derivation of variance of negative binomial distribution , pmf P(X=X)=((x-1) taken (k-1)) (p^k)(q^x-k) and E(X)=k/p. Start form Var(X)=E(X²)-(E(X))²=E(X(X-1)+X)-(E(X))²=E(X(X-1))+E(X)-(E(X))²

The goal is to derive the variance of the negative binomial distribution using the provided probability mass function (PMF) and moment formulas. Let's go step by step:

### Step 1: Recall the PMF of the Negative Binomial Distribution
We are given the PMF as:
\[
P(X = x) = \binom{x-1}{k-1} p^k q^{x-k}, \quad x = k, k+1, k+2, \ldots
\]
where \( p \) is the probability of success, \( q = 1 - p \) is the probability of failure, and \( k \) is the number of successes.

The expected value is given as:
\[
E(X) = \frac{k}{p}.
\]

### Step 2: Variance Definition
The variance of \( X \), \( \text{Var}(X) \), can be expressed in terms of the expectation of \( X^2 \) as:
\[
\text{Var}(X) = E(X^2) - (E(X))^2.
\]
We will now derive \( E(X^2) \).

### Step 3: Breaking Down \( E(X^2) \)
You can express \( E(X^2) \) in terms of the first and second factorial moments:
\[
E(X^2) = E(X(X-1) + X) = E(X(X-1)) + E(X).
\]

Now, we need to compute \( E(X(X-1)) \) and \( E(X) \).

### Step 4: Deriving \( E(X) \)
The expected value \( E(X) \) is given as:
\[
E(X) = \frac{k}{p}.
\]

### Step 5: Deriving \( E(X(X-1)) \)
The second factorial moment \( E(X(X-1)) \) for the negative binomial distribution can be derived as follows:
\[
E(X(X-1)) = \frac{k(k+1)}{p^2}.
\]

### Step 6: Substituting into Variance Formula
Now we have both terms needed for the variance:
\[
E(X^2) = E(X(X-1)) + E(X) = \frac{k(k+1)}{p^2} + \frac{k}{p}.
\]

Thus:
\[
E(X^2) = \frac{k(k+1)}{p^2} + \frac{k}{p}.
\]
We can simplify this expression:
\[
E(X^2) = \frac{k(k+1)}{p^2} + \frac{k}{p} = \frac{k(k+1) + kp}{p^2} = \frac{k(k+1+p)}{p^2}.
\]

### Step 7: Variance Formula
Now, plug \( E(X^2) \) and \( (E(X))^2 \) into the variance formula:
\[
\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{k(k+1+p)}{p^2} - \left( \frac{k}{p} \right)^2.
\]
Simplifying:
\[
\text{Var}(X) = \frac{k(k+1+p)}{p^2} - \frac{k^2}{p^2}.
\]
\[
\text{Var}(X) = \frac{k(k+1+p) - k^2}{p^2} = \frac{k(p+1)}{p^2}.
\]

### Final Answer
Thus, the variance of the negative binomial distribution is:
\[
\text{Var}(X) = \frac{k(1-p)}{p^2}.
\]

This is the standard result for the variance of the negative binomial distribution.

---

Do you have any further questions, or would you like more details on any part of this derivation?

Here are 5 follow-up questions you might consider:

1. How do we derive the expected value for the negative binomial distribution?
2. What is the interpretation of \( p \) and \( k \) in the context of the negative binomial distribution?
3. How does the negative binomial distribution differ from the binomial distribution?
4. Can we derive the moment-generating function of the negative binomial distribution?
5. How does the variance change as \( p \) approaches 1?

**Tip:** In probability theory, the second factorial moment is often easier to work with than the second ordinary moment, especially for discrete distributions.

Derivation of variance of negative binomial distribution, pmf P(X=x) = ((x-1) taken (k-1)) (p^k)(q^(x-k)) and E(X)=k/p. Start from Var(X) = E(X²) - (E(X))² = E(X(X-1) + X) - (E(X))² = E(X(X-1)) + E(X) - (E(X))².

This page provides a detailed derivation of the variance for the negative binomial distribution. Using the PMF, expected value, and second factorial moment, the variance is calculated step by step. Learn how to compute key statistical values for the negative binomial distribution. Suitable for advanced college-level statistics courses.

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