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: $\text{Var}(X) = np(1 - p)$

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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.