →(X × Y )c is equivalent to continuity of C × C ′→(X × Y )c for all compact C ′ ⊂ Y . But on the compact set C × C ′ , the two topologies on X × Y agree, so we are done. (This proof is from [Dugundji 1966].) ⊔⊓ b b Returning to the context of Proposition A.14, part (b) of ... |
they do not behave as well with respect to products. The product of two ordered simplices has a canonical subdivision into ordered simplices using the shuffling operation described in §3.B, and ∆ this allows the product of two ordered complexes to be given a canonical ordered ∆ complex structure. Without orderings this n... |
ions we obtain a collection of disjoint simplices, with one n simplex for each nondegenerate n simplex of Y . Then doing the identifications as g varies over injections attaches these nondegenerate simplices together to form an s complex, which is |Y | . The quotient map from the collection of disjoint simplices to |Y |... |
rlag GTM 62, 1978. J. A. Wolf, Spaces of Constant Curvature, McGraw Hill, 1967. 6th ed. AMS Chelsea 2010. Papers J. F. Adams, On the non-existence of elements of Hopf invariant one, Ann. of Math. 72 (1960), 20–104. J. F. Adams, Vector fields on spheres, Ann. of Math. 75 (1962), 603–632. Bibliography 543 J. F. Adams, On ... |
lex 5, 519 CW pair 7 cycle 106 excess 499 excision 119, 201, 360 excisive triad 476 Ext 195, 316, 317 extension lemma 348 extension problem 415 exterior algebra 213, 284 external cup product 214 548 Index face 103 fiber 375 fiber bundle 376, 431 H–space 281, 342, 419, 420, 422, 428 HNN extension 93 hocolim 460, 462 fiber ... |
ve saved — you money or helped you to make a better decision? (List as many as you can.) 1.1.6 Probabilities are often depicted in popular movies and television programs. List as many examples as you can. Do you think the probabilities were portrayed there in a “reasonable” way? 1.2 Probability Models A formal definitio... |
, P 2 , P 3 , and P 4 . [0 1] be any outcome. What 2 P 2 . Compute P 1 , 1 2 3 4 , and P 1 1 2 3 , and P 1 2 1 2 3 4 , and P 1 1 12, and P 1 2 1 2 3 , and P 1 1 2 3 , and P 1 1 3, and P 2 3 2 3. Compute 1 6, and , and 3 P 3 0 3. Is P 3 P 2 P 2 1 8 10 Section 1.3: Properties of Probability Models Figure 1.2.4: Venn dia... |
the remainder of this section, we consider a few simple combinatorial rules and their application in probability theory when the uniform distribution is appropriate. EXAMPLE 1.4.5 Counting Sequences: The Multiplication Principle Suppose we ip three fair coins and roll two fair sixsided dice. What is the prob abilit... |
EMS 1.4.14 Show that a probability measure defined by (1.4.1) is always additive in the sense of (1.2.1). 1.4.15 Suppose we roll eight fair sixsided dice. What is the probability that the sum of the eight dice is equal to 9? What is the probability that the sum of the eight dice is equal to 10? What is the probability ... |
n and the die are independent in this example, to indicate that the occurrence of one does not have any inuence on the probability of the other occurring. More formally, we make the following definition. Definition 1.5.2 Two events A and B are independent if P A B P A P B Now, because P A B if and only if P A B P B or P... |
s (a) and (b) change in this case? 28 Section 1.6: Continuity of P DISCUSSION TOPICS 1.5.19 Suppose two people each ip a fair coin simultaneously. Will the results of the two ips usually be independent? Under what sorts of circumstances might they not be independent? (List as many such circumstances as you can.) 1.5.... |
rain 0, Y snow rain snow clear , we might define a second random variable Y by 7 8 if it is clear. That is 1 2 if it snows, and Y 0 if it rains, Y 1 2, and Y clear 7 8. EXAMPLE 2.1.3 If the sample space corresponds to ipping three different coins, then we could let X be the total number of heads showing, let Y be the t... |
very real number y. x for every real number x. B , for any subset B of the real numbers. B , for any subset B of the real numbers. for every real number for every real number 2.2.1 Consider ipping two independent fair coins. Let X be the number of heads that appear. Compute P X 2.2.2 Suppose we ip three fair coins, a... |
