Question

uppose the random variable X follows a Bernoulli(p) distribution with unknown parameter p such that 0≤p≤1,X∼Bernoulli(p). Show that the variance of X is equal to p(1−p), V[X]=p(1−p)

Answer 1

To show that the variance of a Bernoulli distributed random variable X is p(1-p), we calculate the mean (μ) and then the expectation of X squared (E[X^2]). The variance is the difference between E[X^2] and μ^2, which yields the result p(1-p).

Step-by-step explanation:

The question asks us to demonstrate that the variance of a Bernoulli distributed random variable X is p(1-p), where p is the probability of success in each trial. The Bernoulli distribution is a special case of the binomial distribution with n=1. Since variance is defined as the expectation of the squared deviation of a random variable from its mean (E[(X - μ)2]), let's deduce the variance of X step-by-step.

If a random variable X follows the Bernoulli distribution, it can take on two values: 1 (success) with probability p, and 0 (failure) with probability 1-p (since there are only two possible outcomes in each trial and the sum of probabilities must be 1).

First, we calculate the mean (μ) of X:

• μ = 1*p + 0*(1-p) = p

Now, we find the variance of X using the formula for variance:

• V[X] = E[X2] - (μ)2
• E[X2] because X takes value 1 (success) or 0 (failure). It is 12*p + 02*(1-p) = p
• V[X] = p - p2 = p(1 - p)