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Suppose we have some general set $X$ with a finite measure $mu$ (WLOG $mu(X)=1$). We assume that there is a random partition on $X$ to "good" and "bad", and each $xin X$ is either good or bad. For each $x$ define the event $A_x=$ $x$ is bad.
Define $A$ to be the event that the set of bad $x$ has measure larger than $epsilon$.
I want to show that there exists some $xin X$ s.t. $mathbbP(A_x | A)>epsilon$.
Intuitively, it should be clear that , since the measure of all the bad $x$ is larger than $epsilon$, but I am stuggling proving it.
probability probability-theory
asked Aug 2 at 9:55
joeyg
19210
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3
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Suppose we have some general set $X$ with a finite measure $mu$ (WLOG $mu(X)=1$). We assume that there is a random partition on $X$ to "good" and "bad", and each $xin X$ is either good or bad. For each $x$ define the event $A_x=$ $x$ is bad.
Define $A$ to be the event that the set of bad $x$ has measure larger than $epsilon$.
I want to show that there exists some $xin X$ s.t. $mathbbP(A_x | A)>epsilon$.
Intuitively, it should be clear that , since the measure of all the bad $x$ is larger than $epsilon$, but I am stuggling proving it.
probability probability-theory
asked Aug 2 at 9:55
joeyg
19210
I think you have some basic concepts of probability mixed up here. An event is a measurable subset of the sample space; it doesn't make sense to define an event $A_x$ for each element $x$ of $X$. Nor does it make sense to define an event in terms of the measure of another event. You've defined $A_x$ and $A$ as random variables - in which case the expression $mathbb P(A_xmid A)$ is not meaningful.
– Math1000
Aug 2 at 12:54
I think you didn't understand the question. $x$ being good or bad, has nothing to do with $mu$, it depends on the random partition on $X$.There is no problem defining $A_x$ as defined. How did you deduce that $A_x$ and $A$ are random variables???
– joeyg
Aug 2 at 15:58
@joeyg Your question is not formulated in the standard Kolomogorov model for Probability Theory and Math 1000 was right is asking all those questions. You are not getting an answer to your question because it is hard to understand the question Mathematically.
– Kavi Rama Murthy
Aug 3 at 7:44
add a comment |Â
up vote
3
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up vote
3
down vote
favorite
Suppose we have some general set $X$ with a finite measure $mu$ (WLOG $mu(X)=1$). We assume that there is a random partition on $X$ to "good" and "bad", and each $xin X$ is either good or bad. For each $x$ define the event $A_x=$ $x$ is bad.
Define $A$ to be the event that the set of bad $x$ has measure larger than $epsilon$.
I want to show that there exists some $xin X$ s.t. $mathbbP(A_x | A)>epsilon$.
Intuitively, it should be clear that , since the measure of all the bad $x$ is larger than $epsilon$, but I am stuggling proving it.
probability probability-theory
asked Aug 2 at 9:55
joeyg
19210
Suppose we have some general set $X$ with a finite measure $mu$ (WLOG $mu(X)=1$). We assume that there is a random partition on $X$ to "good" and "bad", and each $xin X$ is either good or bad. For each $x$ define the event $A_x=$ $x$ is bad.
Define $A$ to be the event that the set of bad $x$ has measure larger than $epsilon$.
I want to show that there exists some $xin X$ s.t. $mathbbP(A_x | A)>epsilon$.
Intuitively, it should be clear that , since the measure of all the bad $x$ is larger than $epsilon$, but I am stuggling proving it.
probability probability-theory
asked Aug 2 at 9:55
joeyg
19210
edited Aug 2 at 10:04
asked Aug 2 at 9:55
joeyg
19210
asked Aug 2 at 9:55
joeyg
19210
asked Aug 2 at 9:55
joeyg
19210
19210
I think you have some basic concepts of probability mixed up here. An event is a measurable subset of the sample space; it doesn't make sense to define an event $A_x$ for each element $x$ of $X$. Nor does it make sense to define an event in terms of the measure of another event. You've defined $A_x$ and $A$ as random variables - in which case the expression $mathbb P(A_xmid A)$ is not meaningful.
– Math1000
Aug 2 at 12:54
I think you didn't understand the question. $x$ being good or bad, has nothing to do with $mu$, it depends on the random partition on $X$.There is no problem defining $A_x$ as defined. How did you deduce that $A_x$ and $A$ are random variables???
– joeyg
Aug 2 at 15:58
@joeyg Your question is not formulated in the standard Kolomogorov model for Probability Theory and Math 1000 was right is asking all those questions. You are not getting an answer to your question because it is hard to understand the question Mathematically.
