X be a random variable which follow binomial distribution with parameters n & p. Derive the expressions for E[X] and Var[X] without using moment generating function

To derive the expressions for the **expected value** (E[X]) and the **variance** (Var[X]) of a random variable \( X \) that follows a binomial distribution with parameters \( n \) and \( p \), we proceed as follows:

### Binomial Distribution Definition:
A random variable \( X \) follows a binomial distribution if it counts the number of successes in \( n \) independent Bernoulli trials, where each trial has a probability of success \( p \). The probability mass function (PMF) of \( X \) is given by:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \quad \text{for} \quad k = 0, 1, 2, \dots, n
\]
where \( \binom{n}{k} \) is the binomial coefficient.

### 1. Deriving \( E[X] \) (Expected Value):

The **expected value** of \( X \) is defined as:

\[
E[X] = \sum_{k=0}^{n} k \cdot P(X = k)
\]

We can take a different approach using **linearity of expectation**. A binomial random variable \( X \) can be viewed as the sum of \( n \) independent Bernoulli random variables:

\[
X = X_1 + X_2 + \dots + X_n
\]
where each \( X_i \) is a Bernoulli random variable with parameter \( p \), i.e., \( P(X_i = 1) = p \) and \( P(X_i = 0) = 1 - p \).

For each Bernoulli random variable \( X_i \), the expected value is:

\[
E[X_i] = 1 \cdot p + 0 \cdot (1 - p) = p
\]

Now, using the linearity of expectation:

\[
E[X] = E[X_1 + X_2 + \dots + X_n] = E[X_1] + E[X_2] + \dots + E[X_n] = p + p + \dots + p = np
\]

Thus, the expected value of a binomial random variable \( X \) is:

\[
E[X] = np
\]

### 2. Deriving \( \text{Var}(X) \) (Variance):

The **variance** of \( X \) is defined as:

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

#### Step 1: Find \( E[X^2] \)

Since \( X = X_1 + X_2 + \dots + X_n \), we use the expansion:

\[
E[X^2] = E\left[\left(X_1 + X_2 + \dots + X_n\right)^2\right]
\]

Expanding the square:

\[
E[X^2] = E[X_1^2 + X_2^2 + \dots + X_n^2 + 2X_1X_2 + 2X_1X_3 + \dots]
\]

Since \( X_i^2 = X_i \) (because \( X_i \) takes values 0 or 1), we have:

\[
E[X^2] = E[X_1 + X_2 + \dots + X_n] + 2 \cdot E[X_1X_2 + X_1X_3 + \dots]
\]

Now, \( E[X_i] = p \) and for \( i \neq j \), the product \( X_i X_j \) is also a Bernoulli random variable, so \( E[X_i X_j] = P(X_i = 1 \text{ and } X_j = 1) = p^2 \). Therefore, we get:

\[
E[X^2] = np + 2 \cdot \binom{n}{2} p^2
\]

Simplifying \( \binom{n}{2} = \frac{n(n-1)}{2} \):

\[
E[X^2] = np + n(n-1)p^2
\]

#### Step 2: Compute \( \text{Var}(X) \)

Now, using \( \text{Var}(X) = E[X^2] - (E[X])^2 \), we substitute the values of \( E[X^2] \) and \( E[X] \):

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

Simplifying:

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

\[
\text{Var}(X) = np + n^2p^2 - np^2 - n^2p^2
\]

\[
\text{Var}(X) = np(1 - p)
\]

Thus, the variance of a binomial random variable \( X \) is:

\[
\text{Var}(X) = np(1 - p)
\]

### Summary:
For a binomial random variable \( X \) with parameters \( n \) and \( p \):

1. **Expected Value**: \( E[X] = np \)
2. **Variance**: \( \text{Var}(X) = np(1 - p) \)

---

Would you like more details or have any questions?

Here are some related questions to deepen your understanding:

1. How would the expected value and variance change if \( p = 0.5 \)?
2. What is the probability that a binomial random variable \( X \) equals its expected value?
3. Can you derive \( E[X^3] \) for a binomial random variable?
4. How does the variance change as the number of trials \( n \) increases?
5. How do you compute the skewness of a binomial distribution?

**Tip:** The binomial distribution converges to a normal distribution as \( n \) becomes large, according to the Central Limit Theorem.

X be a random variable which follows a binomial distribution with parameters n & p. Derive the expressions for E[X] and Var[X] without using moment generating function.

Learn how to derive the expected value (E[X]) and variance (Var[X]) for a binomially distributed random variable X with parameters n and p. This solution uses the linearity of expectation and Bernoulli trials without employing moment generating functions. Ideal for undergraduate-level probability courses.

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