Integration, integrals, MCT & DCT Flashcards
Definition of an integral:
Given f: (Omega, F,mu) -> Rbar where f is measurable.
Integral of f w.r.t. measure mu:
Integral(f)dmu
Integral w.r.t Dirac measure:
mu = Sx0 is a dirac measure.
Sx0(A) = {1, if x0 element of A & 0 otherwise
Integral(f)dSx0 = f(x0)
~if xi is in set A: if yes, Sxi(A)=1, if no Sxi(A) = 0
General rule of integrating w.r.t. dirac measure:
If mu = a1Sx1 + a2Sx2+ …
where ai element of [0,infinity) & xi element of omega:
mu(A) = sum(for i)ai*Sxi(A)
Integral(f)dmu = sum(for i)f(xi)
Integral w.r.t. lebesgue-measure:
f: (R,B(R),lambda)-> Rbar
Integral(f)dlamba = Integral(from -infinity to infinity) (f(x))dx
Integral w.r.t. Lebesgue-Stieltjies measure:
If function F: Real->Real is
increasing (a ≤ b => F(a) ≤ F(b) ) &
right continuous (lim(x->a^+)F(x) = F(a)
musubscripF( (a,b)) = musubscripF([a,b]) = musubscripF((a,b]) = F(b)- F(a)
If X: (Omega,F,Real) -> Real &
F(X) = P(X ≤ x), what is X’s distribution?
The Lebesgue-Stieltjies measure, musubscriptF(.).
Integral w.r.t. pushed-measure:
If G: (Omega,F,Real) -> (S, L, muG^-1) & G is measurable.
(thus B element of L,G^-1(B) element of F because G is measurable)
muG^-1(B) = mu(G^-1(B))
(Push-measure = mu-measure of pullback)
muG^-1 is a measure on (S,L)
If X: (Omega,F,Real) -> Real &
F(X) = P(X ≤ x), what is X?
A random variable
True or False:
i) If a function G is measurable
<=> G is continuous?
ii) If a function G is continuous => G is measurable?
iii) If a function G is measurable => G is continuous?
i) False
ii) True
iii)False
General rule of integrating a simple function:
Given f: (Omega,F,mu) ->(Rbar, B(Rbar) and assuming f is measurable.
If f = sum(for i)ai*Isubscript(Ai)
where ai element of Real &
Ai element of F, then
Integral(f)dmu = sum(for i)ai*mu(Ai)
Integral(Isubscript(A))dmu = mu(A)
Definition of expectation:
Suppose X: (Omega,F,P) -> (Rbar,B(Rbar)) is measurable (thus X is measurable)
then
E[X] = Integral(X)dP
Do integrals have to exist?
No
Properties of integrating simple functions:
If f & g are simple functions , then
1) Integral(f)dmu doesn’t depend on the simple interpretation of f.
2) Integral(Isubscript(A))dmu = mu(A) for A element of F
3) Linearity:
If f,g ≥ 0 & a,b ≥ 0, then
Integral(af + bg)dmu = aIntegral(f)dmu +bIntegral(g)dmu
4)Monotonicity:
If f,g ≥ 0 & f ≤ g, then
Integral(f)dmu ≤ Integral(g)dmu
What do we call measurable function whose integrals exist?
Integrable functions
When is a function f: Omega -> Real
simple?
If it only takes a finite number of different values.
If f: (Omega, F,mu) -> Clo Real, what is:
i) sF ?
ii) sF^+ ?
iii) mF ?
iv) mF^+ ?
i) sF : set of simple functions
ii) sF^+ : set of simple function ≥ 0
iii) mF : set of measurable functions
iv) mF^+ :set of measurable function ≥ 0
If f: (Omega, F,mu) -> Rbar, what is:
i) sF ?
ii) sF^+ ?
iii) mF ?
iv) mF^+ ?
