Every person has a random number of children, drawn from a common distribution with mean 3 and variance 2. The numbers of children of each person are independent. Let M be the number of grandchildren of a certain person. Then:
E[M]=
Var(M)=
E[M] = 9
Var(M) = 24
E[M] = The expected number of grandchildren of a person
Well, it really depends on how enthusiastic the parents are about having children. Some parents might have a lot, others might not have any at all. So, on average, it's hard to say. But let's say the average number of grandchildren is X, then we can definitely say:
E[M] = X, where X is the expected value of the number of grandchildren.
Now, to calculate Var(M) which is the variance of the number of grandchildren, we need to consider the variance of the number of children for each person and how that affects the number of grandchildren. Let's call the variance of the number of children Y, then we can say:
Var(M) = Y * X.
So, the variance of the number of grandchildren depends on the variability in the number of children and the average number of grandchildren. With the given mean and variance of the number of children, we can calculate the variance of the number of grandchildren.
To find the expected value (E[M]) and variance (Var(M)) of the number of grandchildren (M) of a certain person, we can use the properties of expected value and variance.
Given that each person has a random number of children drawn from a common distribution with mean 3 and variance 2, we can assume that the number of children follows a Poisson distribution with parameter λ = 3. The mean and variance of a Poisson distribution are both equal to λ.
Therefore, we have:
E[M] = E[Number of children per person] × E[Number of children per child]
= 3 × 3 (expected number of children per person is 3, and expected number of children per child is also 3 because they follow the same distribution)
= 9
Var(M) = Var[Number of children per person] × E[Number of children per child]^2
+ E[Number of children per person]^2 × Var[Number of children per child]
= 2 × 3^2 + 3^2 × 2
= 18 + 18
= 36
Therefore, E[M] = 9 and Var(M) = 36 for the number of grandchildren (M) of a certain person.
To find the expected number of grandchildren (E[M]) and the variance of the number of grandchildren (Var(M)), we need to use the properties of the mean and variance of independent random variables.
Let's start with finding the expected number of grandchildren (E[M]):
Since the number of children each person has is randomly drawn from a common distribution with a mean of 3, we can consider it as a random variable X with a mean of 3.
The number of grandchildren (M) is the sum of the number of children for each child of the person.
The expected number of grandchildren (E[M]) can be found using linearity of expectation. This property states that the expected value of the sum of independent random variables is equal to the sum of their expected values.
Since the number of children is independent of each other, we can use linearity of expectation to find E[M]:
E[M] = E[X₁ + X₂ + ... + Xₙ]
By linearity of expectation, we can rewrite this as:
E[M] = E[X₁] + E[X₂] + ... + E[Xₙ]
Since each child has the same distribution and the same expected value, we can rewrite this as:
E[M] = n * E[X]
Substituting the mean of X, we get:
E[M] = n * 3
So, E[M] = 3n, where n is the number of children the person has.
Now let's move on to finding the variance of the number of grandchildren (Var(M)):
Since the number of children is randomly drawn from a common distribution with a variance of 2, we can consider it as a random variable X with a variance of 2.
The number of grandchildren (M) is the sum of the number of children for each child of the person.
The variance of the number of grandchildren (Var(M)) can be found using the property of variance.
Var(M) = Var(X₁ + X₂ + ... + Xₙ)
Since the number of children is independent, the variance of the sum can be written as:
Var(M) = Var(X₁) + Var(X₂) + ... + Var(Xₙ)
Since each child has the same distribution and the same variance, we can rewrite this as:
Var(M) = n * Var(X)
Substituting the variance of X, we get:
Var(M) = n * 2
So, Var(M) = 2n, where n is the number of children the person has.
In summary:
E[M] = 3n
Var(M) = 2n