Chapter 2: Probability Spaces And Random Variables

Question 1

Sample space

Answer 1

Ω

Set containing every possible outcome

Question 2

Event

Answer 2

Collection of possible outcomes

Subset of Ω

Subset of sample space

Question 3

Definition 2.1.1

F, set of subsets

Sigma field

Answer 3

Let F be a set of subsets of Ω. We say F is a sigma-field if:

1) empty and sample space are elements
2) complements are elements
3) unions are elements

Question 4

P(Ω)

Answer 4

The power set F= P( Ω )

Question 5

Sub-sigma-field

Answer 5

g is subset of F

g is a sub-sigma-field of F

Ie a more restrictive set of information than F has

Question 6

Measurable subset

Answer 6

A subset A of Ω is measurable/ a measurable event if A is element of F

Ie A is F-measureable

Ie A is an event

Question 7

Probability space

Answer 7

(Ω, F, P)

Sample space
Sigma field
Probability measure

Question 8

Empty set

Answer 8

Test nothing happens

Question 9

Examples of sigma fields for chosen experiments

Coin toss

Answer 9

H or T

Ω = {H,T}

F= { ∅, {H}, {T} , Ω}

Set of all possible subsets

In order

It’s a test we get nothing ( this has chance 0)
A test we get a Head
A test we get a tail
A test we get a head or tail

Question 10

Examples of sigma fields for chosen experiments

Coin toss with 2 coins

Answer 10

Ω = { HH, HT, TH,TT}

How we F (set of subset) only contains events we are interested in..
Suppose that all we care about is whether we get two heads

So define F as set of subsets we care about

For example if interested in whether both coins are heads we have
F = { ∅, {H,H}, Ω{H,H}, Ω}

Ie complement of HH is Ω\HH “not HH”

And empty and sample space for sigma field

Test nothing (chance0)
Test HH
Test not HH
Test two coins

Question 11

Probability measure

Answer 11

P is a function P:F to [0,1] st

1) P[Ω] =1
2) if A_1, A_2,… in F are pair-wise disjoint

(ie A_i intersection A_j = empty set for all I,j st I not equal to j)

Then
P[ Union from I=1 to infinity ] = sigma from I=1 to Infinity of P[A_i] *

*the probability of the Union is the probability that any one of the events happens

• uses the probability of A1 Union A_2 that are disjoint is the sum of the individual probabilities
  P[A_1 Union A_2 ] = P[A_1] + P[A_2]
• this last condition is sigma additivity
  (We have sigma-additivity and * by def 2.1.1 of sigma field)

Question 12

Probability space

Answer 12

Triple ( Ω, F, P)
Ω sample space
Where F is a sigma-field
And P is a probability measure

Examples of F:
* F= {∅, Ω} this is a no information sigma field as we have Ω the event that anything happens (probability 1) and ∅ the event that nothing happens ( always probability 0)

• we care about one event eg if outcome is in a subset A so we use sigma field:
  F= {∅, A, Ω\A, Ω}

These F are sigma fields- we can check the conditions

Question 13

Single coin toss probability measure

Answer 13

Ω ={H,T}

F= { ∅, {H}, {T}, Ω}

And define P[{H}] | = P [{T}] = 0.5 if it’s a fair coin toss ie probability of a set containing H = probability of the set containing T

And define P(Ω) =1 and P(∅) =0

We want to choose F to be smaller than P(Ω)

Question 14

Lemma 2.1.5 intersections of sigma fields

Answer 14

Let I be any set. For any i in I let F_i be a sigma-field. Then F = intersections of i in I of F_i

Is a σ-field

Ie we contain all the information that is common to all
Proof: conditions

1) F_i is a sigma field so empty set is an element of F_i and so empty set is an element of intersections F_i
2) if A is an element of F = intersections of F_i then A is an element of F_i for each i. Since each F_i is a sigma-field then Ω\A is an element of F_i for each i. Hence Ω\A is an element of the intersections of F_i

3)
If A_j in F for all j then A_j is in F_i for all i and j. Since each F_i is a sigma-field, unión of A_j is an element of F_i for all i. Hence union of A_j is an element of the intersections of F_i

Question 15

Corollary 2.1.6

Intersections of sigma-fields σ-fields

Answer 15

If F_1 and F_2 are sigma-fields then so is F_1 intersection F_2

Simple case of lemma

Question 16

Def 2.1.7

Events in sigma fields

Answer 16

Let E_1, E_2,.. be subsets of Ω.
σ(E_1, E_2,..) is the smallest σ -fields containing E_1, E_2,…

(Any event in any of F_i)

Question 17

Defined Big curly F

How do we construct a sigma field without writing it down

Answer 17

For a given Ω (sample space), finite or countable E_1, E_2,.. subset of Ω (events). ( we are interested in)

Let big_curly_F be the set of all sigma-fields that contain E_1,E_2,..

