# Test in the Expectation and the Variance in a Sampling Distribution

Test the expectation and the variance in a sampling distribution with a program.

## The expectation and variance in sampling distribution

Suppose $$X$$ as a random variable in a distribution, and $$X_1 , \cdots , X_m$$ as independent samples of $$X$$, define $$Y$$ as follows.

$Y = \frac{1}{m} \sum _{k=1}^{m} X_k$

Then, $$E(Y)$$, $$V(Y)$$ is written as follows.

\begin{eqnarray*} E(Y) & = & E(X) \\ V(Y) & = & \frac{1}{m} E(X) \end{eqnarray*}

## Explanation

Let’s review the calculation of the expectation and the variance in a sampling distribution.

$$X_i$$ are independent, then $$E(Y) = \frac{1}{m} \sum _{k=1}^{m} E(X_k) = \frac{1}{m} E(X)$$ .

The variance meets $$V(A+B) = V(A)+V(B)$$ when the two random variables, $$A$$ and $$B$$, are independent. Therefour, $$V(Y) = m V \left( \frac{1}{m} X \right) = \frac{1}{m} V(X)$$ .

## Test in Binomial distribution

Test the real value in Binomial distribution with computer.

When $$X \sim \textrm{Binomial}(n, p)$$, $$E(X) = np$$, $$V(X) = np(1-p)$$ .

Suppose $$Y$$ as a mean value of $$m$$ independent samples. Then, $$E(Y) = \frac{1}{m} np$$ and $$V(Y) = \frac{1}{m} np(1-p)$$ .

Now, let’s suppose $$n = 3$$, $$p = 0.5$$, $$m = 10$$ and calculate them. $$E(X) = E(Y) = 0.5$$, $$V(X) = 0.75$$, $$V(Y) = 0.075$$ .

### Code in Kotlin

At first, define n, p and m.

Build a method to simulate each $$X$$ value for each Bernoulli trial. To generate random value which is not less than 0 and less than 1 with Math.random, compare it to p, return 1 if it is not less than $$p$$. Do it $$n$$ times.

Define the method to calculate the mean value of $$m$$ trials. This correspontds to $$Y$$ value.

Calculate $$Y$$ 10,000 times and calculate the expectation and the variance.

The code outputs as follows, and you can see $$E(Y)$$ and $$V(Y)$$ are as we reviewed before.