Increasing sequence of events and the probability of their limit
If $A_1subset A_2subset A_3subsetcdots$ is an increasing sequence of events with limit $A=bigcup_{i=1}^infty A_i$. Prove that $$lim_{nrightarrowinfty} P(A_n)=P(A)$$
My attempt so far:
Since $A_1subset A_2subset A_3subsetcdots$ is increasing and $$bigcup_{i=1}^infty A_i=lim_{nrightarrowinfty} A_n wedge A=bigcup_{i=1}^infty A_i$$ then $lim_{nrightarrowinfty}A_n=A$. I have that $$Pleft(lim_{nrightarrowinfty}A_nright)=P(A)Rightarrow lim_{nrightarrowinfty}P(A_n)=P(A)$$
probability probability-theory
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
asked Oct 6 '16 at 15:34
seeit
113
add a comment |
If $A_1subset A_2subset A_3subsetcdots$ is an increasing sequence of events with limit $A=bigcup_{i=1}^infty A_i$. Prove that $$lim_{nrightarrowinfty} P(A_n)=P(A)$$
My attempt so far:
Since $A_1subset A_2subset A_3subsetcdots$ is increasing and $$bigcup_{i=1}^infty A_i=lim_{nrightarrowinfty} A_n wedge A=bigcup_{i=1}^infty A_i$$ then $lim_{nrightarrowinfty}A_n=A$. I have that $$Pleft(lim_{nrightarrowinfty}A_nright)=P(A)Rightarrow lim_{nrightarrowinfty}P(A_n)=P(A)$$
probability probability-theory
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
asked Oct 6 '16 at 15:34
seeit
113
add a comment |
If $A_1subset A_2subset A_3subsetcdots$ is an increasing sequence of events with limit $A=bigcup_{i=1}^infty A_i$. Prove that $$lim_{nrightarrowinfty} P(A_n)=P(A)$$
My attempt so far:
Since $A_1subset A_2subset A_3subsetcdots$ is increasing and $$bigcup_{i=1}^infty A_i=lim_{nrightarrowinfty} A_n wedge A=bigcup_{i=1}^infty A_i$$ then $lim_{nrightarrowinfty}A_n=A$. I have that $$Pleft(lim_{nrightarrowinfty}A_nright)=P(A)Rightarrow lim_{nrightarrowinfty}P(A_n)=P(A)$$
probability probability-theory
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
asked Oct 6 '16 at 15:34
seeit
113
If $A_1subset A_2subset A_3subsetcdots$ is an increasing sequence of events with limit $A=bigcup_{i=1}^infty A_i$. Prove that $$lim_{nrightarrowinfty} P(A_n)=P(A)$$
My attempt so far:
Since $A_1subset A_2subset A_3subsetcdots$ is increasing and $$bigcup_{i=1}^infty A_i=lim_{nrightarrowinfty} A_n wedge A=bigcup_{i=1}^infty A_i$$ then $lim_{nrightarrowinfty}A_n=A$. I have that $$Pleft(lim_{nrightarrowinfty}A_nright)=P(A)Rightarrow lim_{nrightarrowinfty}P(A_n)=P(A)$$
probability probability-theory
probability probability-theory
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
asked Oct 6 '16 at 15:34
seeit
113
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
asked Oct 6 '16 at 15:34
seeit
113
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
edited Oct 6 '16 at 15:46
Jimmy R.
33k42157
33k42157
asked Oct 6 '16 at 15:34
seeit
113
asked Oct 6 '16 at 15:34
seeit
113
asked Oct 6 '16 at 15:34
seeit
113
113
add a comment |
add a comment |
2 Answers
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Hint: One usually require from a probability to be $sigma$-additive, i.e. when $(B_i)_{igeq 1}$ is a countable disjointed family, then $P(bigcup_i B_i)=sum_i P(B_i)$. So try to rewrite your increasing union as a countable union of disjoint sets.
edited Oct 6 '16 at 15:46
Michael Hardy
1
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
add a comment |
Let $B_1 = A_1$ and $B_{n+1} = A_{n+1}setminus A_n$ for $nge 1$. Then
$$
A_N = bigcup_{n=1}^N B_n
$$
and
$$
bigcup_{n=1}^infty A_n = bigcup_{n=1}^infty B_n,
$$
and
$$
B_n cap B_m = varnothing text{ for } nne m.
$$
So
begin{align}
P(A) & = Pleft( bigcup_{n=1}^infty A_n right) = Pleft( bigcup_{n=1}^infty B_n right) \[10pt]
& = sum_{n=1}^infty P(B_n) qquad left( begin{array}{l} text{by countable additivity of } P text{ and} \ text{pairwise disjointness of } {B_n}_{n=1}^infty end{array} right) \[10pt]
& = lim_{Ntoinfty} sum_{n=1}^N P(B_n) = lim_{Ntoinfty} Pleft( bigcup_{n=1}^N B_n right) = lim_{Ntoinfty} P(A_N).
end{align}
answered Oct 6 '16 at 15:44
Michael Hardy
1
add a comment |
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2 Answers
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2 Answers
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Hint: One usually require from a probability to be $sigma$-additive, i.e. when $(B_i)_{igeq 1}$ is a countable disjointed family, then $P(bigcup_i B_i)=sum_i P(B_i)$. So try to rewrite your increasing union as a countable union of disjoint sets.
edited Oct 6 '16 at 15:46
Michael Hardy
1
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
add a comment |
Hint: One usually require from a probability to be $sigma$-additive, i.e. when $(B_i)_{igeq 1}$ is a countable disjointed family, then $P(bigcup_i B_i)=sum_i P(B_i)$. So try to rewrite your increasing union as a countable union of disjoint sets.
