In the last example, we saw that the conditional distribution of π, which was a uniform over a smaller range (and in some sense, less uncertain), had a smaller variance, i.e., π΅πΊπ(πβ£π΄)β€π΅πΊπ(π). Here is an example where this is not true. Let π be uniform on {0,1,2} and let π΅ be the event that π belongs to {0,2}.
a) What is the variance of π?
π΅πΊπ(π)=
(a) 2/3
(b) 1
Well, well, well! It seems we have a variance question here. Let's see if we can clown our way through it.
Now, we have a uniform distribution π that takes values from {0, 1, 2}. To find the variance of π, we need to know a couple of things.
First, we need to find the mean of π. Since π is uniform, with equal probability for each value, the mean is simply the average of the possible values. In this case, the mean is (0 + 1 + 2)/3 = 1.
Next, to find the variance, we need to calculate the squared difference between each value and the mean, and then take the average of those squared differences. The formula for variance is a bit of a clown fest, but bear with me.
For each value π¦, we calculate (π¦ - ππππ)^2. Let's do this for each value in π:
(0 - 1)^2 = 1
(1 - 1)^2 = 0
(2 - 1)^2 = 1
To find the variance, we take the average of these squared differences. Summing them up, we get 1 + 0 + 1 = 2. Dividing by 3 (the number of values), we get 2/3.
So, the variance of π is π΅πΊπ(π) = 2/3. And that's the end of our clown show!
To calculate the variance of π, we first need to determine the mean of π.
Since π is uniformly distributed on {0, 1, 2}, each value has an equal probability of occurring. Therefore, the mean of π is given by:
πΈ(π) = (0 + 1 + 2)/3 = 3/3 = 1
Next, we can calculate the variance using the formula:
π΅πΊπ(π) = πΈ[(π - πΈ(π))^2]
Substituting the mean value into the formula, we get:
π΅πΊπ(π) = πΈ[(π - 1)^2]
Now, we'll compute the variance by considering each value of π and its respective probability:
π΅πΊπ(π) = [(0 - 1)^2 * 1/3] + [(1 - 1)^2 * 1/3] + [(2 - 1)^2 * 1/3]
= [(-1)^2 * 1/3] + [0^2 * 1/3] + [1^2 * 1/3]
= [1/3] + [0] + [1/3]
= 2/3
Therefore, the variance of π is 2/3.
To calculate the variance of π, we need to know the values of π and their corresponding probabilities. In this case, π is uniform on the set {0, 1, 2}, which means that each value in the set is equally likely to occur.
To find the variance, we can use the formula:
π΅πΊπ(π) = π¬[(π - π¬[π])^2]
where π¬[π] is the expected value of π.
Since π is a uniform random variable, the expected value π¬[π] can be calculated as the average of the possible values:
π¬[π] = (0 + 1 + 2) / 3 = 1
Now, let's calculate the variance:
π΅πΊπ(π) = π¬[(π - π¬[π])^2]
Since π can take on values 0, 1, and 2, we'll substitute these values in the formula:
π΅πΊπ(π) = (0 - 1)^2 * P(π = 0) + (1 - 1)^2 * P(π = 1) + (2 - 1)^2 * P(π = 2)
Since π is uniform, each value has an equal probability of occurring, which is 1/3. Substituting these values, we get:
π΅πΊπ(π) = (0 - 1)^2 * (1/3) + (1 - 1)^2 * (1/3) + (2 - 1)^2 * (1/3)
Simplifying the equation:
π΅πΊπ(π) = (1)^2 * (1/3) + (0)^2 * (1/3) + (1)^2 * (1/3)
π΅πΊπ(π) = 1/3 + 0 + 1/3
π΅πΊπ(π) = 2/3
Therefore, the variance of π is 2/3.