Enumerate
Big_curly_F ={ F_i: i is in I}

And by applying lemme 2.1.5 obtain a sigma-field F which contains all events we want

(Each of these F_i contains events we want And maybe some others)
This means F is the smallest σ-field that has E_1,E_2,.. as events

(Applying lemma 2.1.5)
(F is contained inside any σ-field F which has those events we want

And
It’s the smallest σ -field which contains all the E_i’s
As it’s the intersection of all that contain and thus big_curly_F is smaller than all of F_i ‘s

By the lemma

Ie it EXISTS

Question 18

With Ω as any set and A a subset of Ω

Answer 18

{ empty set, A, Ω\A, Ω} is σ(A)

Question 19

If F_1, F_2 ,.. are sigma fields then

Answer 19

Write sigma( F_1,F_2,..) for the smallest sigma-algebra wrt which all events are measurable

Question 20

Properties of events in F

Answer 20

If A is an element of F then Ω\A element of F and since Ω =A ∪(Ω\A) we have 1=P[A] + P[Ω\A]
=P(Ω)=P(A ∪ (Ω\A))

If A,B in F and A subset of B then B=A Union (B\A)
Which gives

P[B] = P[B\A] + P[A]
as P[B\A] is bigger than or equal to 0
Implying
P[A] is less than or equal to P[B]

Question 21

Lemma 2.1.8

Answer 21

Let A_1, A_2,.. in F where F is a sigma-field. Then intersection from i=1 to infinity of A_i in F

(Countable)

Proof/ we can write

*in general uncountable unions and intersections of measurable sets need not be measurable but lemma may not hold so that F isn’t too large for a Probability measure as harder to define probabilities the bigger F is

Question 22

The empty set

Answer 22

Is an element of F the sigma-field and is the test that nothing happens

Question 23

Set of all subsets of Ω

Answer 23

The set of all subsets of Ω is a sigma field

Question 24

what is the smallest sigma field of unions?

Answer 24

σ(F_1,F_2,…)

Question 25

Random variables pre-image

Answer 25

X:Ω to R. For each outcome ω ∈ Ω X(ω) is a property . X^-1 (A) = {ω ∈ Ω : X(ω) ∈ A} is a PRE-IMAGE ( not inverse) of A under X. and used to find set of outcomes ω ∈ Ω mapping to some set A under X. X-1(a,b) for interval (a,b)

Question 26

EXAMPLE Ω={1,2,3,4,5,6} and F=P(Ω). ``` Property X(ω)= {0 if ω is odd {1 if ω is even ``` ω ∈ Ω

Answer 26

``` X-1({0}) = {1,3,5} X-1({1}) = {2,4,6} X-1({0,1}) = {1,2,3,4,5,6} ```

Question 27

DEF 2.2.1: | Definition of a random variable/measurable function

Answer 27

Let G be a σ-field. A function X:Ω to R is G-measurable if for all subintervals I⊆R we have X-1(I) ∈ G For a probability space (Ω , F,P) we say that X:Ω to R is a RANDOM VARIABLE if X if F-measurable

Question 28

measurable notes:

Answer 28

* X is measurable (using σ-field G) I.e. x∈ mG * P[X∈ A] = P[X-1(A)] (X-1(A) ∈ F) *P[a less than X less than b] = P[X-1(a,b)] = P[ω ∈ Ω ; X(ω) ∈ (a,b) ] *P[X=a]= P[X-1(a)] = P[ω ∈ Ω ; X(ω)=a ]

Question 29

EXAMPLE: toss a coin twice

Answer 29

``` Ω= {HH, HT, TH, TT} F= P(Ω) then every funct X:Ω to R is F-measurable ``` G= {∅, {H,H}, {HT,TT, TH}, Ω } (this is info of "did we get 2 heads" with sigma-field that gives this) if we are interested in #tails: X: Ω to R given by X(ω) = #total tails occurred Then X is NOT G-measurable ie if all we knew was whether or not we had HH, we can't work out exact #tails