he previous example, consider again repeatedly ipping a coin that has probability of coming up tails. Let r be a positive integer, and let Y be the number of tails that appear before the r th head. Then 1 heads (and k tails) on the for k first r k th ip. The probability of this is equal to k ips, and then shows a he... |
ity distribution of the number of white balls observed? 3? What 50 Section 2.3: Discrete Distributions 2.3.18 (Poisson processes and queues) Consider a situation involving a server, e.g., a cashier at a fastfood restaurant, an automatic bank teller machine, a telephone ex change, etc. Units typically arrive for servi... |
roduced by a manufacturer might follow an Exponential . By this we mean that the lifelength X of a distribution for an appropriate choice of randomly selected light bulb from those produced by this manufacturer has probability P X x e z dz x e x of lasting longer than x in whatever units of time are being used. We will... |
f x and 0 otherwise Prove that f is a density function. and 0 otherwise Prove that 0 fixed, the function given by is a density f 2.4.21 (Cauchy distribution) Consider the function given by f x 1 1 1 x 2 x for arctan x ) 2.4.22 (Laplace distribution) Consider the function given by f x Prove that f is a density function. ... |
mpute all these probabilities directly 3 , P X P X 2 5 and P 1 2 X 68 Section 2.5: Cumulative Distribution Functions from FX . We have that FX 3 FX 3 2 4 16 3 1 16 1 n 2 4 16 1 16 3 lim n P X 1 1 2 5 FX 3 4 2 5 2 4 16 FX 1 2 1 FX 2 5 1 0 0625 16 2 4 16 3 4 0 996 0 0 2401 2.5.4 Mixture Distributions Suppose now that F1 ... |
F is right continuous, meaning that F x . (Hint: You will need to use continuity of P 2.5.16 Let F be a cumulative distribution function. Compute (with explanation) the value of limn F n ]. 2.5.17 Let F be a cumulative distribution function. For x F x F x for each x (Theorem 1.6.1).) 2.5.18 Let X be a random variable,... |
. Deter mine the distribution of Y (see Problem 2.4.20). Determine the distribution (see Problem 2.4.19). Determine the distribution Pareto e X X 1 1 CHALLENGES 2.6.21 Theorems 2.6.2 and 2.6.3 require that h be an increasing or decreasing function, at least at places where the density of X is positive (see Theorem 2.6... |
OOF We need to show that, for a b a tinuity of P, we have that P a and f X Y x y dy dx Now, we always have X b P a b dx . Hence, using con lim c d f x y dx dy b d a c f x y dy d x b a f X Y x y dy d x lim c d lim c d as claimed. The result for fY follows similarly. EXAMPLE 2.7.7 (Example 2.7.6 continued) Let X and Y ag... |
and Distributions 95 This prompts the following definition. Definition 2.8.2 Suppose X and Y are two discrete random variables. Then the conditional probability function of Y , given X, is the function pY X defined by pY X y x pX Y x y z pX Y x z pX Y x y pX x defined for all y R1 and all x with pX x 0. 2.8.2 Conditioning... |
hip from the bowl and observe its label s then P s Alternatively, consider a population of students at a university of which a proportion 1 live on campus (denoted by s 1), a proportion 2 live offcampus with their parents (denoted by s 2), and a proportion 3 live offcampus independently (denoted by s 3). If we random... |
0 y 1 0. Suppose P Y g x h y , for some functions g and h. Prove that X and Y are indepen Determine values of C1 and C2, such that f X Y is a valid joint density function, and X and Y are independent. 2.8.18 Let X and Y be discrete random variables. Suppose pX Y x y g x h y , for some functions g and h. Prove that X ... |
Random Variables and Distributions 115 2.9.6 Suppose the joint probability function of X and Y is given by pX otherwise, and B Y , A Y , W X Let Z (a) Compute the joint probability function pZ W z (b) Compute the joint probability function pA B a b . (c) Compute the joint probability function pZ A z a . b . (d) Comput... |
ROOF We begin by noting that P Y smallest value x such that F x i.e., U F y . Therefore, y U . Hence, F 1 U P F 1 U y . But F 1 U is the U , y if and only if But 0 F y 1, and U Uniform[0 1], so P U F y F y . Thus It follows that F is the cdf of Y , as claimed. We note that Theorem 2.10.2 is valid for any cumulative dis... |