i) sF : set of simple functions
ii) sF^+ : set of simple function ≥ 0
iii) mF : set of measurable functions
iv) mF^+ :set of measurable function ≥ 0
Integration for non-negative measurable functions, mF^+:
Given f: (Omega,F,mu) -> Rbar is measurable & f ≥ 0 then
Integral(f)dmu = sup{Integral(T)dmu:
T ≤ f & T element of sF^+}
Integration steps:
Given f: (Omega, F,mu)->(Rbar,B(Rbar)) is measurable
1) If f is simple, i.e.,
f = sumofi(aiIsubscriptAi)
then
Integral(f)dmu = sumofi(aimu(Ai))
2) If f an element of mF^+ & f ≥ 0
then
Integral(f)dmu = sup{Integral(T)dmu:
T ≤ f & T element of sF^+}
Integration for non-negative simple functions, sF^+:
If f simple,f = sumofi(aiIsubscriptAi), & f ≥ 0
then
Integral(f)dmu = sumofi(aimu(Ai))
Facts about if f element of mF^+:
1) Integral(f)dmu always exists
~can be infinity
2) Integral(f)dmu ≥ 0
3) There exists Tn element of sF^+ s.t.:
(i) Tn is increasing (T1 ≤ T2 ≤ …)
(ii)Tn -> T as n -> infinity
(pointwise: lim(n to infinity)Tn(w) = f(w) , for all w element of Omega
(iii) Tn ≤ f for all n
then
Integral(f)dmu = lim(n to infinity)Integral(Tn)dmu
Monotone convergence theorem:
Suppose fn: (Omega,F,mu) -> Rbar is measurable for all n element of N & f ≥0.
Assume that: lim n to infinity(fn) = f & f1 ≤ f2 ≤ …
(thus fn upwards arrow f)
then
lim n to infinity (Integral(fn)dmu)) = Integral(f)dmu
MCT steps
Given fn(x) determine lim n to infinity Integral(fn)dmu
1) Find f(x) = lim n to infinity (fn(x))
2) Check if fn(x) is increasing as a sequence:
fn(x) ≤ fn+1(x) for all n
3) Use MCT results:
lim n to infinity (Integral(fn(x))dmu)) = Integral(f(x))dmu
Integrating general functions:
Given f: (Omega,F,mu) -> Rbar is measurable
& recalling:
i) f^+ = max(x,0) ; f^- = -min(x,0)
thus f^+,f^- element of mF^+
ii) f = f^+ + f^-
what is Integral(f)dmu & when if f intergrable?
Integral(f)dmu = Integral(f^+)dmu + Integral(f^-)dmu
&
f is integrable if Integral(f)dmu exists & finite
When is a function integrable?
If Integral(f)dmu exists and is finite.
When is a function f element of mF mu-integrable?
IFF mu(|f|) = Integral(|f|)dmu < infinity
OR
IFF Integral(f^+)dmu < infinity &
Integral(f^-)dmu < infinity
Definition of mu-integrable functions, Ł^1(Omega,F,mu):
Ł^1(Omega,F,mu) = {f: Integral(|f|)dmu < infinity}
and is a vector space.
If f = g - h & g,h element of mF^+ then Integral(f)dmu=?
Integral(f)dmu = Integral(g)dmu - Integral(h)dmu
True or false:
Given Integral(f+g)dmu
= Integral((f+g)^+)dmu - Integral((f+g)^-)dmu
True
True or false:
(f+g)^+ = f^+ + g^+
False
Dominated Convergence theorem:
Let (fn)n be a sequence of integrable functions s.t.
i)lim n to infinity fn(w) exists for all w element of Omega
ii) There exists a function, g element of Ł^1(Omega,F,mu) s.t. |fn| ≤ g for all n element of N
(thus its bounded by something independent of n)
then
lim n to infinity Integral(fn)dmu = Integral(lim n to infinity)(fn)dmu
Given f is an integrable function and A is a measurable set, what is the integral of f over set A, IntegraloversetA(f)dmu =?
IntegraloversetA(f)dmu = Integral(f*IsubscriptA)dmu
Fatou’s lemma:
If fn element of mF^+ for all n element of N, then
mu(liminf(fn)) ≤ liminf(mu(fn))
Reverse Fatou’s lemma:
If fn element of mF^+ fo all n element of N
then
limsup(mu(fn)) ≤ mu(limsup(fn))
Fatou’s lemma results ordered by size:
mu(liminf(fn)) ≤ liminf(mu(fn)) ≤ limsup(mu(fn)) ≤ mu(limsup(fn))
~terms with limits on the inside is outside
DCT Steps:
Given fn(x) determine lim n to infinity Integral(fn)dmu
1) Find lim n to infinity fn(x)
~get for different cases: x < 0; x = 0 ; x > 0
2) Find binding function g(x) for cases: x < 0; x = 0 ; x > 0
3) Integrate Integral(g(x))dmu
4) Use DCT results:
lim n to infinity Integral(fn(x))dmu ≤ Integral ( lim n to infinity fn(x))dmu
Given measurable function X: (Omega,F,P) -> (Rbar, B(Rbar), PX^-1) & X, what is:
i) PX^-1?
