1 basics Flashcards

Question

Method from lectures Exercise 1.1.7. Show that B(R) contains all sets of the form (a, b), (a, b],[a, b), [a, b], (−∞, a], (−∞, a), (a, +∞), [a, +∞), for a, b ∈ R. Moreover, show that the σ-algebra B(R) is generated by any of these classes

Answer 1

To show ** σ(A_1) = σ(A)** ** σ(A_1) ⊆ σ(A)** Take a ∈ A_1 (a will be an open interval) and is therefore an open set) thus a ∈ A. Thus A_1⊆A. By defn, σ(A) is the smallest σ-algebra containing A. A ∈ σ(A). A_1 ⊆A ⊆σ(A) Claim σ(A_1)⊆ σ(A), by defn A_1 is contained in A thus any sigma algebra containing A_1 will have this σ(A_1) contained. As σ(A) contains A_1 we have this true. ** σ(A) ⊆ σ(A_1)** Take an open set A ∈ A_1 This means A is a countable union of open intervals A= ∪ₙ,∞ (aₙ,bₙ) (aₙ,bₙ) is an open interval, belongs to A_1 and thus σ(A_1). Thus we have a collection of elements that belong to a sigma algebra, since σ(A_1) is a sigma algebra the union is an element of a sigma algebra. ∪ₙ,∞ (aₙ,bₙ)∈ σ(A_1) . Thus A ⊆ σ(A_1). Now we have a class contained in the sigma algebra, by defn σ(A_1) is the smallest such sigma algebra, thus σ(A⊆ σ(A_1)

Answer 2

(i) Recall that B(R) is defined to be the σ-algebra generated by the collection of all open sets of R. Hence B(R) contains all open sets of R. Thus (a, b) ∈ B(R). (ii) (a, b] = ∩_n=1,∞ (a, b +1/n) ∈ B(R), since open sets are Borel, and σ-algebras are closed under countable intersection. (iii) [a, b) =∩_n=1,∞ (a − (1/n), b) ∈ B(R) by the same reasoning. (iv) [a, b] = ∩_{n=1} ∞ (a −(1/n), b + (1/n) ) ∈ B(R) by the same reasoning. (v) (−∞, a] = ∪_{n=1}∞ [−n, a] ∈ B(R) since (by (iv)) closed intervals are Borel and σ-algebras are closed under countable union. (vi) (−∞, a) = ∪_{n=1} ∞ (−n, a) ∈ B(R) since open sets are Borel and σ-algebras are closed under countable union. (vii) (a, ∞) =∪_{n=1} ∞ (a, n) ∈ B(R) by the same reasoning as in (vi). (viii) [a, ∞) = ∪_{n=1} ∞ [a, n] ∈ B(R) by the same reasoning as in (v). Hence B(R) contains all sets of the form(a, b), (a, b], [a, b), [a, b], (−∞, a], (−∞, a), (a, ∞), [a, ∞). We will now show that any of these classes generates B(R)

Answer 3

(I) Since B(R) contains all sets of the form (a, b), and σ({(a, b) : a < b}) is the smallest σ-algebra containing all sets of the form (a, b), we have σ({(a, b) : a < b}) ⊂ B(R). Since every open set in R is a countable union of (disjoint) open intervals, and σ-algebras are closed under countable union, it follows that σ({(a, b) : a < b}) must contain all open sets in R. Since B(R) is the smallest σ-algebra containing all open sets of R, we have B(R) ⊂ σ({(a, b) : a < b}) Therefore we conclude that σ({(a, b) : a < b}) = B(R) (II) Similarly, as B(R) contains all sets of the form (a, b], and σ({(a, b] : a < b}) is the smallest σ-algebra containing all sets of the form (a, b], we have σ({(a, b] : a < b}) ⊂ B(R) Since (a, b) = ∪_n=1 ∞ (a, b − {b − a}/{2n} ] , every open interval is a countable union of intervals of the form (a, b]. So since σ-algebras are closed under countable union, all open intervals must be contained in σ({(a, b] : a < b}). So since σ({(a, b) : a < b}) = B(R) is the smallest σ-algebra containing all open intervals, we have B(R) ⊂ σ({(a, b] : a < b}). Hence we conclude that B(R) = σ({(a, b] : a < b}) (III) Similarly: σ({[a, b) : a < b}) ⊂ B(R) and since (a, b) = ∪_{n=1} ∞ [a + {b − a}/{2n} , b) we also have that B(R) ⊂ σ({[a, b) : a < b}). Therefore σ({[a, b) : a < b}) = B(R). (IV) Similarly: σ({[a, b] : a < b}) ⊂ B(R) and since (a, b) = ∪_{n=1} ∞ [a + (b − a)/(3n), b − {b − a}/3n] we also have B(R) ⊂ σ({[a, b] : a < b}) Therefore σ({[a, b] : a < b}) = B(R) (V) We have σ({(−∞, a] : a ∈ R}) ⊂ B(R). Furthermore (a, b] = (−∞, b] \ (−∞, a], so since σ-algebras are closed under taking differences of sets, σ({(−∞, a] : a ∈ R}) must contain all sets of the form (a, b]. So since σ({(a, b] : a < b}) = B(R) is the smallest σ-algebra containing all sets of the form (a, b], it follows that B(R) ⊂ σ({(−∞, a] : a ∈ R}). Hence we conclude that σ({(−∞, a] : a ∈ R}) = B(R) (VI) Similarly, σ({(−∞, a) : a ∈ R}) ⊂ B(R), furthermore [a, b) = (−∞, b) \ (∞, b) and thus σ({[a, b) : a < b}) ⊂ σ({(−∞, a) : a ∈ R}), i.e. B(R) ⊂ σ({(−∞, a) : a ∈ R}). So we conclude that B(R) = σ({(−∞, a) : a ∈ R}) (VII) Similarly, σ({(a, ∞) : a ∈ R}) ⊂ B(R), furthermore (a, b] = (a, ∞) \ (b, ∞) and thus σ({(a, b] : a < b}) ⊂ σ({(a, ∞) : a ∈ R}) i.e. B(R) ⊂ σ({(a, ∞) : a ∈ R}). So we conclude that σ({(a, ∞) : a ∈ R}) = B(R). (VIII) Similarly, σ({[a, ∞) : a ∈ R}) ⊂ B(R), furthermore [a, b) = [a, ∞) \ [b, ∞) and thus σ({[a, b) : a < b}) ⊂ σ({[a, ∞) : a ∈ R}) i.e. B(R) ⊂ σ({[a, ∞) : a ∈ R}). So we conclude that σ({[a, ∞) : a ∈ R}) = B(R). This finishes the proof. Note: In our proof we used the fact that σ-algebras are closed under taking differences of sets. This can be seen as follows: Let A, B be measurable sets. Then A \ B = A ∩ B^c is measurable, since σ-algebras are closed under complementation and (countable and thus also finite) intersections