: f x y x2 and y1 0 , i.e., if where J is the Jacobian derivative of h, and where h 1 z such that h x y z . is the unique pair x y We must show that whenever a b and c d, we have dz If we let S as [a b] [c d] be the twodimensional rectangle, then we can rewrite this P Z W S f Z W z dz d S Now, using the theory of mult... |
E X 3E X 2. Hence, by Theorem 3.1.2, E 3X 2 2Y 2 2 2 . What is E 3X 2Y ? n 1 and E Y 3n 1 2E Y 1 2 1 EXAMPLE 3.1.15 Let Y Y Binomial n X1 . Then we know (cf. Example 2.3.3) that we can think of 1 if the ith coin is Bernoulli (in fact, Xi Xn, where each Xi 136 Section 3.1: The Discrete Case heads, otherwise Xi Theorem 3... |
e. 1 E Y y fY y dy 1 y 1 y2 dy y 1 y2 dy 1 1 y dy 1 y dy 1 1 which is undefined. Hence, the expected value of Y is undefined (i.e., does not exist) in this case. Theorem 3.1.1 remains true in the continuous case, as follows. 144 Section 3.2: The Absolutely Continuous Case Theorem 3.2.1 (a) Let X be an absolutely continuo... |
alid measure of the average distance of X from X . Furthermore, it would not have the “scale problem” that Var X does. However, we is shall see that Var X has many convenient properties. By contrast, E X very difficult to work with. Thus, it is purely for convenience that we define variance by E X 2 instead of E X X X . ... |
ES 3.3.1 Suppose the joint probability function of X and Y is given by pX otherwise 5 9 5 9 4, E Y 19 3, and E XY with E X (a) Compute Cov X Y . (b) Compute Var X and Var Y . (c) Compute Corr X Y . 3.3.2 Suppose the joint probability function of X and Y is given by 26, as in Example 3.1.16. pX otherwise 0 3 4 0 4 as in... |
nctions Also, taking derivatives, we see3 that m X s E X es X , so m X 0 E X e0X E Xe0 E X Taking derivatives again, we see that m X s E X 2es X , so m X 0 E X 2 e0X E X 2e0 E X 2 Continuing in this way, we obtain the general formula. We now consider an application of Theorem 3.4.3 EXAMPLE 3.4.8 The Mean and Variance o... |
mY to compute the mean of Y . (c) Use mY to compute the variance of Y . 3.4.17 Compute the kth moment of the Weibull Problem 2.4.19). 3.4.18 Compute the kth moment of the Pareto (Hint: Make the transformation u 3.4.19 Compute the kth moment of the Lognormal (Hint: Make the transformation z ln x and use Problem 3.4.15... |
EXAMPLE 3.5.8 Consider again rolling a fair sixsided die, so that S 1 6 and X s s A P X A, and that E X A s for s 1 3 for s S, and with A 1 2 3 4 5 6 , with P s 3 5 6 . We have already computed that 14 3. Hence, Var 14 3 2 A s 14 3 2 P X s A 3 14 3 2 1 3 5 s S 14 3 2 1 3 6 14 3 2 1 3 14 9 1 56 By contrast, because E X... |
small, then it Y . We now consider some is unlikely that Y will be too far from its mean value examples. 3 4 as above. Also, E X 2 18 EXAMPLE 3.6.4 Suppose again that P X Then E X that Var X that P X 4 P X 2 32 2 9, which is true because in fact P X 1 1. On the other hand, setting a 4 2. Hence, setting a 2 12 4 1 42 1 ... |
easure theory, an advanced mathematical subject. However, it is also possible to give a general definition in ele mentary terms, as follows. Definition 3.7.1 Let X be an arbitrary random variable (perhaps neither discrete nor continuous). Then the expected value of X is given by E X P X t dt P X t dt 0 0 provided either... |
we cannot even develop useful approximations for large n due to the difficulty of the problem or perhaps because n is just too small in a particular applica tion. Fortunately, however, we can then use the Monte Carlo approach where the power of the computer becomes available. This is discussed in Section 4.5. In Chapte... |
n Hence, again P Zn converges in probability to Y . Y 0 as n for all 0, so the sequence Zn 4.2.1 The Weak Law of Large Numbers One of the most important applications of convergence in probability is the weak law is a sequence of independent random variables of large numbers. Suppose X1 X2 that each have the same mean ... |
ave Xn to 0 for any particular value of U . Thus, P limn converge to 0 with probability 1. Xn 0 Theorem 4.3.1 and Example 4.3.2 together show that convergence with probability 1 is a stronger notion than convergence in probability. Now, the weak law of large numbers (Section 4.2.1) concludes only that the av erages Mn... |