ii) E[X] = ?
i) PX^-1 : the distribution of X
ii) E[X] = Integral over Rbar (x)dPX^-1(x)
= Integral(X)dP
Given X: (Omega,F,P) -> (Rbar,B(Rbar),PX^-1)
& g: (Rbar,B(Rbar),PX^-1)-> (Rbar,B(Rbar),PX^-1)
thus (Omega,F,P) -> (Rbar,B(Rbar),PX^-1) -> (Rbar,B(Rbar),PX^-1),
change the variables of the integral & find E[g(X)]:
Change of variables:
Integral(g)dPX^-1 = Integral(g o X)dP
In general:
E[g(X)] = Integral(g(x))dPX^-1(x)
Inequality properties of Expectation:
1) Markov’s inequality:
If g ≥ 0 & increasing then
E[g(X)] ≥ g(c)*P({X ≥ c}) for any c element of R
2) Chebyshev’s inequality:
For v > 0, then
P({|X-E[X]|≥ v }) ≤ Var(X)/(v^2)
where Var(X) = E[ (X-E[X])^2 ]
If f is integrable, what is mu{f = ±infinity} ?
mu{f = ±infinity} = 0
thus f is finite mu-a.e.
True or false:
If f is integrable, |Integral(f)dmu| ≤ Integral(|f|)dmu
True
Definition of changing the measure with density:
Given (Omega,F,mu) -> Rbar is measurable and f ≥ 0.
Assume f element of Ł^1(Omega,F,mu),
thus Integral(f)dmu < infinity
Then define v = fmu : F -> [0,infinity] s.t.
v(A) = (fmu)(A) = Integral over A(f)dmu = Integral(f*IsubscriptA)dmu.
v is a measure on (Omega,F)
& v «mu
f is the density of v w.r.t. mu: f = dv/dmu
Chain rule:
Let v = f*mu & g: (Omega,F) -> Rbar s.t. g element of Ł^1(Omega,F,mu) then
Integral(g)dv = Integral(gf)dmu
since f = dv/dmu => dv = f*dmu
Change of variables:
Given f: (Omega,F,mu) -> (T,S,mu(f^-1))
& g: (Rbar,B(Rbar),mu(f-1))-> Rbar & f,g are measurable.
Then
Integral(g)dmuf^-1 = Integral(g o f)dmu
True or false:
g o f = g(f)?
True
Given (Omega,F,mu) is a measure space & A subset Omega:
i) When is A a mu-null (a null set)?
ii)When is the space (Omega,F,mu) complete?
iii) What is the completion of F, Fbar?
i) If there exists a B element of F s.t. A subset B & mu(B)=0;
ii) The space (Omega,F,mu) is complete if every mu-null set is in F.
iii) Fbar = sigma( F or {A subset Omega: A is mu-null })
pg.8 notes L27-L29
If (Omega, F, mu) is a measure space, when does a statement, I0(w), hold mu-almost everywhere (mu-a.e.)?
If { w element of Omega: IO(w) is false} is mu-null
If (Omega, F, P) is a probability space, when does a statement, I0(w) is true P-almost surely (P-a.s.)?
If { w element of Omega: IO(w) is false} is P-null
True or false:
i) If F is complete, Fbar, then
mu({w element of Omega: IO(w) false}) = 0
=> IO(w) true a.e.
ii) If F is complete, Fbar, then
IO(w) true a.e.