Question 30

We need to be able to deduce the info from the given info

Answer 30

We can take the preimage of an interval eg [0,1] we can find set is not an element of G ( X-1([0,1]) = {HH,HT,TH} is not an element of G) so X is not G measurable

Question 31

information and sigma-fields:

Answer 31

σ-field: G chooses which info we care about X is G-measurable ie X∈mG: X depends only on the information in G / information we care about *** given an outcome ω of our experiment, but not knowing which ω ∈ Ω it was, as each event "G" in G represents a piece of info this info is whether or not ω ∈ G ( ie whether or not event "G" has occurred), If this info allows us to deduce the exact value of X(ω) and if this is the case for any ω ∈ Ω then X is G-measurable

Question 32

G and "G"

Answer 32

in notes G is written curly eg similar to g (as is F) | writing G as "G" in flashcards

Question 33

σ-field generated by random variables

Answer 33

X is RV A σ-field is σ(X), containing information given by X

Question 34

LEMMA 2.2.5 σ(X)

Answer 34

X is σ(X)-measurable σ(X) is a σ-field, containing information given by X

Question 35

σ(X)

Answer 35

to construct: include ∅ and Ω look at preimages of X look at complements look at unions

Question 36

remember

Answer 36

operations on random variables produce random variables

Question 37

Lemma 2.2.2: If X is a discrete RV: suppose we have (Ω, F, P) and a RV X:Ω to R and we want to check that X is measurable wrt some smaller σ-field

Answer 37

Let G be a σ-field on Ω. Let X:Ω to R and suppose X takes values {x_1,x_2,...} ( a countable set). Then X is measurable wrt G/ X∈ mG ⇔ for all j, {X=x_j} ∈ G

Question 38

PROOF lemma 2.2.2: countable set measurable wrt G ⇔ each element in G

Answer 38

..

Question 39

example: take Ω={1,2,3,4,5,6}m rolling a dice F= P(Ω) consider G_1 ={∅, {1,3,5}, {2,4,6}, Ω} G_1 ={∅, {1,2,3}, {4,5,6}, Ω}

Answer 39

G_1 ={∅, {1,3,5}, {2,4,6}, Ω} "test is Ω even or odd" G_1 ={∅, {1,2,3}, {4,5,6}, Ω} "test is Ω less than or equal to 3 or bigger than 2" Here, G_1 contains the info of whether roll even or odd etc. We can check both are σ-fields DEFINE X_1(ω) = {0 if ω is odd {1 if ω is even X_2(ω) = {0 if ω is less than or equal to 3 {1 if ω is bigger than 3. X_1 tests if X is even X_2 tests is X is odd we expect that X_1 is measurable wrt G_1 but not wrt G_2 etc * X_1^-1(0) ={1,3,5} * X_1^-1(1) ={2,4,6} both are in G_1 but not in G_2 so X_1 is measurable wrt G_1 but not wrt G_2 ``` similarly * X_2^-1(0) ={1,2,3} * X_2^-1(1) ={4,5,6} both are in G_2 but not in G_1 so X_2 is measurable wrt G_2 but not wrt G_1 ```

Question 40

Example: consider generated σ-field and smallest contained events

Answer 40

G_3 = σ( {1,3}. {2}, {4}, {5}, {6}) has 32 elements but the information given cant tell 1 from 3 X-1(0) = {1,3,5} = { {1}, {1,3}, {3}} ∪ {5} as {1,3} is an element of G_3 and {5} is we also have { {1}, {1,3}, {3}} ∪ {5} is an element of G_3 X-1(1) = {2,4,6} = {2}∪ {4} ∪ {6} each is an elemnt of G_3 and so union is an elemnt of G_3 by def 2.1.7 we thus have X_1 is measurable wrt G_3 and X_2 is also

Question 41

σ-field are associated to

Answer 41

σ-field are associated to each function X:Ω to R

Question 42

DEF 2.2.4 σ-field generated by X

Answer 42

The σ-field generated by X, denoted σ(X) is σ(X)= σ( X-1(I) : I is a subinterval of R)

Question 43

the smallest σ-field that contains events F=P(Ω) Ω=(1,2,3,4,5,6} X(ω) = {1 if ω is odd {2 if ω is even

