Drive the expression of MGF for the Binomial distribution and also find its Mean and Variance?
Drive the expression of MGF for the Binomial
distribution and also find its Mean and Variance?
Solution
The binomial distribution models the number of successes in a fixed number of Bernoulli trials, where each trial has a probability p of success and a probability of (1-p) of failure. The random variable X represents the number of successes in n trials. The probability mass function (PMF) of X is given by:
`p(x) = (n choose x) * p^x * (1-p)^(n-x) for x = 0, 1, 2, ..., n`
where (n choose x) = `n! / (x! * (n-x)!)` is the binomial coefficient.
The moment generating function (MGF) of a random variable X is defined as `MGF(t) = E(e^(tx))`, where t is a real number and E is the expected value operator. To find the MGF of a binomial random variable X, we substitute the PMF into the definition of MGF and simplify:
`MGF(t) = E(e^(tx)) = SUM(x=0 to n) (e^(tx) * p(x)) = SUM(x=0 to n) (e^(tx) * (n choose x) * p^x * (1-p)^(n-x))`
`= SUM(x=0 to n) ( (e^t)^x * (1-p + pe^t)^(n-x) * (n choose x))`
`= (1-p + pe^t)^n`
This MGF tells us that the binomial distribution is completely determined by two parameters: n, the number of trials, and p, the probability of success.
The mean of a binomial random variable X is given by the expected value of X, which is the first moment of the distribution. We can find the mean by taking the derivative of the MGF with respect to t, and then setting t = 0:
`E(X) = d/dt MGF(t) | t=0 = d/dt ( (1-p + pe^t)^n ) | t=0 = n * pe^0 * (1-p + pe^0)^(n-1) = np`
So the mean of a binomial random variable X is equal to np, where n is the number of trials and p is the probability of success.
The variance of a binomial random variable X is given by `Var(X) = E(X^2) - [E(X)]^2`. The second moment of the distribution can be found by taking the second derivative of the MGF with respect to t and then setting t = 0:
`E(X^2) = d^2/dt^2 MGF(t) | t=0 = d^2/dt^2 ( (1-p + pe^t)^n ) | t=0 = n * (n-1) * p^2 e^0 * (1-p + pe^0)^(n-2) = n * (n-1) * p^2 + np`
Therefore, the variance of a binomial random variable X is
`Var(X) = E(X^2) - [E(X)]^2 = n * (n-1) * p^2 + np - (np)^2 = np(1-p)`
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