Test in the Expectation and the Variance in a Sampling Distribution

Table of Contents

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

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.

val n = 3
val p = 0.5
val m = 10

val n = 3

val p = 0.5

val m = 10

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.

fun exec(n: Int, p: Double): Int {
var result = 0
for (i in 1..n) if (Math.random() &gt;= p) result++;
return result
}

fun exec(n: Int, p: Double): Int {

var result = 0

for (i in 1..n) if (Math.random() >= p) result++;

return result

}

Define the method to calculate the mean value of \(m\) trials. This correspontds to \(Y\) value.

fun exec(m: Int, n: Int, p: Double): Double {
var result = 0.0
for (i in 1..m) result += exec(n, p)
return result / m
}

fun exec(m: Int, n: Int, p: Double): Double {

var result = 0.0

for (i in 1..m) result += exec(n, p)

return result / m

}

Calculate \(Y\) 10,000 times and calculate the expectation and the variance.

val timeCount = 100000
var mean = 0.0
var variance = 0.0
repeat(timeCount) {
val result = exec(m, n, p)
mean += result
variance += Math.pow(exec(m, n, p) - 1.5, 2.0)
}
// np = 1.5
println(mean / timeCount)
// np(1-p)/m = 0.075
println(variance / timeCount)

val timeCount = 100000

var mean = 0.0

var variance = 0.0

repeat(timeCount) {

val result = exec(m, n, p)

mean += result

variance += Math.pow(exec(m, n, p) - 1.5, 2.0)

}

// np = 1.5

println(mean / timeCount)

// np(1-p)/m = 0.075

println(variance / timeCount)

以上をまとめると次のようになります。

package com.improve_future.sample
object Main {
@JvmStatic
fun main(args: Array<String>) {
val n = 3
val p = 0.5
val m = 10
val timeCount = 100000
var mean = 0.0
var variance = 0.0
repeat(timeCount) {
val result = exec(m, n, p)
mean += result
variance += Math.pow(exec(m, n, p) - 1.5, 2.0)
}
// np = 1.5
print("Expectation: ")
println(mean / timeCount)
// np(1-p)/m = 0.075
print("Variance   : ")
println(variance / timeCount)
}
fun exec(n: Int, p: Double): Int {
var result = 0
for (i in 1..n) if (Math.random() >= p) result++;
return result
}
fun exec(m: Int, n: Int, p: Double): Double {
var result = 0.0
for (i in 1..m) result += exec(n, p)
return result / m
}
}

package com.improve_future.sample

object Main {

@JvmStatic

fun main(args: Array<String>) {