edited Oct 6 '16 at 15:46
Michael Hardy
1
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
add a comment |
Hint: One usually require from a probability to be $sigma$-additive, i.e. when $(B_i)_{igeq 1}$ is a countable disjointed family, then $P(bigcup_i B_i)=sum_i P(B_i)$. So try to rewrite your increasing union as a countable union of disjoint sets.
edited Oct 6 '16 at 15:46
Michael Hardy
1
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
Hint: One usually require from a probability to be $sigma$-additive, i.e. when $(B_i)_{igeq 1}$ is a countable disjointed family, then $P(bigcup_i B_i)=sum_i P(B_i)$. So try to rewrite your increasing union as a countable union of disjoint sets.
edited Oct 6 '16 at 15:46
Michael Hardy
1
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
edited Oct 6 '16 at 15:46
Michael Hardy
1
edited Oct 6 '16 at 15:46
Michael Hardy
1
edited Oct 6 '16 at 15:46
Michael Hardy
1
1
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
answered Oct 6 '16 at 15:42
H. H. Rugh
23.2k11134
23.2k11134
add a comment |
add a comment |
Let $B_1 = A_1$ and $B_{n+1} = A_{n+1}setminus A_n$ for $nge 1$. Then
$$
A_N = bigcup_{n=1}^N B_n
$$
and
$$
bigcup_{n=1}^infty A_n = bigcup_{n=1}^infty B_n,
$$
and
$$
B_n cap B_m = varnothing text{ for } nne m.
$$
So
begin{align}
P(A) & = Pleft( bigcup_{n=1}^infty A_n right) = Pleft( bigcup_{n=1}^infty B_n right) \[10pt]
& = sum_{n=1}^infty P(B_n) qquad left( begin{array}{l} text{by countable additivity of } P text{ and} \ text{pairwise disjointness of } {B_n}_{n=1}^infty end{array} right) \[10pt]
& = lim_{Ntoinfty} sum_{n=1}^N P(B_n) = lim_{Ntoinfty} Pleft( bigcup_{n=1}^N B_n right) = lim_{Ntoinfty} P(A_N).
end{align}
answered Oct 6 '16 at 15:44
Michael Hardy
1
add a comment |
Let $B_1 = A_1$ and $B_{n+1} = A_{n+1}setminus A_n$ for $nge 1$. Then
$$
A_N = bigcup_{n=1}^N B_n
$$
and
$$
bigcup_{n=1}^infty A_n = bigcup_{n=1}^infty B_n,
$$
and
$$
B_n cap B_m = varnothing text{ for } nne m.
$$
So
begin{align}
P(A) & = Pleft( bigcup_{n=1}^infty A_n right) = Pleft( bigcup_{n=1}^infty B_n right) \[10pt]
& = sum_{n=1}^infty P(B_n) qquad left( begin{array}{l} text{by countable additivity of } P text{ and} \ text{pairwise disjointness of } {B_n}_{n=1}^infty end{array} right) \[10pt]
& = lim_{Ntoinfty} sum_{n=1}^N P(B_n) = lim_{Ntoinfty} Pleft( bigcup_{n=1}^N B_n right) = lim_{Ntoinfty} P(A_N).
end{align}
answered Oct 6 '16 at 15:44
Michael Hardy
1
add a comment |
Let $B_1 = A_1$ and $B_{n+1} = A_{n+1}setminus A_n$ for $nge 1$. Then
$$
A_N = bigcup_{n=1}^N B_n
$$
and
$$
bigcup_{n=1}^infty A_n = bigcup_{n=1}^infty B_n,
$$
and
$$
B_n cap B_m = varnothing text{ for } nne m.
$$
So
begin{align}
P(A) & = Pleft( bigcup_{n=1}^infty A_n right) = Pleft( bigcup_{n=1}^infty B_n right) \[10pt]
& = sum_{n=1}^infty P(B_n) qquad left( begin{array}{l} text{by countable additivity of } P text{ and} \ text{pairwise disjointness of } {B_n}_{n=1}^infty end{array} right) \[10pt]
& = lim_{Ntoinfty} sum_{n=1}^N P(B_n) = lim_{Ntoinfty} Pleft( bigcup_{n=1}^N B_n right) = lim_{Ntoinfty} P(A_N).
end{align}
answered Oct 6 '16 at 15:44
Michael Hardy
1
Let $B_1 = A_1$ and $B_{n+1} = A_{n+1}setminus A_n$ for $nge 1$. Then
$$
A_N = bigcup_{n=1}^N B_n
$$
and
$$
bigcup_{n=1}^infty A_n = bigcup_{n=1}^infty B_n,
$$
and
$$
B_n cap B_m = varnothing text{ for } nne m.
$$
So
begin{align}
P(A) & = Pleft( bigcup_{n=1}^infty A_n right) = Pleft( bigcup_{n=1}^infty B_n right) \[10pt]
& = sum_{n=1}^infty P(B_n) qquad left( begin{array}{l} text{by countable additivity of } P text{ and} \ text{pairwise disjointness of } {B_n}_{n=1}^infty end{array} right) \[10pt]
& = lim_{Ntoinfty} sum_{n=1}^N P(B_n) = lim_{Ntoinfty} Pleft( bigcup_{n=1}^N B_n right) = lim_{Ntoinfty} P(A_N).
end{align}
answered Oct 6 '16 at 15:44
Michael Hardy
1
edited Oct 9 '16 at 23:15
answered Oct 6 '16 at 15:44
Michael Hardy
1
answered Oct 6 '16 at 15:44
Michael Hardy
1
answered Oct 6 '16 at 15:44
Michael Hardy
1
1
add a comment |
add a comment |
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