Answer 43

``` X-1(1) = {1,3,5} X-1(2) = {2,4,6} ``` thus the smallest σ-field that contains events is σ(X) = {∅, {1,3,5}, {2,4,6}, Ω} ie we don't need any more elements as all unions and complements are contained

Question 44

the smallest σ-field that contains events F=P(Ω) Ω=(1,2,3,4,5,6} Y(ω) = {1 if ω=1 {2 if ω=2 {3 if ω=3

Answer 44

``` Y-1(1) = {1} Y-1(2) = {2} Y-1(3) = {3} ``` thus the smallest σ-field that contains events is σ(Y) = {∅, {1}, {2},{3}, {2,3,4,5,6}, {1,3,4,5,6}, {1,2,4,5,6}, {1,3}, {1,2}, {2,3} , {2,4,5,6}, {3,4,5,6}, {1,4,5,6}, {1,2,3},{4,5,6} Ω} taking unions and complements etc to include the 3 events, empty set, sample space and all their complements and unions

Question 45

LEMMA 2.2.5: σ(X) measurable

Answer 45

Let X:Ω to R. Then X is σ(X) measurable

Question 46

Let X:Ω to R. Then X is σ(X) measurable PROOF:

Answer 46

...

Question 47

Define σ(X_1, X_2,...)

Answer 47

σ(X_1, X_2,...) = σ(X_i(I)) : X_i is element of sequence and I is a subinterval of R this is the σ-field containing the info given by X_1, X_2,...

Question 48

2.2.4 gives us: Considering a collection of RVs its better to consider a sub-σ-field G ⊆F Let α∈R and let X, Y, and X_1,_X_2,... be G-measurable functions Ω to R

Answer 48

Then α, αX, X+Y, XY, 1/X are all G-measurable. Further, if X_∞ given by X_∞ (ω) = limit as n tends to infinity of X_n(ω) exists for all ω, then X_∞ is G-measurable

Question 49

EXAMPLE: if X∈mG then

Answer 49

Then (X^2 + X)/ 2 ∈mG Y= e^X then Y ∈mG because Y(ω)= sum from n=0 to infinity X(ω)^n/(n!) we know this limit exists as e^x is defined for all x in the reals consider the partial sums Y_N (ω) = sum from n=0 to N of X(ω)^n /n! ∈mG by (1) and Y(ω) = limit as N to infinity Y_N(ω) exists so Y ∈mG IF X ∈mG and g: R to R is any SENSIBLE FUNCTION then Y=g(X) ∈mG ie all powers, trig functs, e^x limit, polynomial, monotone functions, all piecewise linear functs

Question 50

RECALL FOR INDEPENDENT EVENTS

Answer 50

Events E_1, E_2 are independent if P( E_1 ∩ E_2) = P(E_1) P(E_2) ie change that E_1 and E_2 happens is chance that E_1 * chance that E_2

Question 51

2.2.7 INDEPENDENCE | we define independence of events using sigma fields

Answer 51

Sub- σ-fields G_1, G_2 of F are indep if (note notation F is curly F, G is curly G , "G" is G) P("G"_1 ∩ "G"_2) = P("G"_1)P("G"_2) for all "G_1" in G_1 and "G_2" in G_2 Events E_1 and E_2 are independent if σ(E_1) and σ(E_2) are independent Random vars X_1 and X_2 are indep if σ(X_1) and σ(X_2) are independent

Question 52

2.3 INFINITE SAMPLE SPACE for probability space (Ω, F, P)

Answer 52

Recall if Ω= {x_1, x_2,...,x_n} is finite. We normally take F=P(Ω). Since F contains every subset of Ω, any σ-field on Ω is a sub-σ-field of F. We have seen how it is possible to construct other σ-fields on Ω too. In this case we can define a probability measure on Ω We can define P by choosing some sequence (a_1, a_2,..,a_n) st a_i ∈ [0,1] and sum from i=1 to n of a_i = 1 and set P({x_i}) = a_i This naturally extends to defining P[A] for any subset A ⊆ Ω, by setting more generally P(A) = sum from i=1 to n of P({x_i}) * (indicator function of x_i ∈ A) If Ω is countable we have Ω = {x_1,x_2,..} can replace n with infinity. Infinite sequence transformed into infinite series which is bounded above by 1.