=> mu({w element of Omega: IO(w) false}) = 0
iii) If F is complete, Fbar, then
IO(w) true a.e. <=>
mu({w element of Omega: IO(w) false}) = 0
i) TRUE
ii) TRUE
iii) TRUE
Radon-Nikodym theorem:
If v & mu are sigma-finite measure on (Omega,F) &
v«mu (v abs cont w.r.t. mu), then
there exists a function f: omega-> Rbar s.t.
f = dv/dmu <=> dv = f.dmu
f: density/Radon-Nikodym derivative
Definition of a measure, v, being absolutely continuous w.r.t. another measure, mu
(v«mu)
A measure v on (Omega,F) is absolutely continuous w.r.t. mu
if mu(A) = 0 => v(A) =0;
thus v«mu
Using Radon-Nikodym theorem:
Find f(x) = lim h to o^+ (mu([x, x+h])/(lambda([x, x+h])
Definition of product of measure spaces:
Let (Omega1,F1) & (Omega2,F2) be measurable spaces.
Define projections maps:
pie1 = Omega1 x Omega2 -> Omega1
( (w1,w2) -> pie1(w1,w2) = w1 )
&
pie2 = Omega1 x Omega2 -> Omega2
( (w1,w2) -> pie2(w1,w2) = w2 )
Define F1 circle* F2 = sigma(pie1,pie2) to be the smallest sigma-algebra that makes pie1 & pie2 measurable.
~(w1,w2): sample point in a space of combined outcomes (w1 element of Omega1 occurred & w2 element of Omega2 occurred)
~thus given combined outcome (w1,w2),
pie1(w1,w2)=w1 measures which outcome occurred in Omega1 &
pie2(w1,w2)=w2 measures which outcome occurred in Omega2
~projection maps pie1 & pie2 are measurable given that if we know a combined outcome (w1,w2), we should know the component outcomes, w1 & w2.
True or false:
B(R) circlex B(R) = B(R^2)
True
Definition of product of measures:
Let (S,F1,mu1) & (T,F2,mu2) be sigma-finite measure spaces.
Define map
mu1 circlex mu2: F1 circlex F2 ->Rbar^+
Then product measure of mu1 & mu2 on
(S circlex T, F1 circlex F2) for B element of F1 circlex F2:
mu1 circlex mu2 = Int(Int (IsubB(s,t)) mu2(dt))mu1(ds)
Fubini’s theorem:
Assume (Omega1,F1,mu1) & (Omega2,F2,mu2) are sigma-finite measure spaces then
if f element of Ł^1(Omega1 x Omega2, F1 cricx F2, mu1 circx mu2)
(mu1 circx mu2)(f) = Int(f)dmu1circx mu2
= Int( Int(f(x,y)dmu1(x)) dmu2(y)
=Int( Int(f(x,y)dmu2(y)) dmu1(x)
~ Int( Int(f)dv)du = Int( Int(f)du)dv
~f element of Ł^1 <=> abs. conv. Int(|f|)dmu1dmu2 < inf
Jensen’s inequality for E[X]:
If g: R->R is convex (function as a + 1st derivative: bend upwards), then
E[g(X)] ≥ g(E[X])
Properties of expectation:
1) If X ≤ Y a.s. then E[X] ≤ E[Y]
2) E[aX+bY] = aE[X] + bE[Y]
3) If X = Y a.s. then E[X]=E[Y]
4) Continuity:
Assume X -> X a.s. , then
(i) Xn≥ 0 & Xn upwardsarrow X a.s. (non-neg & increasing)
then E[Xn] -> E[X]
(ii) There exists Y s.t. E[Y] < inf & |Xn| ≤ Y a.s. then
E[Xn] -> E[X]
(iii) & Xn is bounded, exists a constant c s.t. |Xn| ≤ c then
E[Xn] -> E[X] & E[Xn^k] -> E[X^k] for k element of N
5) E[c] = c if c a constant
Definition of correlation:
Let X,Y element of Ł^1(omega,F,P) then correlation
px,y = cov(X,Y)/(sd(X)*sd(Y))
&
-1 ≤ px,y ≤ 1
~X,Y are correlated if cov(X,Y) = 0 or px,y = 0.
Definition of covariance:
Let X,Y element of Ł^1(omega,F,P) then covariance,
cov(X,Y) = E[ (X-E[X]) (Y-E[Y]) ) = E[XY]-E[X]E[Y]
True or false:
Uncorrelated ≠ Independent
True
True or false:
Cov & correlation measures only LINEAR dependence (Y = aX+b)
True
