For X ∼ Poi(λ), show that E(X) = λ and V ar(X) = λ. (Hint: Find E[X(X − 1)] = E(X2

) − E(X).)

To show that E(X) = λ and Var(X) = λ for a random variable X following a Poisson distribution with parameter λ, we can use the fact that the expected value and variance of a Poisson distribution are equal to its parameter.

First, let's find the expected value of X, denoted as E(X).

To find E(X), we can use the fact that for any random variable Y, we have E(Y) = ∑ y P(Y = y), where the sum is taken over all possible values of Y.

For a Poisson distribution, the probability mass function is given by P(X = k) = (e^(-λ) * λ^k) / k!, where k is a non-negative integer.

Using this, we can calculate E(X) as follows:
E(X) = ∑ k k * P(X = k)
= ∑ k k * (e^(-λ) * λ^k) / k!

We can simplify this expression by factoring out the constant e^(-λ) from the sum:
E(X) = e^(-λ) * ∑ k (k * λ^k) / k!

Now, let's examine the term ∑ k (k * λ^k) / k!.
Notice that this is the expected value of X(X - 1), denoted as E[X(X - 1)].
To see this, consider that each term in the sum represents the product of two factors:
- The k term represents one X value, and
- The (λ^k) / k! term represents the probability of X taking that particular value.

Using the fact that E[X(X - 1)] = E(X^2) - E(X), we can rewrite the expression for E(X) as:
E(X) = e^(-λ) * [E(X^2) - E(X)]

Now, let's calculate the second moment E(X^2).
Using the same method as before, we have:
E(X^2) = ∑ k (k^2 * λ^k) / k!

We can rewrite this as:
E(X^2) = ∑ [(k * (k - 1) + k) * λ^k / k!]
= ∑ [(k * (k - 1) * λ^k / k!) + ∑ (k * λ^k) / k!]
= E[X(X - 1)] + E(X)

Therefore, we can rewrite the expression for E(X) as:
E(X) = e^(-λ) * [E[X(X - 1)] + E(X) - E(X)]
= e^(-λ) * E[X(X - 1)]

Since E(X) = e^(-λ) * E[X(X - 1)], we can substitute this into the previous expression for E(X) to get:
E(X) = e^(-λ) * [E(X) + E(X) - E(X)]
= e^(-λ) * [2 * E(X) - E(X)]
= λ

Therefore, we have shown that E(X) = λ.

Now, let's move on to finding the variance of X, denoted as Var(X).

We define the variance Var(X) as Var(X) = E[(X - E(X))^2].
Expanding this expression, we have:
Var(X) = E(X^2 - 2 * X * E(X) + E(X)^2)
= E(X^2) - 2 * E(X) * E(X) + E(X)^2
= E(X^2) - E(X)^2

From earlier, we already know that E(X^2) = E[X(X - 1)] + E(X).
Substituting this into the expression for Var(X), we get:
Var(X) = E[X(X - 1)] + E(X) - E(X)^2
= E[X(X - 1)] + λ - λ^2
= λ + λ - λ^2
= λ

Therefore, we have shown that Var(X) = λ.

Hence, we have proven that for X following a Poisson distribution with parameter λ, E(X) = λ and Var(X) = λ.