If Ω is finite and we have a discrete uniform probability law, and if B≠∅, then the conditional probability law on Ω, given that B occurred, is also discrete uniform.
False
To show that the conditional probability law on Ω, given that B occurred, is also discrete uniform, we need to establish two things:
1. The conditional probability law is a probability distribution.
2. The conditional probability distribution is uniform.
Let's go through each step:
1. The conditional probability law is a probability distribution:
To show this, we need to demonstrate that the conditional probabilities satisfy the axioms of probability, namely non-negativity, normalization, and countable additivity.
- Non-negativity: The conditional probabilities are defined as the ratio of the joint probability to the probability of event B. Since probabilities are non-negative, the conditional probabilities are also non-negative.
- Normalization: The sum of conditional probabilities over the sample space Ω, given that B occurred, should be equal to 1. This can be proven by using the law of total probability, which states that the probability of an event A is the sum of the probabilities of A occurring under various mutually exclusive and exhaustive conditions. In this case, the conditions correspond to subsets of B. Since B is non-empty, the total probability over B is 1. Therefore, the conditional probabilities sum up to 1.
- Countable additivity: This property holds because we are dealing with finite sample spaces.
2. The conditional probability distribution is uniform:
To show this, we need to demonstrate that all the events in the conditional probability law have equal probabilities.
Since Ω is finite and we have a discrete uniform probability law, every event in Ω has the same probability. Given that B occurred, the conditional probability of each event is calculated as the ratio between the probability of the intersection of the event with B and the probability of B. Since the probability of B is non-zero (as B is non-empty), and every event in Ω has the same probability, it follows that all events in the conditional probability law have the same probability.
Hence, we can conclude that if Ω is finite and we have a discrete uniform probability law, and if B≠∅, then the conditional probability law on Ω, given that B occurred, is also discrete uniform.
To show that the conditional probability law on Ω, given that B occurred, is also discrete uniform, we need to understand the concept of conditional probability and its relationship with the discrete uniform probability law.
Conditional probability is the probability of an event occurring, given that another event has already occurred. In this case, we want to find the conditional probability law on Ω, given that B has occurred. Let's denote this probability law as P(· | B), where "·" represents any event in Ω.
First, let's recall the definition of conditional probability. For any two events A and B, the conditional probability of A given B is defined as:
P(A | B) = P(A ∩ B) / P(B)
where P(A ∩ B) represents the probability of both A and B occurring, and P(B) is the probability of event B occurring.
Now, let's consider the discrete uniform probability law on Ω. Since Ω is finite and the probability law is discrete uniform, it means that each element in Ω has an equal probability of occurring, denoted by 1/|Ω|, where |Ω| represents the cardinality (i.e., the number of elements) of Ω.
Given that B ≠ ∅ (i.e., B is not an empty set), we know that P(B) is not equal to zero. This ensures that the denominator in the conditional probability formula is not zero, allowing us to proceed.
To show that the conditional probability law on Ω, given that B occurred, is also discrete uniform, we need to demonstrate that each event in Ω, conditioned on B, has an equal probability of occurring.
In other words, we need to show that for any event A in Ω, P(A | B) = 1/|Ω|.
To do this, we can use the definition of conditional probability:
P(A | B) = P(A ∩ B) / P(B)
Since the intersection of A and B is a subset of B, A ∩ B ⊆ B. This implies that P(A ∩ B) ≤ P(B). Moreover, since B is non-empty, P(B) > 0.
Therefore, P(A | B) = P(A ∩ B) / P(B) ≤ P(B) / P(B) = 1.
We also know that P(A | B) ≥ 0, as probabilities are non-negative. Hence, 0 ≤ P(A | B) ≤ 1.
To determine the specific value of P(A | B), we need to show that P(A | B) is equal to 1/|Ω|. We can do this by demonstrating that P(A | B) = 1/|Ω| satisfies the conditions of a probability measure: 0 ≤ P(A | B) ≤ 1 and ∑ P(A | B) = 1, where the sum is taken over all events A in Ω.
To satisfy the first condition 0 ≤ P(A | B) ≤ 1, we have already shown that P(A | B) is non-negative and less than or equal to 1.
To satisfy the second condition ∑ P(A | B) = 1, we need to show that the sum of all the conditional probabilities is equal to 1. This can be done by summing P(A | B) over all possible events A in Ω, given that B has occurred.
Since there are |Ω| events in Ω, we can write:
∑ P(A | B) = ∑ (1/|Ω|) = (1/|Ω|) ∑ 1 = (1/|Ω|) × |Ω| = 1
Thus, we have shown that for every event A in Ω, P(A | B) = 1/|Ω|, satisfying the conditions for a probability measure.
Therefore, the conditional probability law on Ω, given that B occurred, is indeed discrete uniform, with each event having an equal probability of occurring, which is 1/|Ω|.