Proof of Total Probability Theorem for Conditional Probability

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The law of total probability states:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A in Sigma: Pr left(Aright) = sum_i Pr left(A mid B_iright) Pr left(B_iright)$

I want to prove that this is true also for conditional probabilities. So basically I want to prove the following:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A, C in Sigma: Pr left(A mid Cright) = sum_i Pr left(A mid C cap B_iright) Pr left(B_iright)$

---

This is how I attempted it:
$$Pr(Amid C) = Pr(A|Ccap Omega) = Pr(Amid Ccapleft(bigcup_iB_iright))$$ because it is a partition. Then, using the fact that intersection distributes over union I got: $$Pr(Amid Ccapleft(bigcup_iB_iright)) = Pr(A mid bigcup_ileft(Ccap B_iright))$$

I can't go any further. I know that in a probability space we have that the probability measure $Pr$ is countably additive. I know that if we have a probability space $(Omega, Sigma, Pr)$ then the triplet $(Omega, Sigma, Qr)$ with $$Qr: Qr(A) := Pr(A | C)$$ is a probability space as well. But I have no idea how to use these two information to finish the proof.

probability probability-theory

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edited Jul 29 at 22:11

Andrés E. Caicedo

63.1k7151235

asked Jul 29 at 20:48

Euler_Salter

2,0061331

add a comment | 

up vote
1
down vote

favorite

1

The law of total probability states:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A in Sigma: Pr left(Aright) = sum_i Pr left(A mid B_iright) Pr left(B_iright)$

I want to prove that this is true also for conditional probabilities. So basically I want to prove the following:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A, C in Sigma: Pr left(A mid Cright) = sum_i Pr left(A mid C cap B_iright) Pr left(B_iright)$

---

This is how I attempted it:
$$Pr(Amid C) = Pr(A|Ccap Omega) = Pr(Amid Ccapleft(bigcup_iB_iright))$$ because it is a partition. Then, using the fact that intersection distributes over union I got: $$Pr(Amid Ccapleft(bigcup_iB_iright)) = Pr(A mid bigcup_ileft(Ccap B_iright))$$

I can't go any further. I know that in a probability space we have that the probability measure $Pr$ is countably additive. I know that if we have a probability space $(Omega, Sigma, Pr)$ then the triplet $(Omega, Sigma, Qr)$ with $$Qr: Qr(A) := Pr(A | C)$$ is a probability space as well. But I have no idea how to use these two information to finish the proof.

probability probability-theory

share|cite|improve this question

edited Jul 29 at 22:11

Andrés E. Caicedo

63.1k7151235

asked Jul 29 at 20:48

Euler_Salter

2,0061331

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

The law of total probability states:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A in Sigma: Pr left(Aright) = sum_i Pr left(A mid B_iright) Pr left(B_iright)$

I want to prove that this is true also for conditional probabilities. So basically I want to prove the following:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A, C in Sigma: Pr left(A mid Cright) = sum_i Pr left(A mid C cap B_iright) Pr left(B_iright)$

---

This is how I attempted it:
$$Pr(Amid C) = Pr(A|Ccap Omega) = Pr(Amid Ccapleft(bigcup_iB_iright))$$ because it is a partition. Then, using the fact that intersection distributes over union I got: $$Pr(Amid Ccapleft(bigcup_iB_iright)) = Pr(A mid bigcup_ileft(Ccap B_iright))$$

I can't go any further. I know that in a probability space we have that the probability measure $Pr$ is countably additive. I know that if we have a probability space $(Omega, Sigma, Pr)$ then the triplet $(Omega, Sigma, Qr)$ with $$Qr: Qr(A) := Pr(A | C)$$ is a probability space as well. But I have no idea how to use these two information to finish the proof.

probability probability-theory

share|cite|improve this question

edited Jul 29 at 22:11

Andrés E. Caicedo

63.1k7151235

asked Jul 29 at 20:48

Euler_Salter

2,0061331

The law of total probability states:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A in Sigma: Pr left(Aright) = sum_i Pr left(A mid B_iright) Pr left(B_iright)$

I want to prove that this is true also for conditional probabilities. So basically I want to prove the following:

Let $left(Omega, Sigma, Prright)$ be a probability space.

