1 basics Flashcards
Let f : R to R
x_0 ∈R
consider
x’(t)=f(x(t)) for t ∈ [0,T]
x(0)=x_0
Give a
A sol
A sol is a func x: [0,T] to R of class C^1 and satisfies equalities
e.g. let r>0 x’(t) = rx(t) t ∈ [0,T]
x(0)=1
unique sol
x(t) = exp(rt) t ∈ [0,T]
linear differential equation:
solve by multiplying both sides by exp(-rt) and then noting its integrating by part form to solve
exp(-rt) x(t) = 1
x(t) = exp(rt)
chain rule and ftoc
exp growth
is brownian motion differentiable?
links to avg value of stock, but the value of stock is more stochastic in nature
Brownian motion/wiener process is an example of a stochastic process
brownian motion is non differentiable at “corners” nowhere differentiable, everywhere continuous
2d brownian motion
modelled in 2 components
W_1 and W_2
A union B
{x: x∈A or x∈B}
A intersection B
{x: x∈A and x∈B}
A complement
Aᶜ:= Ω \ A ∈ F
sample space Ω
Non empty set of outcomes
subsets of the sample space Ω are EVENTS collection of all EVENTS is a SIGMA ALGEBRA
Ω COUNTABLE
X={1,2,3}
P(X)
how many elements
P(X)=
{∅,{1,2,3}, {1},{2},{3}, {1.2}, {1,3}, {2,3}}
2^3 = 8 elements incl empty set
the power set is an example of sigma algebra and is the largest example for a given X
Definition 1.1.1. A σ-algebra
Definition 1.1.1. A σ-algebra F on Ω is a collection of subsets of Ω such that
(i) Ω ∈ F (WHOLE SPACE IS AN ELEMENT)
(ii) If A ∈ F then Aᶜ:= Ω \ A ∈ F COMPLEMENTS
(iii) If Aₙ ∈ F for n ∈ N, then ∪ₙ₌₁ ∞ Aₙ ∈ F. (COUNTABLE UNIONS)
(EMPTY SET WILL BE IN)
∅ ∈ F
F σ-algebra
(EMPTY SET WILL BE IN)
Let (Ω, F) be a measurable space. Since F is a σ-algebra which is closed under complementation Ωᶜ∈ F ∅ ∈ F
σ-algebra
closed under intersections?
Show that if Aₙ ∈ F for n ∈ N, then ∩ₙ₌₁ ∞ Aₙ ∈ F
If Aₙ ∈ F for n ∈ N then Aₙᶜ ∈ F
By De-Morgan’s identity
∩ₙ Aₙ = ∩ₙ (Aₙᶜ)ᶜ = (∪ₙ Aₙᶜ )ᶜ
which is in F because σ-algebras are closed under complementation and countable union.
The pair (Ω, F)
The pair (Ω, F) is called a measurable space
measurable sets
elements of the sigma algebra
summary sigma algebra
(If i have an event A in sigma algebra I want the complement of my event also in)
(for two events A and B i want unions and intersections to be in the sigma algebra
sequences of complements and unions also contained
any sigma algebra on A is either {complement of A,A} or bigger up to P(A)
Exercise 1.1.3. Consider the set of real numbers R.
(1) Find the smallest σ-algebra on R.
{∅,R}
∅, R ∈ F by the def. of σ-algebras
Exercise 1.1.3. Consider the set of real numbers R.
(2) Find the smallest σ-algebra on R that contains the interval (0, 1).
{∅,R, (0,1), (−∞, 0] ∪ [1, ∞)}
∅, R ∈ F by the def. of σ-algebras
since σ-algebras are closed under complementation
Def σ(A)
1.1.4
Let Ω be a non empty set and let A be a collection of subsets of Ω. We denote
by σ(A) the intersection of all σ-algebras on Ω that contain A, that is
σ(A) := ∩_{B∈Sₐ}B,
where Sₐ = {B : B is a σ-algebra on Ω and
A ⊂ B}.
This is a sigma algebra
The σ-algebra generated by A
σ(A)
intersection of all s.t they contain A
smallest σ-algebra that contains A.
