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







share|cite|improve this question




















  • 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















up vote
13
down vote

favorite
8












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







share|cite|improve this question




















  • 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













up vote
13
down vote

favorite
8









up vote
13
down vote

favorite
8






8





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







share|cite|improve this question















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






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edited Nov 29 '16 at 8:14









Therkel

1,132921




1,132921










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














  • 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










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Think of it like this:
Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.






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    Think of it like this:
    Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.






    share|cite|improve this answer

























      up vote
      1
      down vote













      Think of it like this:
      Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.






      share|cite|improve this answer























        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 $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.






        share|cite|improve this answer












        Think of it like this:
        Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=int E(1_{A}|X=x)f(x)dx=int P(A|X=x)f(x)dx$.







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        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 14:09









        John_Wick

        1,134111




        1,134111






























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