Let $leftB_1, B_2, ldotsright$ be a partition of $Omega$ such that $forall i: Pr left(B_iright) > 0$.

Then:
$displaystyle forall A, C in Sigma: Pr left(A mid Cright) = sum_i Pr left(A mid C cap B_iright) Pr left(B_iright)$

---

This is how I attempted it:
$$Pr(Amid C) = Pr(A|Ccap Omega) = Pr(Amid Ccapleft(bigcup_iB_iright))$$ because it is a partition. Then, using the fact that intersection distributes over union I got: $$Pr(Amid Ccapleft(bigcup_iB_iright)) = Pr(A mid bigcup_ileft(Ccap B_iright))$$

I can't go any further. I know that in a probability space we have that the probability measure $Pr$ is countably additive. I know that if we have a probability space $(Omega, Sigma, Pr)$ then the triplet $(Omega, Sigma, Qr)$ with $$Qr: Qr(A) := Pr(A | C)$$ is a probability space as well. But I have no idea how to use these two information to finish the proof.

probability probability-theory

share|cite|improve this question

edited Jul 29 at 22:11

Andrés E. Caicedo

63.1k7151235

asked Jul 29 at 20:48

Euler_Salter

2,0061331

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 22:11

Andrés E. Caicedo

63.1k7151235

edited Jul 29 at 22:11

Andrés E. Caicedo

63.1k7151235

edited Jul 29 at 22:11

Andrés E. Caicedo

63.1k7151235

63.1k7151235

asked Jul 29 at 20:48

Euler_Salter

2,0061331

asked Jul 29 at 20:48

Euler_Salter

2,0061331

asked Jul 29 at 20:48

Euler_Salter

2,0061331

2,0061331

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

You didn't state the result for conditional probabilities correctly. Here's an easy way to see how the correct result is derived.

The conditional probability $P_C:=P(cdot mid C)$ is a probability measure, so apply the law of total probability to it to get
$$P_C(A) = sum_i P_C(A mid B_i)P_C(B_i).$$
Now show that $P_C(A mid B_i) = P(A mid B_i cap C)$. Thus,
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i mid C).$$
That's the general form of the law of total probability for conditional probabilities. If, in addition, we assume that $B_i$ and $C$ are independent, so that $P(B_i mid C) = P(B_i)$, then the general law reduces to what you wrote, namely
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i).$$

share|cite|improve this answer

edited Jul 29 at 21:11

answered Jul 29 at 21:01

aduh

4,30031238

• 
  
  
  

  
  

  
  
  
  
  
  

  
  According to Wikipedia here (en.wikipedia.org/wiki/Law_of_total_probability#Statement) we have that the latter should be $P(B_i)$
  – Euler_Salter
  Jul 29 at 21:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Euler_Salter I think you're missing the claim about independence.
  – aduh
  Jul 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'd rephrase that a bit: if there is independence, then your statement looks great. My answer is general and holds with or without independence assumptions.
  – aduh
  Jul 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The expression $P((A mid B_i) mid C)$ has no meaning, but if you simply delete it, then what you wrote is okay.
  – aduh
  Jul 29 at 21:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $mid$ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function. It would be somewhat more correct to say $$P_C(Amid B_i)=P_B_icap C(A)$$...because it is the probability measure of event $A$ under the joint conditions of $B_i$ and $C$
  – Graham Kemp
  Jul 30 at 3:23
  
  

  
  
  
  

 | 
show 3 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
2
down vote



accepted

You didn't state the result for conditional probabilities correctly. Here's an easy way to see how the correct result is derived.

The conditional probability $P_C:=P(cdot mid C)$ is a probability measure, so apply the law of total probability to it to get
$$P_C(A) = sum_i P_C(A mid B_i)P_C(B_i).$$
Now show that $P_C(A mid B_i) = P(A mid B_i cap C)$. Thus,
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i mid C).$$
That's the general form of the law of total probability for conditional probabilities. If, in addition, we assume that $B_i$ and $C$ are independent, so that $P(B_i mid C) = P(B_i)$, then the general law reduces to what you wrote, namely
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i).$$

share|cite|improve this answer

edited Jul 29 at 21:11

answered Jul 29 at 21:01

aduh

4,30031238

• 
  
  
  

  
  

  
  
  
  
  
  

  
  According to Wikipedia here (en.wikipedia.org/wiki/Law_of_total_probability#Statement) we have that the latter should be $P(B_i)$
  – Euler_Salter
  Jul 29 at 21:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Euler_Salter I think you're missing the claim about independence.
  – aduh
  Jul 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'd rephrase that a bit: if there is independence, then your statement looks great. My answer is general and holds with or without independence assumptions.
  – aduh
  Jul 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The expression $P((A mid B_i) mid C)$ has no meaning, but if you simply delete it, then what you wrote is okay.
  – aduh
  Jul 29 at 21:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $mid$ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function. It would be somewhat more correct to say $$P_C(Amid B_i)=P_B_icap C(A)$$...because it is the probability measure of event $A$ under the joint conditions of $B_i$ and $C$
  – Graham Kemp
  Jul 30 at 3:23
  
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote



accepted

You didn't state the result for conditional probabilities correctly. Here's an easy way to see how the correct result is derived.

The conditional probability $P_C:=P(cdot mid C)$ is a probability measure, so apply the law of total probability to it to get
$$P_C(A) = sum_i P_C(A mid B_i)P_C(B_i).$$
Now show that $P_C(A mid B_i) = P(A mid B_i cap C)$. Thus,
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i mid C).$$
That's the general form of the law of total probability for conditional probabilities. If, in addition, we assume that $B_i$ and $C$ are independent, so that $P(B_i mid C) = P(B_i)$, then the general law reduces to what you wrote, namely
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i).$$

share|cite|improve this answer

edited Jul 29 at 21:11

answered Jul 29 at 21:01

aduh

4,30031238

• 
  
  
  

  
  

  
  
  
  
  
  

  
  According to Wikipedia here (en.wikipedia.org/wiki/Law_of_total_probability#Statement) we have that the latter should be $P(B_i)$
  – Euler_Salter
  Jul 29 at 21:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Euler_Salter I think you're missing the claim about independence.
  – aduh
  Jul 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'd rephrase that a bit: if there is independence, then your statement looks great. My answer is general and holds with or without independence assumptions.
  – aduh
  Jul 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The expression $P((A mid B_i) mid C)$ has no meaning, but if you simply delete it, then what you wrote is okay.
  – aduh
  Jul 29 at 21:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $mid$ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function. It would be somewhat more correct to say $$P_C(Amid B_i)=P_B_icap C(A)$$...because it is the probability measure of event $A$ under the joint conditions of $B_i$ and $C$
  – Graham Kemp
  Jul 30 at 3:23
  
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

You didn't state the result for conditional probabilities correctly. Here's an easy way to see how the correct result is derived.

The conditional probability $P_C:=P(cdot mid C)$ is a probability measure, so apply the law of total probability to it to get
$$P_C(A) = sum_i P_C(A mid B_i)P_C(B_i).$$
Now show that $P_C(A mid B_i) = P(A mid B_i cap C)$. Thus,
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i mid C).$$
That's the general form of the law of total probability for conditional probabilities. If, in addition, we assume that $B_i$ and $C$ are independent, so that $P(B_i mid C) = P(B_i)$, then the general law reduces to what you wrote, namely
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i).$$

share|cite|improve this answer

edited Jul 29 at 21:11

answered Jul 29 at 21:01

aduh

4,30031238

You didn't state the result for conditional probabilities correctly. Here's an easy way to see how the correct result is derived.

The conditional probability $P_C:=P(cdot mid C)$ is a probability measure, so apply the law of total probability to it to get
$$P_C(A) = sum_i P_C(A mid B_i)P_C(B_i).$$
Now show that $P_C(A mid B_i) = P(A mid B_i cap C)$. Thus,
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i mid C).$$
That's the general form of the law of total probability for conditional probabilities. If, in addition, we assume that $B_i$ and $C$ are independent, so that $P(B_i mid C) = P(B_i)$, then the general law reduces to what you wrote, namely
$$P(A mid C) = sum_i P(A mid B_i cap C)P(B_i).$$

share|cite|improve this answer

edited Jul 29 at 21:11

answered Jul 29 at 21:01

aduh

4,30031238

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 21:11

edited Jul 29 at 21:11

edited Jul 29 at 21:11

answered Jul 29 at 21:01

aduh

4,30031238

answered Jul 29 at 21:01

aduh

4,30031238

answered Jul 29 at 21:01

aduh

4,30031238

4,30031238

• 
  
  
  

  
  

  
  
  
  
  
  

  
  According to Wikipedia here (en.wikipedia.org/wiki/Law_of_total_probability#Statement) we have that the latter should be $P(B_i)$
  – Euler_Salter
  Jul 29 at 21:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Euler_Salter I think you're missing the claim about independence.
  – aduh
  Jul 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'd rephrase that a bit: if there is independence, then your statement looks great. My answer is general and holds with or without independence assumptions.
  – aduh
  Jul 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The expression $P((A mid B_i) mid C)$ has no meaning, but if you simply delete it, then what you wrote is okay.
  – aduh
  Jul 29 at 21:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $mid$ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function. It would be somewhat more correct to say $$P_C(Amid B_i)=P_B_icap C(A)$$...because it is the probability measure of event $A$ under the joint conditions of $B_i$ and $C$
  – Graham Kemp
  Jul 30 at 3:23
  
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  According to Wikipedia here (en.wikipedia.org/wiki/Law_of_total_probability#Statement) we have that the latter should be $P(B_i)$
  – Euler_Salter
  Jul 29 at 21:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Euler_Salter I think you're missing the claim about independence.
  – aduh
  Jul 29 at 21:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'd rephrase that a bit: if there is independence, then your statement looks great. My answer is general and holds with or without independence assumptions.
  – aduh
  Jul 29 at 21:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The expression $P((A mid B_i) mid C)$ has no meaning, but if you simply delete it, then what you wrote is okay.
  – aduh
  Jul 29 at 21:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $mid$ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function. It would be somewhat more correct to say $$P_C(Amid B_i)=P_B_icap C(A)$$...because it is the probability measure of event $A$ under the joint conditions of $B_i$ and $C$
  – Graham Kemp
  Jul 30 at 3:23
  
  

  
  
  
  

According to Wikipedia here (en.wikipedia.org/wiki/Law_of_total_probability#Statement) we have that the latter should be $P(B_i)$
– Euler_Salter
Jul 29 at 21:05

According to Wikipedia here (en.wikipedia.org/wiki/Law_of_total_probability#Statement) we have that the latter should be $P(B_i)$
– Euler_Salter
Jul 29 at 21:05

2

2

@Euler_Salter I think you're missing the claim about independence.
– aduh
Jul 29 at 21:06

@Euler_Salter I think you're missing the claim about independence.
– aduh
Jul 29 at 21:06

1

1

I'd rephrase that a bit: if there is independence, then your statement looks great. My answer is general and holds with or without independence assumptions.
– aduh
Jul 29 at 21:09

I'd rephrase that a bit: if there is independence, then your statement looks great. My answer is general and holds with or without independence assumptions.
– aduh
Jul 29 at 21:09

1

1

The expression $P((A mid B_i) mid C)$ has no meaning, but if you simply delete it, then what you wrote is okay.
– aduh
Jul 29 at 21:39

The expression $P((A mid B_i) mid C)$ has no meaning, but if you simply delete it, then what you wrote is okay.
– aduh
Jul 29 at 21:39

1

1

$mid$ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function. It would be somewhat more correct to say $$P_C(Amid B_i)=P_B_icap C(A)$$...because it is the probability measure of event $A$ under the joint conditions of $B_i$ and $C$
– Graham Kemp
Jul 30 at 3:23

$mid$ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function. It would be somewhat more correct to say $$P_C(Amid B_i)=P_B_icap C(A)$$...because it is the probability measure of event $A$ under the joint conditions of $B_i$ and $C$
– Graham Kemp
Jul 30 at 3:23

 | 
show 3 more comments

 

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