Question 53

EXAMPLE: toss a coin countably many times. Outcome is ω= ω_1ω_2.... = (ω_1, ω_2,...) where each ω_i is in {H,T} the set of ω is uncountable : Ω= {H,T}^N Define X_n(ω) = ω_n = result of the nth toss σ(X_1) σ(X_1, X_2)

Answer 53

``` Define X_n(ω) = ω_n = result of the nth toss X_n: Ω to {H,T} this isn't a subset of R but we can imagine as 0 or 1 etc ``` We want F = σ(X_1, X_2,..) = σ(X^{-1}_i (I): i in N, I is a subinterval of R) contains all info generated by coin tosses NOTE σ(X_1) = ( { ∅, {H****...}, {T****...} , Ω} ie if first toss is H or T followed by anything else σ(X_1) = ( { ∅, {H****...}, {T****...} , Ω} σ(X_1, X_2) = σ( {HH**..}, {TH**...}, {HT**..}, {TT**...}) = {∅, Ω, {HH**...}, {TH**..}, {HT**..}, {TT**...}, {H*..}, {T*...}, {*H*..}, {*T*..}, {HH**.., TT**..}, {HT**.., TH**..}, {HH**..}^c, {TH**...}^c, {HT*..}^c, {TT*..}^c } ie considering first two tosses unions and complements if 2 ω have same 1st and 2nd outcomes they fall into same subset of σ(X_1, X_2) if a RV dep on anything more than 1st or 2nd outcomes will not be σ(X_1, X_2) measurable

Question 54

HOW TO DEFINE P?

Answer 54

when using infinite Ω will be given probability measure P and properties

Question 55

EXAMPLE HERE GIVEN: P: F to [0,1] 1) X_n are indep RVs 2) P(X_n = H) = P(X_n = T) = 0.5 for all n in N

Answer 55

from this we work with P without being constructed. Don't need to know which subsets of Ω in F as 2.2.6 allows us if we try to take F= P(Ω) there is no probability measure such that 1 and 2 are satisfied

Question 56

2.3.2 Ω={H,T} sequence if fair independent coin tosses P(X_i = T) = P(X_2 =H) = 0.5 P(X_1 = ω_1, X_2 = ω_2, X_3 = ω_3,....) P[X_1 =H] P[ eventually throw a head] P[never throw a head ]

Answer 56

P(X_1 = ω_1, X_2 = ω_2, X_3 = ω_3,...) = P(X_1 =ω_1)P(X_2= ω_2)P(X_3 = ω_3)...... = 0.5 *0.5* 0.5 *... = 0 e.g the probability that we never throw a head So P[X_1 =H] = P[ {ω_i} in Ω : ω_1 =H] = 0.5 P[ eventually throw a head] = P[for some n, X_n =H] = 1- P[for all n X_n =T]=1-0=1 event has probability 1 but not equal to whole sample space P[never throw a head ] = P[for all n X_n =T]= 0.5*0.5*...=0

Question 57

DEF 2.3.2 | ALMOST SURELY

Answer 57

if the event E has P(E) =1 we say E occurs ALMOST SURELY ie coin will eventually throw a headie Y less than or equal to 1 almost surely IFF P[Y less than or equal to 1] =1

Question 58

DEFINE | proportions of heads and tails in the first n tossesX_1,... X_n are q_n ^H and q_n ^T

Answer 58

q_n ^H = (1/n) sum from i=1 to n of (indicator funct of X_i =H) q_n ^T = (1/n) sum from i=1 to n of (indicator funct of X_i =T) q_n ^T + q_n ^H = 1 the random vars indicator of {X_i=H} are iid with E[1_{X_i=H}] =0.5 hence by thm 1.1.1 we have P[q_n ^T tends to 0.5 as n tends to infinity] = 1 by strong law of large # & P[q_n ^H tends to 0.5 as n tends to infinity] = 1 *** event that half tosses H and half are T is event E = {limit as n tends to infinity of q_n ^T = 0.5 and limit as n tends to infinity of q_n ^H = 0.5} occurs almost surely as P[q_n ^H tends to 0.5 , q_n ^T tends to 0.5 ] =1

Question 59

EXPECTATION

Answer 59

if X is continuous RV E[X] = integral from -infinity to infinity of [xf_x(alpha).dx] (pdf) if X is discrete RV E[X] = sum of x in R_x of [xP[X=x]] (R_x is range of x)

Question 60