– Kavi Rama Murthy
Aug 3 at 7:44
add a comment |Â
I think you have some basic concepts of probability mixed up here. An event is a measurable subset of the sample space; it doesn't make sense to define an event $A_x$ for each element $x$ of $X$. Nor does it make sense to define an event in terms of the measure of another event. You've defined $A_x$ and $A$ as random variables - in which case the expression $mathbb P(A_xmid A)$ is not meaningful.
– Math1000
Aug 2 at 12:54
I think you didn't understand the question. $x$ being good or bad, has nothing to do with $mu$, it depends on the random partition on $X$.There is no problem defining $A_x$ as defined. How did you deduce that $A_x$ and $A$ are random variables???
– joeyg
Aug 2 at 15:58
@joeyg Your question is not formulated in the standard Kolomogorov model for Probability Theory and Math 1000 was right is asking all those questions. You are not getting an answer to your question because it is hard to understand the question Mathematically.
– Kavi Rama Murthy
Aug 3 at 7:44
I think you have some basic concepts of probability mixed up here. An event is a measurable subset of the sample space; it doesn't make sense to define an event $A_x$ for each element $x$ of $X$. Nor does it make sense to define an event in terms of the measure of another event. You've defined $A_x$ and $A$ as random variables - in which case the expression $mathbb P(A_xmid A)$ is not meaningful.
– Math1000
Aug 2 at 12:54
I think you have some basic concepts of probability mixed up here. An event is a measurable subset of the sample space; it doesn't make sense to define an event $A_x$ for each element $x$ of $X$. Nor does it make sense to define an event in terms of the measure of another event. You've defined $A_x$ and $A$ as random variables - in which case the expression $mathbb P(A_xmid A)$ is not meaningful.
– Math1000
Aug 2 at 12:54
I think you didn't understand the question. $x$ being good or bad, has nothing to do with $mu$, it depends on the random partition on $X$.There is no problem defining $A_x$ as defined. How did you deduce that $A_x$ and $A$ are random variables???
– joeyg
Aug 2 at 15:58
I think you didn't understand the question. $x$ being good or bad, has nothing to do with $mu$, it depends on the random partition on $X$.There is no problem defining $A_x$ as defined. How did you deduce that $A_x$ and $A$ are random variables???
– joeyg
Aug 2 at 15:58
@joeyg Your question is not formulated in the standard Kolomogorov model for Probability Theory and Math 1000 was right is asking all those questions. You are not getting an answer to your question because it is hard to understand the question Mathematically.
– Kavi Rama Murthy
Aug 3 at 7:44
@joeyg Your question is not formulated in the standard Kolomogorov model for Probability Theory and Math 1000 was right is asking all those questions. You are not getting an answer to your question because it is hard to understand the question Mathematically.
– Kavi Rama Murthy
Aug 3 at 7:44
add a comment |Â
1 Answer
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Provided that $mathbbP(A) > 0$, the answer is yes. In fact, the set of $x$ for which $mathbbP(A_x | A) > epsilon$ has positive measure.
Suppose $mathbbP$ is a probability measure on the space of $sigma$-measurable subsets of $X$, where $mu$ is a measure on $sigma$. Then, by definition,
$$mathbbP(A_x | A) = fracint_A 1_B(x);mathbbP(dB)mathbbP(A)$$
where $1_B$ is the indicator function of $B$, and the bad set $B$ of the partition ranges over all sets where $mu(B) > epsilon$. Integrating with respect to the measure $mu$ gives
$$int_X mathbbP(A_x | A);mu(dx) = frac1mathbbP(A) int_A int 1_B(x);mu(dx) mathbbP(dB) = frac1mathbbP(A) int_A mu(B);mathbbP(dB) > frac1mathbbP(A)epsilonmathbbP(A).$$
It follows that $mathbbP(A_x | A) > epsilon$ on a set of positive measure.
answered Aug 3 at 19:50
Strants
5,06421636
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Provided that $mathbbP(A) > 0$, the answer is yes. In fact, the set of $x$ for which $mathbbP(A_x | A) > epsilon$ has positive measure.
Suppose $mathbbP$ is a probability measure on the space of $sigma$-measurable subsets of $X$, where $mu$ is a measure on $sigma$. Then, by definition,
$$mathbbP(A_x | A) = fracint_A 1_B(x);mathbbP(dB)mathbbP(A)$$
where $1_B$ is the indicator function of $B$, and the bad set $B$ of the partition ranges over all sets where $mu(B) > epsilon$. Integrating with respect to the measure $mu$ gives
$$int_X mathbbP(A_x | A);mu(dx) = frac1mathbbP(A) int_A int 1_B(x);mu(dx) mathbbP(dB) = frac1mathbbP(A) int_A mu(B);mathbbP(dB) > frac1mathbbP(A)epsilonmathbbP(A).$$
It follows that $mathbbP(A_x | A) > epsilon$ on a set of positive measure.
answered Aug 3 at 19:50
Strants
5,06421636
add a comment |Â
up vote
0
down vote
Provided that $mathbbP(A) > 0$, the answer is yes. In fact, the set of $x$ for which $mathbbP(A_x | A) > epsilon$ has positive measure.
