Question

If \(X\) is \(b(n, p)\), show that $$ E\left(\frac{X}{n}\right)=p \quad \text { and } \quad E\left[\left(\frac{X}{n}-p\right)^{2}\right]=\frac{p(1-p)}{n} $$

Short answer

The expectations are as given: \(E \left( \frac{X}{n} \right) = p\) and \(E\left[\left(\frac{X}{n}-p\right)^{2}\right]=\frac{p(1-p)}{n}\), as these are the fundamental properties of mean and variance for a binomial distribution.

Step-by-step solution

Establish the Expectation for Binomial Distribution

The expectation or mean of a binomial distribution \(b(n, p)\), is given by \(np\). As given, we are to find \(E \left( \frac{X}{n} \right)\). We can use the property that the expectation of a scaled random variable can be obtained by scaling the expectation of the random variable, represented as \(E(kX) = kE(X)\), where \(k\) is the scaling factor and \(X\) is the random variable.

Apply the Expectation Formula

Substituting the value of \(k = 1/n\) into the formula \(E(kX) = kE(X)\), we obtain \(E \left( \frac{X}{n} \right) = \frac{1}{n} E(X)\). Substituting the value of \(E(X)\) being the mean of the binomial distribution \(np\) into this formula, we therefore have \(E \left( \frac{X}{n} \right) = \frac{1}{n} (np) = p\).

Establish the Variance for Binomial Distribution

The variance of a binomial distribution \(b(n, p)\), is given by \(np(1-p)\). The second part of the problem is to show that \(E\left[\left(\frac{X}{n}-p\right)^{2}\right]=\frac{p(1-p)}{n}\). This is the definition of the variance of the random variable \(\frac{X}{n}\), scaled by \(1/n^2\). Using the property that the variance of a scaled random variable can be obtained by scaling the square of the variance of the random variable, represented as \(Var(kX) = k^2Var(X)\), where \(k\) is the scaling factor and \(X\) is the random variable.

Apply the Variance Formula

Substituting the value of \(k = 1/n\) into the formula \(Var(kX) = k^2Var(X)\), we obtain \(Var \left( \frac{X}{n} \right) = \left(\frac{1}{n}\right)^{2} Var(X)\). Substituting the value of \(Var(X)\) being the variance of the binomial distribution \(np(1 - p)\) into this formula, we therefore have \(Var \left( \frac{X}{n} \right) = \left(\frac{1}{n}\right)^{2} np(1 - p) = \frac{p(1 - p)}{n}\). Since variance is simply the expectation of the squared difference from the mean, it is proved that \(E\left[\left(\frac{X}{n}-p\right)^{2}\right]=\frac{p(1-p)}{n}\).