WHEN X IS AN F-MEASURABLE FUNCTION FROM Ω to R, RVS MIGHT NOT BE DISCRETE OR CONTINUOUS .

Answer 60

we use Lebesgue integration to define E(X) in this case

Question 61

E[X] defined:

Answer 61

E[X} is defined for all X se either: 1) X is bigger than or equal to 0, in which case its possible that E[X] = infinity 2) General X for which E[ |X|] less than infinity * we haven't got E[X] = - infinity * we use the modulus to avoid "infinity take away infinity" ***if X is bigger than or equal to 0 and P[X= infinity] bigger than 0, this implies E[X] = infinity (the chance of X being infinite will outweigh all of the finite possibilities and make E[X] Infinite)

Question 62

PROPOSITION 2.4.1 Expectations for RVs X and Y

Answer 62

For random variables X, Y such that E[X] and E[Y] exist: LINEARITY if a,b in R then E[ aX +bY] = a E[X] + bE[y] ``` INDEPENDENCE if X, Y are indep then E[XY] = E[X]E[Y] ABSOLUTE VALUES |E[X]| is less thn or equal to E[|X|] MONOTONICITY if X is less than or equal to Y then E[X] is less than or equal to E[Y] ```

Question 63

indicator function for events

Answer 63

for event A in F, 1_A: Ω to R 1_A(ω) = {1 if ω is in A {0 if ω isn't in A E(1_A) = 1P(A) + 0(P(A^c)) =P(A) * sometimes write 1_{event} = 1{event}

Question 64

LEMMA 2.4.2 | indicator function for events

Answer 64

Let A∈G (remember G is curly G similar to g) Then function 1_A is G-measurable ``` proof: Range of 1_A is {0,1}. Pre-images are: 1_A -1 (0) =Ω\A ∈G 1_A -1 (1) = A ∈G by2.2.2, Y is G-measurable ```

Question 65

CONDITIONING

Answer 65

breaking RV into cases for example, given any random var X we write: X= X 1_{X≥1} + X1_{X less than 1} ( as only one of the RHS terms is non-zero)

Question 66

for example, given any random var X we write: X= X 1_{X≥1} + X1_{X less than 1} ( as only one of the RHS terms is non-zero) we use this to prove an inequality:

Answer 66

putting |X| in place of X then taking the expectation to obtain: E[|X|] = E[|X| 1_{|X|≥1} ] + E[|X| 1_{|X|less than 1} ] ≤ E[X^2 1_{|X|≥1} ] + 1 ≤ E[X^2] +1 (to reduce second key point we can only use |x| is less than or equal to x^2 if x is bigger than or equal to 1) ie we want to prove E[|X|] ≤E[X^2] +1 |X| = |X| 1_{|X|≥1} + |X| 1_{|X| less thn 1} BY MONOTONICITY: FIRST TERM |X| 1_{|X|≥1} ≤ X^2 1_{|X|≥1} assume |X| ≥ 1 and for reals X ≤ X^2 SECONT TERM |X| 1_{|X|less thn 1}≤ 1_{|X| Less thn 1} assume |X| less than 1 and for reals X ≤ 1 |X| = |X| 1_{|X|≥1} + |X| 1_{|X| less thn 1} ≤ X^2* 1_{|X|≥1} + 1 *1_{|X|less thn 1} ≤ X^2 + 1 by monotonicity of expectation: E(|X|) ≤ E [1 +X^2] -1 + E[X^2]

Question 67

DEF 2.4.3 L^p

Answer 67

Let p∈ [1, infinity) we say that X ∈ L^p if E[|X|^p] is less than infinity we usually care about cases p = 1 or p = 2

Question 68

L^1 and L^2

Answer 68

*by definition L^1 is the set of random variables such that E[|X|] is less than infinity ie that its finite * L^2 is the set of random variables such that Var(X) is less than infinity Var(X) = E(X^2) - E(X)^2 ie finite var * from 2.7, if X ∈L^2 then X∈L^1

Question 69

DEF 2.4.4 | bounded

Answer 69

We say that a random variable X is bounded if there exist a deterministic constant c ∈R such that |X| ≤ c if X is bounded then using monotonicity: E[|X|^p] ≤ E[C^p] =c^p less than infinity meaning that X ∈ L^p for all p