Can we prove the law of total probability for continuous distributions?
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If we have a probability space $(Omega,mathcal{F},P)$ and $Omega$ is partitioned into pairwise disjoint subsets $A_{i}$, with $iinmathbb{N}$, then the law of total probability says that $P(B)=sum_{i=1}^{n}P(B|A_{i})P(A_i{})$. This law can be proved using the following two facts:
begin{align*}
P(B|A_{i})&=frac{P(Bcap A_{i})}{P(A_{i})}\
Pleft(bigcup_{iin mathbb{N}} S_{i}right)&=sum_{iinmathbb{N}}P(S_{i})
end{align*}
Where the $S_{i}$'s are a pairwise disjoint and a $textit{countable}$ family of events in $mathcal{F}$.
However, if we want to apply the law of total probability on a continuous distribution $f$, we have (like here):
$$P(A)=int_{Omega}P(A|x)f(x)dx$$
which is the law of total probabillity but with the summation replaced with an integral, and $P(A_{i})$ replaced with $f(x)dx$. The problem is that we are conditioning on an $textit{uncountable}$ family. Is there any proof of this statement (if true)?
probability probability-distributions
edited Nov 29 '16 at 8:14
Therkel
1,132921
asked Nov 29 '16 at 6:20
adiselann
351212
add a comment |
up vote
13
down vote
favorite
If we have a probability space $(Omega,mathcal{F},P)$ and $Omega$ is partitioned into pairwise disjoint subsets $A_{i}$, with $iinmathbb{N}$, then the law of total probability says that $P(B)=sum_{i=1}^{n}P(B|A_{i})P(A_i{})$. This law can be proved using the following two facts:
begin{align*}
P(B|A_{i})&=frac{P(Bcap A_{i})}{P(A_{i})}\
Pleft(bigcup_{iin mathbb{N}} S_{i}right)&=sum_{iinmathbb{N}}P(S_{i})
end{align*}
Where the $S_{i}$'s are a pairwise disjoint and a $textit{countable}$ family of events in $mathcal{F}$.
However, if we want to apply the law of total probability on a continuous distribution $f$, we have (like here):
$$P(A)=int_{Omega}P(A|x)f(x)dx$$
which is the law of total probabillity but with the summation replaced with an integral, and $P(A_{i})$ replaced with $f(x)dx$. The problem is that we are conditioning on an $textit{uncountable}$ family. Is there any proof of this statement (if true)?
probability probability-distributions
edited Nov 29 '16 at 8:14
Therkel
1,132921
asked Nov 29 '16 at 6:20
adiselann
351212
5
Fortunately it is true; otherwise we would also have trouble with conditional expectation for continuous variables. I was not able to find the proof in my measure theory book this morning, skimming through it, but going the long way, you may be convinced by the proof of the law of total expectation together with the relation that $P(Amid B) = E[1_Amid B]$.
– Therkel
Nov 29 '16 at 7:54
add a comment |
up vote
13
down vote
favorite
up vote
13
down vote
favorite
If we have a probability space $(Omega,mathcal{F},P)$ and $Omega$ is partitioned into pairwise disjoint subsets $A_{i}$, with $iinmathbb{N}$, then the law of total probability says that $P(B)=sum_{i=1}^{n}P(B|A_{i})P(A_i{})$. This law can be proved using the following two facts:
begin{align*}
P(B|A_{i})&=frac{P(Bcap A_{i})}{P(A_{i})}\
Pleft(bigcup_{iin mathbb{N}} S_{i}right)&=sum_{iinmathbb{N}}P(S_{i})
end{align*}
Where the $S_{i}$'s are a pairwise disjoint and a $textit{countable}$ family of events in $mathcal{F}$.