A is a subset of the power set on Ω
Exercise 1.1.5. Show that S_A is non empty.
S_A:= {B : B is a σ-algebra and A ⊂ B}
Let Ω be nonempty set and let A be a collection of subsets of Ω.
Let S_A:= {B : B is a σ-algebra and A ⊂ B}
Then P(Ω) ∈ S_A, because P(Ω) is a σ-algebra, and A ⊂ P(Ω). Hence S_A is non empty.
ex 1.1.5 Show that intersection of σ-algebras is a σ-algebra. Consequently, σ(A) is indeed a σ-algebra.
Let H be a family of σ-algebras.
Then
1) by first axiom Ω ∈ F for all F ∈ H. Therefore
Ω ∈ ∩_{F∈H} F
2) Let A ∈ ∩_{F∈H} F
Then A ∈ F ∀F ∈ H.
So since σ-algebras are closed under complementation:
A^c ∈ F ∀F ∈ H.
Hence
A^c ∈ ∩_{F∈H}F.
So ∩_{F∈H} F is closed under complementation.
3) Let A_1,A_2,… ∈ ∩_{F∈H} F
Then
A_1,A_2,… ∈ F ∀F ∈ H.
So since σ-algebras are closed under countable union:
∪ₙ Aₙ ∈ F ∀F ∈ H.
Hence
∪ₙ Aₙ ∈ ∩{F∈H} F.
So ∩{F∈H} F is closed under countable union.
By 1,2,3: ∩_{F∈H} F is a σ-algebra.
e.g Ω ={1,2,3,4,5}
A={{1} {1,2}, {4,5}}
σ(A)
is there a sigma algebra F s.t A contained in F and if G is another sigma algebra s.t contains A then F contained in G
σ-algebra generated by A
{∅,{1,2,3,4,5},
{1} {1,2}, {4,5},
{2,3,4,5}, {3,4,5}, {1,2,3},
{1,2,4,5}}
{3} {1,4,5}
e.g Ω =R
A={[0,1]}
is there a sigma algebra F s.t A contained in F and if G is another sigma algebra s.t contains A then F contained in G
σ-algebra generated by A
{∅,R, (0,1), (−∞, 0] ∪ [1, ∞)}
Definition 1.1.6. The Borel σ-algebra
The Borel σ-algebra on Rᵈ, denoted by B(Rᵈ) is the σ-algebra generated by the
collection of open sets of Rᵈ
“Consider a sigma algebra generated by A”
A ={u in R s.t U is open}
countable unions of open intervals
The borel sigma algebra is the smallest sigma algebra containing open sets
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
A_1: collection of all open intervals on R
A_2 collection of closed intervals
A_3: collection of (−∞, a]
A_4: collection of (−∞, a)
A_5: collection of (a,+∞)
A_6: collection of [a,+∞)
A_7: collection of closed subsets of R
Then we can show the sigma algebra generated by open sets is the same as the one generated by all of these.
We show σ(A_1)= σ(A_2)……
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
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)
Method in notes
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
(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)
Method in notes
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
continued
(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
σ-algebras are closed under taking differences of sets?
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
Borel sigma algebra considered
On R^d we will always consider the Borel σ-algebra B(R^d).
we choose whichever generates the set most convenient for us
Def 1.1.8
F/F’ measurable
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}
e.g. suppose we have function f mapping: from c to Ω’
ω₁→1
ω₂→3
ω₃→2
ω₄→1
f⁻¹({1,2}) =
when is it measurable F/F’ measurable
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’
F-measurable.
Ω’ = R^d,
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.
e.g Let Ω={ω₁,ω₂,ω₃,ω₄}
σ-algebra F ={∅, Ω, {ω₁},{ω₂,ω₃,ω₄}}
what is this?