Answer 4

Note: In our proof we used the fact that σ-algebras are closed under taking differences of sets. This can be seen as follows: Let A, B be measurable sets. Then A \ B = A ∩ B^c is measurable, since σ-algebras are closed under complementation and (countable and thus also finite) intersections

Answer 5

On R^d we will always consider the Borel σ-algebra B(R^d). we choose whichever generates the set most convenient for us

Answer 6

Let (Ω, F) and (Ω', F') be two measurable spaces. A function f : Ω → Ω' is called **F/F'-measurable** if f⁻¹(B) ∈ F for all B ∈ F' (If I take any set in F',B, the inverse image of this set under f is the set) f⁻¹(B) ={w∈Ω| f(w) ∈B}

Answer 7

f⁻¹({1,2}) ={ω₁,ω₃,ω₄} For the function to be measurable we want this preimage to be a subset of the original Ω for every subset of Ω'

Answer 8

If Ω' = R^d, then as already mentioned we will always consider F' = B(R^d). Moreover, in this case, for a F/B(R^d)-measurable function we will simply say it is F-measurable.

Answer 9

σ({ω₁}) ie the smallest sigma algebra generated by {ω₁}

Answer 10

Checking every set: f⁻¹(∅)= ∅ f⁻¹(Ω)= f⁻¹({1,2})= {w∈Ω| f(w) ∈Ω') } ={ω₁,ω₂,ω₃,ω₄}= Ω f⁻¹({1})= {ω₁} f⁻¹({2})= {ω₂,ω₃,ω₄} These we are elements in F thus the function f is measurable

Answer 11

No, clearly always g⁻¹(∅)= ∅ g⁻¹(Ω)= f⁻¹({1,2})= {w∈Ω| f(w) ∈Ω') } ={ω₁,ω₂,ω₃,ω₄}= Ω g⁻¹({1})= {ω₁, ω₂} but doesn't belong to F not an element of F

Answer 12

Let (Ω, F) and (Ω', F') be two measurable spaces. Let A' be a collection of subsets of Ω' and assume that F' = σ(A'). (the sigma algebra is generated by a collection of subsets) Then, a function f : Ω → Ω' is F/F' -measurable **if and only if** f⁻¹(B) ∈ F for all B ∈ A' (A' the generating class) ------------------------------------------ if inverse image is a subset then is measurable but in this case F' is generated by class A so we don't need to check the inverse for all elements of F but only of B ∈ A', the generating class A' In particular If target space is R and borel sigma algebra, suffices to check inverse image of one of our prev A_i is a measurable set .

Answer 13

We know g ◦ f : Ω₁ → Ω₃ is defined on Ω₁.To show that g ◦ f : Ω₁ → Ω₃ is F₁/F₃-measurable: Let A∈F₃. We want to show that inverse image of A belongs in F₁. (g ◦ f)⁻¹(A) = f⁻¹(g⁻¹(A)). We know g is F₂/F₃-measurable. Thus g⁻¹(A)∈F₂. Since f is F₁ /F₂-measurable f⁻¹(g⁻¹(A))∈F₁. Thus the composition is F₁/F₃-measurable:

Answer 14

Went through (for students with some background with analysis, "don't get too frustrated" We need to show that f is B(R)/B(R)-measurable. Since B(R) is generated by the collection of open sets in R, recalling Lemma 1.1.9 it suffices to show that f⁻¹(A) ∈ B(R) ∀A ⊂ R such that A is open. To this end, let A ⊂ R be open. Since f is continuous, the preimage f⁻¹(A) is also open. So since open sets are Borel, f⁻¹ (A) ∈ B(R), as required.

Answer 15

(Ω, F) to (R, B(R)) the Borel sigma algebra is considered here on R To check measurable look at inverse image f⁻¹(A) for A ∈B(R) and check it is in F. Or from the lemma check if A in the form A=(a,b). f⁻¹(A) = {w∈Ω| f(w) ∈A } = {w∈Ω| 1 ∈A } = {Ω if 1 ∈A {∅ if 1 not in A constant functions are measurable (swap 1 to constant c)

Answer 16

Let (Ω, F) be a measurable space and let f, g : Ω → R be F-measurable functions. Then f + g and f · g are F-measurable. (Here we are using borel as on R)

Answer 17

Similarly, since the collection of sets of the form [a, ∞) generates B(R), it suffices to show that (infₙ fₙ)⁻¹ ([a, ∞)) ∈ F ∀a ∈ R. Now, (infₙ fₙ)⁻¹ ([a, ∞)) = {ω ∈ Ω : infₙ fₙ ∈ [a, ∞)} = {ω ∈ Ω : ∀n ∈ N, fₙ(ω) ≥ a} =∩_{n∈N}{ω ∈ Ω : fn(ω) ≥ a} =∩_{n∈N} fₙ⁻¹ ([a, ∞)) which is in F since [a, ∞) is Borel, fₙ is F/B(R)-measurable, and σ-algebras are closed under countable intersection. Hence infₙ fₙ is also measurable

Answer 18

𝒻~ caligraphic A I want to check the inverse image of a borel set is a borel set Checking : supₙ (fₙ⁻¹ (A)) From the lemma it suffices to check smaller class A∈B(R)= σ(𝒻) We choose A ⊂𝒻 most convenient 𝒻 = {(-∞,α]| α∈R} (supₙ fₙ)⁻¹ ((-∞,α]) = {ω ∈ Ω : supₙ fₙ(ω) ∈(-∞,α] } = {ω ∈ Ω : supₙ fₙ(ω) ≤ α } (sup< then all are less than) ={ω ∈ Ω : supₙ fₙ(ω) ≤ α for all n∈N } =∩ₙ {ω ∈ Ω : supₙ fₙ(ω) ≤ α } =∩ₙ fₙ⁻¹ ((-∞,α]) ((-∞,α]⊂B(R) is a borel set; the inverse image of a borel set via a measurable function is a measurable set so fₙ⁻¹ ((-∞,α])∈F for all n in N) Thus for a sequence of elements in the sigma algebra, the intersection will also be in ∩ₙ fₙ⁻¹ ((-∞,α]) ∈F thus measureable

Answer 19

provided each fₙ is measurable supₙ fₙ is measurable and infₙ fₙ is measurable Thus the limit lim inf fₙ is measurable (limₙ inf fₙ = supₖ(inf _ₙ≥ₖ fₙ ) take the infimum of fₙ for n≥k looking at infimum of less and less things means its a increasing sequence in k for limits of sequences of fₙ is measurable functions and the limits exists then the limit is measurable function too limits, compisitions of measurable functions are measurable!

Answer 20

didnt go through in particular but mentioned it If for each ω ∈ Ω the limit limn→∞ fₙ(ω) exists, then f(ω) = limₙ→∞ fₙ(ω) = lim inf ₙ→∞ fₙ(ω) = sup_{n≥0} inf_{m≥n} fₘ(ω) which is measurable by part 1. of the exercise.

Answer 21

Let (Ω, F) be a measurable space. A non-negative function P : **F** → **[0, 1]** is called a **probability measure** if 1. P(∅) = 0 2. If Aₙ ∈ F, for n ∈ N, and Aₙ ∩ Aₘ = ∅ for n ≠ m, then P(∪ₙ₌₁∞ Aₙ) = Σₙ₌₁∞ P(Aₙ) (disjoint union = sums of individual probabilities) 3. P(Ω) = 1 (Funct P is defined for arguments which are sets)

Answer 22

PROBABILITY SPACE

Answer 23

A non-negative function µ : F → [0, +∞] satisfying 1 and 2 of the above definition will be called a measure A probability space is a particular case of a measure space

Answer 24

if there exists B ∈ F such that P(B) = 0 and A ⊂ B

Answer 25

The probability space (Ω, F, P) is called complete if F contains all the null sets.

Answer 26

The probability measure gives properties for disjoint unions so rewrite as a disjoint set union We have that B= A∪ (B\A) Since A and (B \ A) are disjoint, we get P(B) = P(A) + P(B \ A). Since P(B \ A) ≥ 0, we get P(B) ≥ P(A).

Answer 27

Ω= {1,2,3,4,5,6} F= P(Ω)={A| A⊆Ω} power set = {{1,2,3,4,5,6}, ∅, {1}, {2,3,4,5,6},....} 2^6 elements P(A)= #A/6 P({2,4,6}) = 3/6=1/2 take disjoint sets union prob= sum P( {2,3} U {5,6})= P ({2,3,5,6})= 4/6 or sum of them

Answer 28

(Ω, F, P) Let Aₙ, Bₙ ∈ F, for n ∈ N such that Aₙ ⊂ Aₙ₊₁ and Bₙ₊₁ ⊂ Bₙ for all n ∈ N. Then, i) P(∪ₙ Aₙ) = lim_{n→∞} P(Aₙ) Sequence of events which is increasing (largerset cont original set) ii) P(∩ₙ Bₙ) = lim{n→∞} P(Bₙ) Sequence of events which is decreasing (smaller and smaller subsets of a set)

Answer 29

In general not true, consider the real line Ω,=[0,1] P((a,b)) = b-a B_1= [1/2,1] B_n = [0,1] for all n >=2 then intersection of all B_n's is [1/2,1] which has measure 1/2 but the limit of the final probability should be one (its one for all B_n where n =2,3,4,.... thus this doesnt work as don't decrease

Answer 30

remark: A₁ and A₂ are not Ω; might have something smaller but not equal to Ω; there exists which have measure 0 without being the empty set e.g signleton set A={1/2} in measure [1/2 -e, 1/2 + e] = 2e will have measure 0} (We know as A₁⊆ A₁∩A₂ andA₂ ⊆ A₁∩A₂ the probabilities will be ≤) proof will differ from notes Sol: Let us prove that **P (A₁∪A₂)= P(A₁)+P(A₂)- P(A₁∩A₂)** We don't know if they are disjoint so we express as disjoint union A₁ ∪ A₂ = (A₁∩A₂) ∪ (A₁\A₂) ∪(A₂\A₁) Thus by the properties of prob measures with this sum of disjoint sets P(A₁∪A₂) = P(A₁∩A₂) +P(A₁\A₂)+P(A₂\A₁) (1) Write A₁ = (A₁\A₂) ∪ (A₁∩A₂) disjoint P(A₁) = P(A₁\A₂) + P(A₁∩A₂) Rearranges to P(A₁\A₂) = P(A₁)- P(A₁∩A₂) Write A₂ = (A₂\A₁) ∪ (A₁∩A₂) similarly rearranges to P (A₂\A₁) = P(A₂ )- P(A₁∩A₂) sub into (1) P(A₁∪A₂) = P(A₁∩A₂) +[ P(A₁)- P(A₁∩A₂)]+[ P(A₂ )- P(A₁∩A₂)] =P(A₁)+P(A₂)- P(A₁∩A₂) Thus by using this formula if P(A₁) = P(A₂) = 1 We know 1 ≥P(A₁∪A₂) ≥ P(A₁) =1 thus P(A₁∪A₂)=1 P(A₁∪A₂) = 2- P(A₁∩A₂) (≤ 1 (its a probability he didnt use that fact) Thus 1≤ P(A₁∩A₂). 1 = 2- P(A₁∩A₂) Thus P(A₁∩A₂) = 1

Answer 31

events such that each have probability 1 their infinite intersection will also: Claim that ∀k∈N P( ∩ₙ₌₁ᵏ Aₙ) = 1 Base case: k=1 P( A₁ ) = 1 Induction hypothesis: Assume P( ∩ₙ₌₁ᵏ Aₙ) = 1 Inductive step: Consider P( ∩ₙ₌₁ᵏ⁺¹ Aₙ) = P( (∩ₙ₌₁ᵏ Aₙ) ∩ Aₖ₊₁ ) We know that P(Aₖ₊₁ )=1 by given sequence and also by induction hypothesis we have that P( ∩ₙ₌₁ᵏ Aₙ) = 1 And the previous exercise shows P( ∩ₙ₌₁ᵏ⁺¹ Aₙ) = P( (∩ₙ₌₁ᵏ Aₙ) ∩ Aₖ₊₁ ) = 1 Therefore we have shown by induction that this is true for all k in N. Dealing with infinity: Define new set Bₖ = ∩ₙ₌₁ᵏ Aₙ B₁ =A₁ B₂ = A₁∩A₂ ⊆ A₁= B₁ B₃= A₁∩A₂∩A₃ ⊆ A₁∩A₂ = B₂ In general Bₖ₊₁ ⊆ Bₖ decreasing seq of sets we know that for a decreasing seq the probability of the intersection is the limit of the probabilities ∩ₙ₌₁∞ Bₖ = ∩ₙ₌₁∞ ∩ₙ₌₁ᵏ Aₙ = ∩ₙ₌₁∞ Aₙ an intersection of decreasing sets thus P(∩ₙ₌₁∞ Aₙ) = P(∩ₙ₌₁∞ Bₖ) = lim_{ₖ to ∞} P(Bₖ) = 1 we showed by induction so limit is 1

Answer 32

e.g Aₙ ∪ Aₙ₊₁ ∪ Aₙ₊₂= Bₙ then take ∩ₙ₌₁ ∞ Bₙ Take w∈{Aₖ i.o.} ⟺ w∈ ∩ₙ₌₁ ∞ ∪ⱼ₌ₙ∞ Aⱼ ⟺ w∈ ∪ⱼ₌ₙ∞ Aⱼ for all n in N so there exists m>n s.t w∈ A_m w belongs to infinitely many A_n consider A₁,A₂,A₃,A₄,A₅,A₆,A₇,... Give me and n =4 say, i can find m >n=4 s.t w ∈A_m but if this is true for every n then I can always find a later set s.t w belongs, true for any n meaning w belongs to infinitely many.

Answer 33

Ω={H,T} flip once sample space Flip twice then sample space Ω={(H,H), (H,T), (T,H) ,(T,T)} NOTATION differs to {H,H} thus Ω_2 = { (a₁,a₂) | a_i∈{H,T}} Ωₙ = { (aₙ)ₙ₌₁∞ | a_i∈{H,T}} Ω = { (aᵢ)ᵢ₌₁∞ | aᵢ∈{H,T}} countable as its the # of reals

Answer 34

Ω = { (aᵢ)ᵢ₌₁∞ | aᵢ∈{H,T}} Lets consider set/event Aₙ= I get H nth time Aₙ={(a₁,a₂,a₃,..,aₙ₋₁Haₙ₊₁,...l) |aᵢ∈{H,T} for i ≠n} e.g A₂= {(H,H), (T,H)} = {(a₁,a₂)| |a₁∈{H,T} a₂ =H} ??? e.g A₂= {(H,H), (T,H), (T,H,T), (H,H,H), (H,H,T), (T,H,H),....} If I choose w ∈ Aₙ I know w= (a₁,a₂,a₃,..,aₙ₋₁H........) If we take w ∈ {Aₖ i.o.} ⟺ w∈ ∩ₙ₌₁ ∞ ∪ⱼ₌ₙ∞ Aⱼ ⟺ w∈ ∪ⱼ₌ₙ∞ Aⱼ for all n in N so there exists m>n s.t w∈ A_m w belongs to infinitely many A_n ie for n=1 there is an m_1 s.t w∈ A_m_1 n=2 there is an m_2 > m_1+1 s.t w∈ A_m_2.... So our H will be at position m_1,m_2,... true for any m thus we have infinitely many heads?

Answer 35

If we have a subset A of B then P(A)≤ P(B) B increasing sequence then intersection: i) P(∪ₙ Aₙ) = lim_{n→∞} P(Aₙ) Sequence of events which is increasing (largerset cont original set) ii) P(∩ₙ Bₙ) = lim{n→∞} P(Bₙ) Sequence of events which is decreasing (smaller and smaller subsets of a set)

Answer 36

complement of having infinitely many heads we will have finitely many heads we have heads up to some point, but after that there are only tails {Aₖ i.o.}ᶜ = (∩ₙ₌₁ ∞ ∪ₙ₌ₘ∞ Aₙ)ᶜ de Morgans twice gives = ∪ₙ₌₁ ∞ (∩ₙ₌ₘ∞ (Aₘᶜ)) Taking w∈ ∪ₙ₌₁ ∞ (∩ₙ₌ₘ∞ (Aₘᶜ)) = {Aₘᶜ eventually} thus there exists an n s.t w∈ ∩ₙ₌ₘ∞ (Aₘᶜ)) this means there exists n in N s.t for all m≥n w∈ Aₘᶜ ie for all positions greater than n we have w not in A_m so the position coords are not heads ie they are all tails

Answer 37

then {Aₖ i.o.} = ∩ₙ₌₁ ∞ ∪ⱼ₌ₙ∞ Aⱼ {Aₙ eventually}= ∪ₙ₌₁ ∞ (∩ₙ₌ₘ∞ (Aₘᶜ))

Answer 38

Consider (Ω, F, P) Aₖ⊆ F k in N Let (Aₖ)ₖ₌₁∞ be a sequence of events. (so they lie in our sigma algebra) If sumₖ₌₁∞ P(Aₖ) < ∞, then P(Aₖ i.o.) = 0, where {Aₖ infinitely open} {Aₖ i.o.} = ∩ₙ₌₁ ∞ ∪ⱼ₌ₙ∞ Aⱼ -------------------------------- If this is 0 then P({Aₖ i.o.}ᶜ)= P({ Aₖ eventually})= 1 the sum of the probabilities of the events {En} is finite then the probability that infinitely many of them occur is 0, that is,

Answer 39

12 feb 1.2 expectations? missed some?

Answer 40

Let (Ω, F, P) probability space function X:Ω → R is a **random variable* if X is F -measurable (F/B(R)- measurable) (meaning every time we take an element in B(R), then the pre-image of X is in F)

Answer 41

σ(X, Y ) := σ({X−1(A) : A ∈ B(R)} ∪ {Y −1(B) : B ∈ B(R)}). Notice that the collection {X−1(A) : A ∈ B(R)} ∪ {Y−1(B) : B ∈ B(R)} is not a σ-algebra in general(!). In order to ease the notation, most of the times that we deal with a random variable X we will drop the argument ω. For example, the set {ω ∈ Ω : X(ω) ≤ 1} will simply be denote by {X ≤ 1} of the probability P({ω ∈ Ω : X(ω) ≤ 1}) will simply be denoted by P(X ≤ 1)

Answer 42

Q_X is called the distribution of X or the law of X

Answer 43

Let (Ω, F, P) be a probability space and let X : Ω → R be a random variable. The distribution of X is a probability measure Q_X on the (R, B(R)), which is given by Q_X(A) = P(X ∈ A) for all A ∈ B(R) . (=P(w∈Ω, X(w)∈A)) (if X is not measurable not defined)

Answer 44

F_X(x)= P( X≤ x) =P(X in (-infinity,x]) is a particular case of Q_X remember those sets generate the whole B(R)

Answer 45

A random variable X is called simple if there exist c_1, ..., c_n ∈ R and events A_1, ..., A_n ∈F (sigma algebra) such that X =Σᵢ₌₁ⁿ cᵢ1_A_ᵢ

Answer 46

If X is a **simple** with representation defined prev, then we define the **expectation of X** by E[X] : =Σᵢ₌₁ⁿ cᵢP(Aᵢ) It turns out that the expectation does not depend on the representation of X

Answer 47

if X, Y are simple random variables and a, b ∈ R, then E[aX + bY ] = aE[X] + bE[Y ].

Answer 48

It is also easy to see that if X, Y are simple random variables eith X ≤ Y , then E[X] ≤ E[Y ].

Answer 49

(If using that X is a non-negative RV then there exists a sequence random variables X_n, n ∈ N simple RVs such that for all ω ∈ Ω, X_n(ω) ↑ X(ω), as n ↑ ∞. Indeed, we can set X_n = h_n(X), where h_n(x) = min{[x2^n]2^{−n}, 2^n} where [a] denotes the lower integer part of a ∈ R.

Answer 50

Take X_n= h_n(X) where H_n(X)= min {x2^n} 2^{-n}, 2^n} (lower integer part?) (idea in notes)

Answer 51

E(X) =ᵈᵉᶠ = limₙ→ ∞ E(X_n) where X_n is a sequence converging to X as n tends to infinity X_n is an increasing sequence well defined as X_n is a non decreasing seq

Answer 52

E[X+] < ∞, E[X−] < ∞, where X+=max{X,0} X-= max{-X,0} equivalently X is integrable IFF if E{|X|}< ∞

Answer 53

Expectations are well-defined bc X+ X- and |X| are non-negative

Answer 54

denote the set of all integrable random variables by L₁(Ω). L₁(Ω)={all integrable r.vs} ={X:Ω→R r.vs s.t E(|X|)< ∞ } AND For X ∈ L₁(Ω), we set E[X] = E[X⁺] − E[X⁻]

Answer 55

Let X, Y ∈L₁(Ω),and a, b ∈ R. Then the following hold: 1. E[aX + bY ] = aE[X] + bE[Y ] linearity 2. If X ≤ Y , then E[X] ≤ E[Y ]. monotonicity

Answer 56

E(X)= Let X be a random variable. Then, for any non-negative B(R)-measurable g : R → R we have E[X)] =∫_Ω X dP or ∫_Ω X(w) dP(w) y the integral of X with respect to the measure P (P is a measure, lebesgue measure wrt to P)

Answer 57

E(X)= ∫_R xfₓ(x) .dx E(g(X))= ∫_R g(x)fₓ(x) .dx can you always find pdf? For these cases we work with the law of distribution of the RV (for discrete is sum and probability mass funct)

Answer 58

Let X be a random variable. Then, for any non-negative B(R)-measurable g : R → R we have E[g(X)] =∫_R g(x) Qₓ dx. where Qₓ is the distribution function (probably measure) valid for functs g taking values with different signs as long as well defined g+ and g- using g=g+ -g-

Answer 59

E(g(X))= ∫_R g(x)fₓ(x) .dx Actually, this is not a definition but a consequence of Theorem 1.2.3, since, the expression "X has a density f" means exactly that the distribution of X satisfies Qₓ(dx) = fₓ(x) dx.

Answer 60

= (E[|X|ᵖ])¹/ᵖ For p ∈ [1, ∞) (when p=1 this norm is the expectation of value X; collection of all random values in L_p space forms a linear subspace, 0 is constant therefore measurable and this tells you how far from **0**) (links to integral expectation E.G p=2)

Answer 61

= inf{K ≥ 0 : P(|X| ≤ K) = 1}. For p ∈ [1, ∞) so taking the K for which our RV absolute Probability is within 1

Answer 62

For p ∈ [1, ∞) denote Lₚ(Ω) the linear space of all random variables X : Ω → R such that ||X||ₗₚ < ∞. s,t the L_p norm is finite

Answer 63

Let X be a random variable. Then, for any ε > 0 we have P(|X| ≥ ε) ≤ (1/ε)E[|X|] probability greater than epsilon is bounded by a bound which depends on expectation

Answer 64

Let p, q ∈ [1, ∞] with p⁻¹ + q⁻¹ = 1. Then for any X,Y rvs ||XY ||_L₁ ≤ ||X||_Lₚ ||Y ||_L_q L₁ norm is bounded for the product (generalisation of cauchys inequality) (E[|Xy|]) ≤ (E[|X|ᵖ])¹/ᵖ , (E[|Y|ᵠ])¹/ᵠ where power of q)

Answer 65

Let p ∈ [1, ∞]. For any random variables X and Y we have ||X + Y ||_Lp ≤ ||X||_Lp + ||Y ||_Lp (L_p spaces are normed vector spaces) .

Answer 66

Hint: using Holders inequality Let p, q ∈ [1, ∞] with p⁻¹ + q⁻¹ = 1. Then for any X,Y rvs ||XY ||_L₁ ≤ ||X||_Lₚ ||Y ||_L_q Then ||1.X ||_L₁ ≤ ||1||_Lₚ ||X ||_L_q = (E[|1.X|]) ≤ (E[|1|ᵖ])¹/ᵖ . (E[|X|ᵠ])¹/ᵠ Suppose p ≥ q (ᵠ is ^q here) ||X||_Lq ᵠ = E[|X|^q] by defn = E[1 |X|^q] by holder's ≤ E[1][E[(|X|ᵠ)ᵖ/ᵠ]) ᵠ/ᵖ = ||X||_Lp ^q Hence, raising both sides of the inequality to the 1/q finishes the proof

Answer 67

Let p ∈ [1, ∞]. Let X_n ∈ LLₚ (Ω), for n ∈ N such that lim_{n,m→∞} ||X_n − X_m||Lₚ = 0. (Cauchy) Then there exists X ∈ L(Ω), such that lim_n→∞ ||Xn − X||Lₚ = 0. (Every cauchy seq in the space converges to some value in the space)

Answer 68

As we have seen, random variables are functions defined on a set Ω. As such, we have different modes of convergence.

Answer 69

For a sequence of RVs Let (Xₙ)_n∈N, be a sequence of random variables. We say that Xₙ converges to a random variable X almost surely if P{ω ∈ Ω : lim_{n→∞} Xₙ(ω) = X(ω)}= 1. We usually write Xₙ → X almost surely (This is a measurable set, we apply the probability measure to the measurable set)

Answer 70

Let (Xₙ)_n∈N, be a sequence of random variables. We say that Xₙ converges to a random variable X in probability if for any ε > 0 lim {ₙ→∞} {ω ∈ Ω : |Xₙ(ω) − X(ω)| ≥ ε}= 0. We usually write Xₙ → X in probability. (probability: no matter how arbitrary small epsilon the limit)

Answer 71

Let (Xₙ)_n∈N, be a sequence of random variables that belong to ₚ(Ω). We say that Xn converges to a random variable X ∈Lₚ(Ω) in Lₚ if lim_{n→∞} ||Xₙ − X||_Lₚ = 0 We usually write Xₙ → X in Lₚ

Answer 72

Let (Xₙ)_ₙ∈N, X be random variables. The following hold. (i) If Xₙ → X almost surely, then Xₙ → X in probability. (ii) If Xₙ → X in probability, then there exists a subsequence (X_nₖ)k∈N such that X_nₖ → X almost surely. (iii) If Xₙ → X in Lp, then Xₙ→ X in probability.

Answer 73

Let ε > 0. By Markov’s inequality, we have P(|Xₙ − X| > ε) = P(|Xₙ − X|ᵖ> εᵖ) ≤ (1/εᵖ) E[|Xₙ − X|ᵖ] = (1/εᵖ) ||Xₙ − X||ₗₚᵖ Taking limits: The right hand side of the above inequality tends to zero as n → ∞, by assumption. Hence, so does the left hand side. This shows that X_n → X in probability ignore: Let (Xₙ)_n∈N, be a sequence of random variables that belong to ₚ(Ω). Suppose Xₙ → X in Lₚ for X ∈Lₚ(Ω) in Lₚ THEN lim_{n→∞} ||Xₙ − X||_Lₚ = 0 by defn of convergence for all ε > 0 there is an N s.t for all n>N ||Xₙ − X||_Lₚ < ε thus... we need to show that for any ε > 0 lim {ₙ→∞} {ω ∈ Ω : |Xₙ(ω) − X(ω)| ≥ ε}= 0.

Answer 74

Let Xₙ, X : Ω → R, for n ∈ N, be non-negative random variables such that **almost surely** Xₙ ↑ X as n ↑ ∞. Then E[X] = limn→∞ E[Xₙ]

Answer 75

Let Xₙ : Ω → R, for n ∈ N, be non-negative random variables. Then E[lim inf n→∞ Xₙ] ≤ lim_n→∞ inf E[Xₙ]

Answer 76

Suppose that Xₙ → X **almost surely** and that **there exists** Y ∈ L₁(Ω) such that |Xₙ| ≤ Y . (bounded) Then X ∈ L_1(Ω) and E[X] = lim_n→∞ E[Xₙ].

Answer 77

consider a probability space (Ω, F, P) The events A₁, ..., An are called independent if P(∩ ᵢ₌₁ ⁿ Aᵢ₌₁) = ∏ᵢ₌₁ ⁿ P(A ᵢ). probability is multiplies for intersections of indep events

Answer 78

sigma algebras G_1, ..., G_n are called independent if for any A_1 ∈ G1, ..., A_n ∈ G_n, the events A_1, ..., A_n are independent.

Answer 79

The random variables X_1, ..., X_n are independent if the σ-algebras σ(X_1), ...σ(X_n) are independent

Answer 80

the elements of a family of events, σ-algebras, random variables are independent if the element of any finite sub-family are independent.

Answer 81

Let G_1, G_2 ⊂ F be σ-algebras. Let curly_A_1 ⊂ G_1 and curly_A_2 ⊂ G_2 be π-systems (classes that are closed under finite intersection) such that σ(curly_A_1) = G_1 and σ(curly_A_2) = G_2. (sigma algebras gen by these events) Then G_1 and G_2 are independent if and only if P(A_1 ∩ A_2) = P(A_1)P(A_2) for all A_1 ∈ curly_A1, A2 ∈ curly_A2.

Answer 82

XY ∈ L_1(Ω) and E[XY ] = E[X]E[Y ]

Answer 83

Let X₁, ..., Xₙ be **independent** random variables, m₁ + ... + mₖ = n and f₁, ..., fₖ be measurable functions of Rᵐ₁, ..., R ᵐₖ respectively. Then the random variables Y₁ = f₁(X₁, ..., Xₘ₁), Y₂ = f₂(Xₘ_₁ ₊₁ , ..., Xₘ_₁ ₊ₘ_₂ ), ..., Yk = f(Xₘ_₁+...+ₘ_{ₖ−₁}₊₁, ..., Xₙ) are independent.

Answer 84

Let X ∈ L_1(Ω) and let G ⊂ F be a sub-σ-algebra of F. One can show that there exists a unique Y ∈ L1(Ω) with the following two properties: 1. Y is G-measurable 2. E[**1**_A X] = E[**1**_A Y ] for all A ∈ G (indicator funct for A) Y ~ conditional expectation of X given G Y = E[X|G].

Answer 85

For (Ω, F, P) and RV X X:Ω to R Let X ∈ L_1(Ω) ( IFF E[|x|] < infinity) and let 𝒢 ⊂ F be a sub-σ-algebra of F. There exists a unique RV Z∈ L_1(Ω) (z dep on Z and 𝒢) s,t E[**1**_A X] = E**1**_A Z] for all A ∈𝒢 So **conditional expectations** given a sigma algebra are a random variable themselves Z= E[X|𝒢]

Answer 86

elements of sigma algebra 𝒢 Aᵢ∈𝒢 unions ∪_{i in I} Aᵢ ∈𝒢 complements e.g complement of A_1 will be a union of other partition elements intersections and complements of these turn out to be unions turns out all elments are unions of the A_is

Answer 87

We want to show that almost surely on Ω, we have E(X|𝒢) = Y , i.e. that E[**1**_G X] = E[**1**_G Y ] ∀G ∈ 𝒢 We want to show that for all A∈𝒢 E[ **1**_A x]= E[ **1**_A Σᵢ₌₁ ⁿ cᵢ 1_Aᵢ] Suppose A= Aⱼ j∈{1,...,n} Then E[ **1**_Aⱼ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ] =E[ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ **1**_Aⱼ ] has value 1 only if we are considering the intersection, both 1 = E[ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ ∩_Aⱼ ] = E[cⱼ **1**_Aⱼ ] =(***) ------------------------------- since A₁, . . . , Aₙ are disjoint, so **1**_Aᵢ ∩_Aⱼ = {1_∅ if i ≠ j (empty set has no elements) {1Aᵢ if i = j = {0 if i ≠ j {1_Aⱼ if i = j. ------------------------------------ Replacing cⱼ (***)= = E[cⱼ **1**_Aⱼ ] = E[(1/P(Aⱼ)) E[**1**_Aⱼ X] **1**_Aⱼ ] = cⱼ E[**1**_Aⱼ ] by linearity of expectation, multiplying by RV =(1/P(Aⱼ))) E[**1**_Aⱼ X] E[**1**_Aⱼ] by expectations of simple RVs (E[**1**_Aⱼ] =P(Aⱼ) =E[**1**_Aⱼ X] ------------- Now checking the case for all A∈𝒢 : Suppose A= ∪_{j∈I} Aⱼ j∈{1,...,n} E[ **1**_∪_{j∈I}Aⱼ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ] = E[ Σⱼ **1**_Aⱼ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ] expectation of sum =Σⱼ∈I E[**1**_Aⱼ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ] singular A_j =Σⱼ∈I E[**1**_Aⱼ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ] =Σⱼ∈I E[**1**_Aⱼ X] =E[**1**_{∪_{j∈I}Aⱼ} C ] -------------- characteristic of a union **1**_∪_{j∈I} =Σⱼ **1**_Aⱼ -------

Answer 88

since A₁, . . . , Aₙ are disjoint, so **1**_Aᵢ ∩_Aⱼ = {1_∅ if i ≠ j (empty set has no elements) {1Aᵢ if i = j = {0 if i ≠ j {1_Aⱼ if i = j.

