derivation of variance of negative binomial distribution , pmf P(X=X)=((x-1) taken (k-1)) ((pk)(qx-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. Show step by step especially E(X(X-1))

The variance of a negative binomial distribution can be derived from its definition, and in this case, we'll break it down step by step, especially focusing on \(E[X(X-1)]\).

### Step 1: Define the Negative Binomial Distribution
The Negative Binomial distribution has the following probability mass function (pmf):

\[
P(X = x) = \binom{x-1}{k-1} p^k q^{x-k}
\]
where:
- \(X\) is the number of trials until \(k\) successes.
- \(p\) is the probability of success on a single trial.
- \(q = 1 - p\) is the probability of failure on a single trial.
- \(x \geq k\), meaning \(X\) counts the total number of trials until the \(k\)-th success.

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

### Step 2: Expression for Variance
The variance \( \text{Var}(X) \) can be written as:
\[
\text{Var}(X) = E[X^2] - (E[X])^2
\]
Using the identity \( X^2 = X(X-1) + X \), we can rewrite this as:
\[
\text{Var}(X) = E[X(X-1)] + E[X] - (E[X])^2
\]
So, we need to compute \(E[X(X-1)]\), \(E[X]\), and \(E[X^2]\).

### Step 3: Calculate \(E[X(X-1)]\)

First, we focus on \(E[X(X-1)]\). This expectation is calculated as:

\[
E[X(X-1)] = \sum_{x=k}^{\infty} x(x-1) P(X=x)
\]

Substitute the pmf for \( P(X=x) \):

\[
E[X(X-1)] = \sum_{x=k}^{\infty} x(x-1) \binom{x-1}{k-1} p^k q^{x-k}
\]

This sum represents the second factorial moment of the negative binomial distribution. However, the key insight comes from using known properties of the negative binomial distribution. For a negative binomial random variable \(X\), the second factorial moment \(E[X(X-1)]\) is:

\[
E[X(X-1)] = \frac{k(k+1)}{p^2}
\]

This result can be derived through generating functions or a detailed combinatorial analysis, but it's a well-known result.

### Step 4: Final Expression for Variance

Now that we have \(E[X(X-1)]\), we can plug it into the formula for variance. Recall that:

\[
\text{Var}(X) = E[X(X-1)] + E[X] - (E[X])^2
\]

Substituting \(E[X(X-1)] = \frac{k(k+1)}{p^2}\) and \(E[X] = \frac{k}{p}\), we get:

\[
\text{Var}(X) = \frac{k(k+1)}{p^2} + \frac{k}{p} - \left(\frac{k}{p}\right)^2
\]

Simplifying the terms:

\[
\text{Var}(X) = \frac{k(k+1)}{p^2} + \frac{k}{p} - \frac{k^2}{p^2}
\]

Combine the terms:

\[
\text{Var}(X) = \frac{k(k+1) - k^2}{p^2} = \frac{k}{p^2}
\]

Thus, the variance of the negative binomial distribution is:

\[
\text{Var}(X) = \frac{kq}{p^2}
\]
where \(q = 1 - p\).

### Conclusion

The variance of a negative binomial distribution is \( \frac{kq}{p^2} \). Key steps involved computing \(E[X(X-1)]\) as \( \frac{k(k+1)}{p^2} \), then plugging this into the variance formula and simplifying.

Would you like to go over any details or have any other questions?

Here are 5 related questions to expand the topic:
1. How is the moment generating function (MGF) of the negative binomial distribution derived?
2. What is the difference between a binomial and a negative binomial distribution?
3. Can you explain the real-world scenarios where the negative binomial distribution is used?
4. How does the variance of the negative binomial change with different values of \(p\)?
5. How would the derivation of variance change if \(X\) follows a geometric distribution (special case)?

**Tip:** Understanding factorial moments and their use in variance calculations is crucial for working with distributions like the negative binomial.

derivation of variance of negative binomial distribution, pmf P(X=X)=((x-1) taken (k-1)) ((pk)(qx-k)) and E(X)=k/p. Start from Var(X)=E(X²)-(E(X))²=E(X(X-1)+X)-(E(X))². Show step by step especially E(X(X-1))

Explore the step-by-step derivation of the variance of a negative binomial distribution. Learn how to compute key components such as E(X(X-1)) and use the PMF, E(X)=k/p, and variance formula Var(X)=E(X²)-(E(X))². Suitable for undergraduate students in probability and statistics.

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