σ({ω₁})
ie the smallest sigma algebra generated by {ω₁}
Let Ω={ω₁,ω₂,ω₃,ω₄}
σ-algebra F ={∅, Ω, {ω₁},{ω₂,ω₃,ω₄}}
Let Ω’={1,2}
σ-algebra F’ ={∅, Ω, {1},{2}}
define function f mapping Ω to Ω’:
ω₁→1
ω₂→2
ω₃→2
ω₄→2
check if F/F’ measurable
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
Let Ω={ω₁,ω₂,ω₃,ω₄}
σ-algebra F ={∅, Ω, {ω₁},{ω₂,ω₃,ω₄}}
Let Ω’={1,2}
σ-algebra F’ ={∅, Ω, {1},{2}}
Define function g mapping Ω to Ω’:
ω₁→1
ω₂→1
ω₃→2
ω₄→2
check if g is F/F’ measurable
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
Lemma 1.1.9
for two measurable spaces
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
.
Exercise 1.1.10. Let (Ω₁, F₁), (Ω₂, F₂), and (Ω₃, F₃) be measurable spaces.
Let f : Ω₁ → Ω₂ be F₁ /F₂-measurable and
let g : Ω₂ → Ω₃ be F₂/F₃-measurable.
Show that g ◦ f : Ω₁ → Ω₃ is
F₁/F₃-measurable COMPOSITION
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:
Exercise 1.1.11. Let f : R → R be a continuous function. Show that it is B(R)-measurable (Borel measurable)
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.
Consider
f:R to R
f(x)= 1
is this measurable?
(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)
Proporition 1.1.12
f+g
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)
Exercise 1.1.13. CONSIDERING SEQUENCES OF FUNCTIONS
Let (Ω, F) be a measurable space and let fₙ : Ω → R be F-measurable functions (borel measurable sequence of functs) for n ∈ N.
inf fₙ are measurable.
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
Exercise 1.1.13. CONSIDERING SEQUENCES OF FUNCTIONS
Let (Ω, F) be a measurable space and let fₙ : Ω → R be F-measurable functions (borel measurable sequence of functs) for n ∈ N.
(1) Show that supₙ fₙ,
Will be F measurable
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
Summary from exercises for sequences of measurable functions fₙ
sup and inf
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!
Exercise 1.1.13. CONSIDERING SEQUENCES OF FUNCTIONS
Let (Ω, F) be a measurable space and let fₙ : Ω → R be F-measurable functions (borel measurable sequence of functs) for n ∈ N.
(2) Show that lim supv→∞ fₙ is a measurable functionf
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.
Def 1.1.1.4
PROBABILITY MEASURE
Let (Ω, F) be a measurable space. A non-negative function P : F → [0, 1] is called a probability measure if
1. P(∅) = 0
- 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) - P(Ω) = 1
(Funct P is defined for arguments which are sets)
(Ω, F, P)
PROBABILITY SPACE
A measure
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
Let (Ω, F, P) be a probability space. A set A ⊂ Ω is called null, if
if there exists B ∈ F such
that P(B) = 0 and A ⊂ B
The probability space (Ω, F, P) is called complete if
The probability space (Ω, F, P) is called complete if F contains all the
null sets.
Exercise 1.1.15. Show that if A, B ∈ F and A ⊂ B, then P(A) ≤ P(B).
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).
Lemma 1.1.19 (Borel-Cantelli)
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,
Definition 1.1.20.
random variable
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)
Definition 1.1.23.
DISTRIBUTION OF X
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)
Lemma 1.2.4 (Markov’s inequality).
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
Lemma 1.2.5 (Hölders inequality).
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)
Lemma 1.2.6 (Minkowski’s inequality or triangle inequality)
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)
.
Exercise 1.2.7. Show that if p ≥ q, then ||X||_Lq ≤ ||X||_Lp
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
Definition 1.3.3 (convergence in L_p).
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
Lemma 1.4.2 (Fatou’s Lemma)
Let Xₙ : Ω → R, for n ∈ N, be non-negative random variables.
Then
E[lim inf n→∞ Xₙ] ≤ lim_n→∞ inf E[Xₙ]
Theorem 1.4.3 (Lebesgue’s theorem on Dominated Convergence).
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].
Lemma 1.5.2.
independence
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.
scaling mean and variance