Answer 89

E[ Σᵢ₌₁ ⁿ **1**_Aᵢ] =Σᵢ₌₁ ⁿ P(Aᵢ) E[ Σᵢ₌₁ ⁿ cᵢ **1**_Aᵢ] =Σᵢ₌₁ ⁿ cᵢ P(Aᵢ) takes finitely many values

Answer 90

**1**_∪_{j∈I}Aⱼ =Σⱼ **1**_Aⱼ = beacuse w belongs to exactly one of them, disjoint thus others are 0

Answer 91

In general conditional expectations are random variables not # they can be numbers if sigma algebra is trivial

Answer 92

Let X, Y ∈ L₁(Ω) (by defn can take the cond exp) and 𝒢 ⊂ F be a σ-algebra. (by defn its a RV so is F-measurable) The following hold 1. **Linearity**: If a, b are constants, then E[aX + bY |𝒢] = aE[X|𝒢] + bE[Y |𝒢]. 2. Taking out what is known: If X is 𝒢-measurable and XY is integrable, then E[XY |𝒢] = XE[Y |𝒢]. **In particular, if X is G-measurable, then E[X|𝒢] = X.** 3. **Conditioning** on independent information: If W ⊂ F is a σ-algebra independent of σ(σ(X), 𝒢), then E[X|σ(G, W)] = E[X|𝒢]. In particular, if σ(X) is independent of W, then E[X|W] = E[X]. 4. **Tower property**: If W ⊂ 𝒢 is a σ-algebra, then we have E[X|W] = E[E[X|𝒢]|W]. 5. **Monotonicity**: If X ≥ 0 then E[X|𝒢] ≥ 0. 6. **Conditional Jensen’s inequality**: If φ : R → R is convex, such that φ(X) ∈ L₁(Ω), then φ(E[X|𝒢]) ≤ E[φ(X)|𝒢].

Answer 93

sequence of RVS converge in L₁ then conditional exp converges in L₁ Xₙ converges to X E[Xₙ|G] converges to E[X|G]| ------------- Let (Ω, F, P) be a probability space, and let 𝒢 ⊂ F be a sub-σ-algebra. Suppose that Xₙ, X ∈ L_1(Ω, F, P) and that limn→∞ ||Xₙ − X||_L₁Ω = 0. Then lim_n→∞ ||E[Xₙ|𝒢] − E[X|𝒢]||L₁(Ω) = as E[·|𝒢] is a linear operator: =lim_n→∞ ||E[Xₙ-X|𝒢]||_L₁Ω By L_1-norm abs value =lim_n→∞ E[ | E[Xₙ-X|𝒢] | ] since | · | is convex, by Jensen’s inequality: (absolute value inside) ≤ lim_n→∞ E[ E[ | Xₙ-X | |𝒢] ] by the law of total expectation(or tower property) lim_n→∞ E[ | Xₙ-X | ] by the def. of the L1(Ω)-norm: lim_n→∞ || Xₙ-X ||L₁(Ω) by assumption =0

Answer 94

E[ X |𝒢] =E[X] ------------ suppose we take a trivial sigma algebra W= {Ø,Ω} Then E[ X |W] =E[X] It suffices to check that E[**1**_A E[X]]= E[**1**_AX] for all A ∈W sigma algebra W if this is true then LHS = E[ X |W] is true as can either by empty set or Ω easy to check for 2 tower property (E[E[X |𝒢] = E[ E[X |𝒢] |W] = E[ X |W]= E[X] , as the trivial sigma algebra is always contained in 𝒢

Answer 95

By defn the conditional expectation E[X|G] is G-measurable E[X|G] ∈L₂ᴳ(Ω). (check: E[X|G] is G-measurable and finite uder L2 norm) Moreover, by the conditional Jensen’s inequality we see that E[|E[X|G]|²] ≤ E[E[|X|²|G]] = E[|X|²] < ∞. Consequently, E[X|G] ∈ L₂ᴳ and by definition of the minimum we get ||X − E[X|G]||_L₂ ≥ min_{Z∈L₂ᴳ} ||X − Z||_L₂ (by definition I am taking the smallest from the set, for each RV Z i have a real # each one is above min) tbc .

Answer 96

On the other hand, for any Z ∈ L₂ᴳ ||X - Z||ₗ₂ ² = ||X - Z − E[X|G] + E[X|G] ||ₗ₂ ² By expanding ² = ||X − E[X|G]||ₗ₂ ² + E[(X − E[X|G])(E[X|G] − Z)] +||X - Z||ₗ₂ ² ≥ ||X − E[X|G]||ₗ₂² + E[(X − E[X|G])(E[X|G] − Z)] We claim the last term on the RHS is zero. Indeed, since (E[X|G] − Z) and E[X|G] are G-measurable, we have E[(X − E[X|G])(E[X|G] − Z)]= we can take an extra conditional expectation inside the expectation bc the expectation of the conditional expectation is the expection itself = E[E[(X − E[X|G])**(E[X|G] − Z)**|G]] "(E[X|G] − Z)is G measurable something times something g-measurable given g we can use the property to pull it out" = **E[(E[X|G] − Z)**E[(X − E[X|G])|G]] "claim (E[X|G] − E[X|G])] is 0,as conditional exp are linear , by defn conditional exp is G measurable so cond exp wrt g again doesnt change" = E[(E[X|G] − Z)(E[X|G] − E[X|G])] = E[(E[X|G] − Z) · 0] = 0. showing for any such Z the inequality holds

Answer 97

L₂ᴳ (Ω)= {X: Ω→R| X is 𝒢-measurable , ||X||_L₂ < ∞.} L₂(Ω)= {X: Ω→R| X is F-measurable , ||X||_L₂ < ∞.} L₂ᴳ (Ω)⊆L₂(Ω) this is because if I choose an X in L₂ᴳ (Ω) it is 𝒢-measurable thus inverse images of borel sets belong to 𝒢: ∀A∈B(R), X⁻¹(A)∈𝒢⊆F if I take any 𝒢-measurable it is automatically F-measurable due to 𝒢 ⊂ F being a sub-σ-algebra of F.

Answer 98

consider R^2 plane taking vector in R^2 **x,y** in R^2 also operation +:closed under + **x+y** also in R^2 closed under scalar multiplication: a**x** in R ^2 if X is F-measurable , ||X||_L₂ < ∞: im claiming that for two RVS X,Y X+Y is also a RV in L_2 F-measurable ||X+Y||_L₂ by triangle =< ||X||_L₂ +||Y||_L₂ < ∞ is also finite aX is also a RV ||aX||_L₂ < a||X||_L₂ < ∞ another interpretation is thatL₂ᴳ (Ω)⊆L₂(Ω) lies on the line, any Z in L₂ᴳ (Ω) lies along the line, we can scale etc and maintain closure under these operations

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