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Chapter 1: Problem 4

If the variance of the random variable \(X\) exists, show that $$ E\left(X^{2}\right) \geq[E(X)]^{2} $$

Short Answer

Expert verified
By using the definition of variance and the property that variance is always non-negative, the statement \(E[X^2] \geq (E[X])^2\) can be shown to be true.

Step by step solution

01

Write down the definition of variance

The variance of a random variable \(X\), denoted as \(Var(X)\), is defined as the expected value of the squared deviation of \(X\) from its expected value, i.e., \(Var(X) = E[(X - E[X])^{2}]\). