# An upper bound for a conditional probability

• sabbagh80
In summary: P(A| B\cup C) = P(A|B) + P(A|C) - P(A| B \cap C) I'm taking the OPs question at face value. As posed, this is question is about the summation of (conditional) probabilities, not a calculation of posterior probabilities. It seems clear that if B and C are disjoint sets then: P(A|B\cup C)= P(A|B)+P(A|C).This would be the maximum value of the left side of the equation which I understand to be the OP's question. It's not clear to me why you replaced the OP's notation with the notation involving posterior probabilities.
sabbagh80
Hi everyone,

The problem:
Is this relation true? If so, how (or maybe where) it could be proved?P(A│B∪C)≤P(A│B)+P(A│C)-P(A|BC)

and what about its possible generalization?

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$P(x)= 1/8$ for $x = 1,2,..,8$

$$B = \{ 1,2,3 \}$$
$$C = \{ 3,4,5 \}$$
$$A = \{2,3,4,6,7,8 \}$$

so you mean that this inequality is not always true. But of course we have: P(A│B∪C)≤P(A│B)+P(A│C)
Can we find a better upper bound for P(A│B∪C)?

sabbagh80 said:
so you mean that this inequality is not always true. But of course we have: P(A│B∪C)≤P(A│B)+P(A│C)
Can we find a better upper bound for P(A│B∪C)?

The upper bound of $$P(A|B\cup C)$$ is when $$P(A|B) + P(A|C)\geq P(A)$$. Draw Venn diagrams to see why.

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sabbagh80 said:
Can we find a better upper bound for P(A│B∪C)?

You'll have to explain what you are trying to do before anyone will know what "better"means in this context.

Using the idea:

$$P(X|Y) = P(Y|X) \frac{P(X)}{P(Y)}$$

and not worrying about the possibility of division by zero, you could try:

$$P(A| B \cup C) = P(B \cup C | A) \frac{P(A)}{P(B \cup C)}$$
$$= \Bigg( P(B|A) + P(C|A) - P(BC| A) \Bigg) \frac{P(A)}{P(B \cup C)}$$
$$= \Bigg( P(A|B) \frac{P(B)}{P(A)} + P(A|C) \frac{P(C)}{P(A)} - P(A|BC) \frac{P(BC}{P(A)} \Bigg) \frac{P(A)}{P(B \cup C)}$$
$$= P(A|B) \frac{P(B)}{P(B \cup C)} + P(A|C)\frac{P(C)}{P(B \cup C)} - P(A|BC)\frac{P(BC)}{P(B \cup C)}$$
$$\leq P(A|B) + P(A|C) - P(A|BC) \frac{P(BC)}{P(B \cup C)}$$

Another approach is the algebraic realization of a Venn diagram. The possible sets in this problem are: $$ABC, ABC^c, AB^cC,A^cBC,A^cBC^c,A^cB^cC,A^cB^cC^c$$
Each probability, such as $$P(A|B)$$ can be expressed as a fraction involving sums of the probabilities of those sets.

by "better", I mean "tighter" than the derived upper bound. can we further improve this bound?

sabbagh80 said:
by "better", I mean "tighter" than the derived upper bound. can we further improve this bound?

That doesn't clarify what you want because it doesn't say what quantities are desired on the right hand side.

The equality:
$$P(A|B \cup C) = P(A|B) \frac{P(B)}{P(B \cup C)} + P(A|C)\frac{P(C)}{P(B \cup C)} - P(A|BC)\frac{P(BC)}{P(B \cup C)}$$
is "tighter" than any inequality.

Stephen Tashi said:
That doesn't clarify what you want because it doesn't say what quantities are desired on the right hand side.

The equality:
$$P(A|B \cup C) = P(A|B) \frac{P(B)}{P(B \cup C)} + P(A|C)\frac{P(C)}{P(B \cup C)} - P(A|BC)\frac{P(BC)}{P(B \cup C)}$$
is "tighter" than any inequality.

you are right, I have not explain it precisely.
What I mean is this:
If I want to have both P(A|B) and P(A|C) on the right hand side of inequality and I want to find a coefficient for P(A|BC) such that the obtained expression be the tightest upper bound for the left side expression, what is it?
in other words,
$$P(A|B \cup C) \leq P(A|B)+P(A|C)-?P(A|B \cap C)$$

See if I have rephrased you problem correctly:

To find
Max $K$ such that $P(A|B \cup C) \leq P(A|B)+P(A|C)- K P(A|B \cap C)$

Define:

$$P_1 = Pr( A B C)$$
$$P_2 = Pr( A B C^c)$$
$$P_3 = Pr( A B^c C)$$
$$P_4 = Pr( A B^c C^c)$$
$$P_5 = Pr( A^c B C)$$
$$P_6 = Pr( A^c B C^c)$$
$$P_7 = Pr( A^c B^c C)$$
$$P_8 = Pr( A^c B^c C^c)$$

The problem becomes: Minimize $K$ subject as a function of the $P_i$ subject to the constraints:

$$0 \leq P_i \leq 1$$
$$\sum_{i=1}^8 P_i = 1$$
$$\frac{P_1 + P_2 + P_3}{P_1 + P_2 + P_3 + P_5 + P_6 + P_7} = \frac{P_1 + P_2}{P_1 + P_2 + P_5 + P_6} + \frac{P_1 + P_3}{P_1 + P_3 + P_5 + P_7} - K \frac{P_1}{P1 + P_5}$$

sabbagh80 said:
you are right, I have not explain it precisely.
What I mean is this:
If I want to have both P(A|B) and P(A|C) on the right hand side of inequality and I want to find a coefficient for P(A|BC) such that the obtained expression be the tightest upper bound for the left side expression, what is it?
in other words,
$$P(A|B \cup C) \leq P(A|B)+P(A|C)-?P(A|B \cap C)$$

If B and C are disjoint sets then $$P(A|B\cup C)=P(A|B)+P(A|C)- (K = P(A|B\cap C))$$ where K=0.

The maximum value of K will be the maximal intersection of B and C. K is not a coefficient. It is simply the value of the last term on the right side of the equation. Note the maximum value of the left side is when K=0.

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$$P(A|B\cup C)=P(A|B)+P(A|C)- (K = P(A|B\cap C))$$ where K=0.
Are we using the same notation? For $P(X|Y)$ meaning "the probability of X given Y" we have:
$$P(B \cup C | A) = P(B|A) + P(C|A) - P(B \cap C | A)$$
but not
$$P(A| B\cup C) = P(A|B) + P(A|C) - P(A| B \cap C)$$

Stephen Tashi said:
Are we using the same notation? For $P(X|Y)$ meaning "the probability of X given Y" we have:
$$P(B \cup C | A) = P(B|A) + P(C|A) - P(B \cap C | A)$$
but not
$$P(A| B\cup C) = P(A|B) + P(A|C) - P(A| B \cap C)$$

I'm taking the OPs question at face value. As posed, this is question is about the summation of (conditional) probabilities, not a calculation of posterior probabilities.

It seems clear that if B and C are disjoint sets then:
$$P(A|B\cup C)= P(A|B)+P(A|C).$$

This would be the maximum value of the left side of the equation which I understand to be the OP's question. It's not clear to me why you replaced the OP's notation with the likelihoods.

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this is question is about the summation of (conditional) probabilities
It's not clear to me why you replaced the OP's notation with the likelihoods.

In terms of conditional probabilities, an example of what I'm talking about is:

$P(x)= 1/8$ for $x = 1,2,..,8$

$$B = \{ 1,2\}$$
$$C = \{ 4,5\}$$
$$A = \{2,3,4,6,7,8 \}$$

$$P(A|B \cup C) = \frac{ P(\{2,4\} }{P(\{1,2,4,5\}} = (2/8)/(4/8) = 1/2$$
$$P(A|B) = \frac{ P(\{2\})}{P(\{1,2\}} = (1/8)/(2/8) = 1/2$$
$$P(A|C) = \frac{ P(\{4\}}{P(\{4,5\})}= (1/8)/(2/8) = 1/2$$

sabbagh80 can say if he means something different.

I don't know what distinction you are making between probabilities and liklihoods.

Stephen Tashi said:
Are we using the same notation? For $P(X|Y)$ meaning "the probability of X given Y" we have:
$$P(B \cup C | A) = P(B|A) + P(C|A) - P(B \cap C | A)$$
but not
$$P(A| B\cup C) = P(A|B) + P(A|C) - P(A| B \cap C)$$

Stephen Tashi said:
I don't know what distinction you are making between probabilities and liklihoods.

The top line of terms above are the likelihoods of the corresponding terms of the bottom line.

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I'm satisfied with both comments #5 and #9 (by "Stephen Tashi").
I also thank "SW VandeCarr" for his/her useful discussions.

## 1. What is an upper bound for a conditional probability?

An upper bound for a conditional probability is the maximum possible value that the probability of an event occurring can have, given certain conditions or information. It is a way to limit the range of potential probabilities and provide a more precise estimation.

## 2. How is an upper bound for a conditional probability calculated?

The upper bound for a conditional probability is typically calculated using mathematical formulas or equations, depending on the specific scenario. It may involve considering the probability of multiple events occurring together, or taking into account previous probabilities or information.

## 3. What is the purpose of determining an upper bound for a conditional probability?

Determining an upper bound for a conditional probability can be useful in many contexts, such as risk assessment, decision-making, and statistical analysis. It allows for a more accurate estimation of the likelihood of an event occurring, and can help in making informed decisions based on this information.

## 4. Can an upper bound for a conditional probability change?

Yes, an upper bound for a conditional probability can change depending on the conditions or information being considered. As more data or information is gathered, the upper bound may shift to reflect a more precise estimation of the probability of an event occurring.

## 5. How does an upper bound for a conditional probability differ from a lower bound?

An upper bound for a conditional probability represents the maximum possible value, while a lower bound represents the minimum possible value. Together, these bounds can provide a range within which the true probability is likely to fall.

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