2 is an i.i.d. sequence of random variables each having finite and finite variance Xn be the sample sum and X1 Sn n be the sample mean. The central limit theorem is concerned with the 2 Let Sn mean Mn distribution of the random variable Zn Sn n Mn n n Mn n 0 and Var Zn 2 We know E Mn 2 n which implies that where 1 The va... |
ause n hence implies that n variance of the sample X1 denote this estimate of the variance by S2 n is consistent for 2 as well, we conclude that 2 This 2 n the sample Xn When the sample size n is fixed, we will often It is common to call 2 n Again, using the above argument, we have that, for large n the interval Mn 3 n ... |
al I , as follows. 1. Select a large positive integer n. 2. Obtain Ui Uniform[0 1], independently for i 1 2 n. 3. Set Ti cos U 2 i sin U 4 i , for i 1 2 n. 4. Estimate I by Mn T1 Tn n. For large enough n, this algorithm will provide a good estimate of the integral I . For example, the following table records the estima... |
nte Carlo approximation of Uniform[0 10] distribution.) 4.5.7 Suppose we repeat a certain experiment 2000 times and obtain a sample average of 5 and a standard error of 17. In terms of this, specify an interval that is virtually certain to contain the experiment’s (unknown) true mean . 4.5.8 Suppose we repeat a certain... |
s in general (see Problem 4.6.13). 236 Section 4.6: Normal Distribution Theory Note that using linear algebra, we can write the equations U n i 1 bi Xi of Theorem 4.6.2 in matrix form as V n i 1 ai Xi and U V A X1 X2 Xn A a1 b1 a2 b2 an bn where (4.6.1) i 1 for all i, we have that Cov U V Furthermore, the rows of A are... |
Y 2 C3 Y Y 3? 2 n N 7 2 be independent. Find values of C1 2 C4 N 0 1 be independent. Prove that X 2 C5 . Y 2 2 m be independent. Prove that X Y 2 n m . X4n be i.i.d. with distribution N 0 1 . Find a value of C such 4n F n 3n Xn 1 be i.i.d. with distribution N 0 1 . Find a value of C such C X 2 2 X1 .6.7 Let X1 X2 that ... |
ent of one another and are also all indepen dent of X. It follows that n 1 S2 2 is independent of X. It also follows, by the definition of the chisquared distribution, that n 1 S2 2 2 n 1 . Proof of Theorem 4.6.7 We want to show that when U t n , then fU u n 1 2 n 2 1 u2 n n 1 2 1 n for all u R1. 250 Section 4.7: Furt... |
mple of this, in Table 5.2 we have the values of both Y and Z for each patient in the treatment group. As it turns out, this additional information, known as covariates, can be used to make our comparisons more accurate. This will be discussed in Chapter 10. 256 Section 5.1: Why Do We Need Statistics 10 11 12 13 14 15 ... |
the same as the predicted lifelength before the machine starts working is a special characteristic of the Exponential distribution. This will not be true in general (see Exercise 5.2.4). The tail probability measuring the plausibility of the value x0 5 is given by P X 5 X 1 e x 1 dx e 4 0 0183, 5 which indicates that x... |
mple from the statistical model , then a sample X1 x1 : f f Note that, wherever possible, we will use uppercase letters to denote an unobserved value of a random variable X and lowercase letters to denote the observed value. So an observed sample X1 Xn will be denoted x1 xn 264 Section 5.3: Statistical Models EXAMPLE 5... |
d you say about the true value of the parameter if you had observed X 5.3.8 Suppose we have a statistical model P1 P2 , where P1 and P2 are probability measures on a sample space S Further suppose there is a subset C S such that 1 Discuss how you would make an inference about the P1 C true distribution of a response s ... |
ht place N chips in a bowl, each with a unique label corresponding to a population ele ment, and then randomly draw n chips from the bowl without replacement. The labels Alterna on the drawn chips identify the individuals that have been selected from tively, for the randomization, we might use a table of random numbe... |
ly discussed in Chapter 10. We can also define f X1 X2 for the joint distribution, and joint density histograms are again useful when X1 and X2 are both continuous quantitative variables. EXAMPLE 5.4.3 Suppose there are four candidates running for mayor in a particular city. A random sample of 1000 voters is selected; t... |
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