However, if we want to apply the law of total probability on a continuous distribution $f$, we have (like here):
$$P(A)=int_{Omega}P(A|x)f(x)dx$$
which is the law of total probabillity but with the summation replaced with an integral, and $P(A_{i})$ replaced with $f(x)dx$. The problem is that we are conditioning on an $textit{uncountable}$ family. Is there any proof of this statement (if true)?
probability probability-distributions
edited Nov 29 '16 at 8:14
Therkel
1,132921
asked Nov 29 '16 at 6:20
adiselann
351212
If we have a probability space $(Omega,mathcal{F},P)$ and $Omega$ is partitioned into pairwise disjoint subsets $A_{i}$, with $iinmathbb{N}$, then the law of total probability says that $P(B)=sum_{i=1}^{n}P(B|A_{i})P(A_i{})$. This law can be proved using the following two facts:
begin{align*}
P(B|A_{i})&=frac{P(Bcap A_{i})}{P(A_{i})}\
Pleft(bigcup_{iin mathbb{N}} S_{i}right)&=sum_{iinmathbb{N}}P(S_{i})
end{align*}
Where the $S_{i}$'s are a pairwise disjoint and a $textit{countable}$ family of events in $mathcal{F}$.
However, if we want to apply the law of total probability on a continuous distribution $f$, we have (like here):
$$P(A)=int_{Omega}P(A|x)f(x)dx$$
which is the law of total probabillity but with the summation replaced with an integral, and $P(A_{i})$ replaced with $f(x)dx$. The problem is that we are conditioning on an $textit{uncountable}$ family. Is there any proof of this statement (if true)?
probability probability-distributions
probability probability-distributions
edited Nov 29 '16 at 8:14
Therkel
1,132921
asked Nov 29 '16 at 6:20
adiselann
351212
edited Nov 29 '16 at 8:14
Therkel
1,132921
asked Nov 29 '16 at 6:20
adiselann
351212
edited Nov 29 '16 at 8:14
Therkel
1,132921
edited Nov 29 '16 at 8:14
Therkel
1,132921
edited Nov 29 '16 at 8:14
Therkel
1,132921
1,132921
asked Nov 29 '16 at 6:20
adiselann
351212
asked Nov 29 '16 at 6:20
adiselann
351212
asked Nov 29 '16 at 6:20
adiselann
351212
351212
5
Fortunately it is true; otherwise we would also have trouble with conditional expectation for continuous variables. I was not able to find the proof in my measure theory book this morning, skimming through it, but going the long way, you may be convinced by the proof of the law of total expectation together with the relation that $P(Amid B) = E[1_Amid B]$.
– Therkel
Nov 29 '16 at 7:54
add a comment |
5
Fortunately it is true; otherwise we would also have trouble with conditional expectation for continuous variables. I was not able to find the proof in my measure theory book this morning, skimming through it, but going the long way, you may be convinced by the proof of the law of total expectation together with the relation that $P(Amid B) = E[1_Amid B]$.
– Therkel
Nov 29 '16 at 7:54
5
5
Fortunately it is true; otherwise we would also have trouble with conditional expectation for continuous variables. I was not able to find the proof in my measure theory book this morning, skimming through it, but going the long way, you may be convinced by the proof of the law of total expectation together with the relation that $P(Amid B) = E[1_Amid B]$.
– Therkel
Nov 29 '16 at 7:54
Fortunately it is true; otherwise we would also have trouble with conditional expectation for continuous variables. I was not able to find the proof in my measure theory book this morning, skimming through it, but going the long way, you may be convinced by the proof of the law of total expectation together with the relation that $P(Amid B) = E[1_Amid B]$.
– Therkel
Nov 29 '16 at 7:54
add a comment |
1 Answer
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Think of it like this:
Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.
answered Nov 22 at 14:09
John_Wick
1,134111
add a comment |
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1 Answer
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1 Answer
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up vote
1
down vote
Think of it like this:
Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.
answered Nov 22 at 14:09
John_Wick
1,134111
add a comment |
up vote
1
down vote
Think of it like this:
Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.
answered Nov 22 at 14:09
John_Wick
1,134111
add a comment |
up vote
1
down vote
up vote
1
down vote
Think of it like this:
Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.
answered Nov 22 at 14:09
John_Wick
1,134111
Think of it like this:
Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.
answered Nov 22 at 14:09
John_Wick
1,134111
answered Nov 22 at 14:09
John_Wick
1,134111
answered Nov 22 at 14:09
John_Wick
1,134111
answered Nov 22 at 14:09
John_Wick
1,134111
1,134111
add a comment |
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Fortunately it is true; otherwise we would also have trouble with conditional expectation for continuous variables. I was not able to find the proof in my measure theory book this morning, skimming through it, but going the long way, you may be convinced by the proof of the law of total expectation together with the relation that $P(Amid B) = E[1_Amid B]$.
– Therkel
Nov 29 '16 at 7:54