Suppose $mathbbP$ is a probability measure on the space of $sigma$-measurable subsets of $X$, where $mu$ is a measure on $sigma$. Then, by definition,
$$mathbbP(A_x | A) = fracint_A 1_B(x);mathbbP(dB)mathbbP(A)$$
where $1_B$ is the indicator function of $B$, and the bad set $B$ of the partition ranges over all sets where $mu(B) > epsilon$. Integrating with respect to the measure $mu$ gives
$$int_X mathbbP(A_x | A);mu(dx) = frac1mathbbP(A) int_A int 1_B(x);mu(dx) mathbbP(dB) = frac1mathbbP(A) int_A mu(B);mathbbP(dB) > frac1mathbbP(A)epsilonmathbbP(A).$$
It follows that $mathbbP(A_x | A) > epsilon$ on a set of positive measure.
answered Aug 3 at 19:50
Strants
5,06421636
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Provided that $mathbbP(A) > 0$, the answer is yes. In fact, the set of $x$ for which $mathbbP(A_x | A) > epsilon$ has positive measure.
Suppose $mathbbP$ is a probability measure on the space of $sigma$-measurable subsets of $X$, where $mu$ is a measure on $sigma$. Then, by definition,
$$mathbbP(A_x | A) = fracint_A 1_B(x);mathbbP(dB)mathbbP(A)$$
where $1_B$ is the indicator function of $B$, and the bad set $B$ of the partition ranges over all sets where $mu(B) > epsilon$. Integrating with respect to the measure $mu$ gives
$$int_X mathbbP(A_x | A);mu(dx) = frac1mathbbP(A) int_A int 1_B(x);mu(dx) mathbbP(dB) = frac1mathbbP(A) int_A mu(B);mathbbP(dB) > frac1mathbbP(A)epsilonmathbbP(A).$$
It follows that $mathbbP(A_x | A) > epsilon$ on a set of positive measure.
answered Aug 3 at 19:50
Strants
5,06421636
Provided that $mathbbP(A) > 0$, the answer is yes. In fact, the set of $x$ for which $mathbbP(A_x | A) > epsilon$ has positive measure.
Suppose $mathbbP$ is a probability measure on the space of $sigma$-measurable subsets of $X$, where $mu$ is a measure on $sigma$. Then, by definition,
$$mathbbP(A_x | A) = fracint_A 1_B(x);mathbbP(dB)mathbbP(A)$$
where $1_B$ is the indicator function of $B$, and the bad set $B$ of the partition ranges over all sets where $mu(B) > epsilon$. Integrating with respect to the measure $mu$ gives
$$int_X mathbbP(A_x | A);mu(dx) = frac1mathbbP(A) int_A int 1_B(x);mu(dx) mathbbP(dB) = frac1mathbbP(A) int_A mu(B);mathbbP(dB) > frac1mathbbP(A)epsilonmathbbP(A).$$
It follows that $mathbbP(A_x | A) > epsilon$ on a set of positive measure.
answered Aug 3 at 19:50
Strants
5,06421636
answered Aug 3 at 19:50
Strants
5,06421636
answered Aug 3 at 19:50
Strants
5,06421636
answered Aug 3 at 19:50
Strants
5,06421636
5,06421636
add a comment |Â
add a comment |Â
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I think you have some basic concepts of probability mixed up here. An event is a measurable subset of the sample space; it doesn't make sense to define an event $A_x$ for each element $x$ of $X$. Nor does it make sense to define an event in terms of the measure of another event. You've defined $A_x$ and $A$ as random variables - in which case the expression $mathbb P(A_xmid A)$ is not meaningful.
– Math1000
Aug 2 at 12:54
I think you didn't understand the question. $x$ being good or bad, has nothing to do with $mu$, it depends on the random partition on $X$.There is no problem defining $A_x$ as defined. How did you deduce that $A_x$ and $A$ are random variables???
– joeyg
Aug 2 at 15:58
@joeyg Your question is not formulated in the standard Kolomogorov model for Probability Theory and Math 1000 was right is asking all those questions. You are not getting an answer to your question because it is hard to understand the question Mathematically.
– Kavi Rama Murthy
Aug 3 at 7:44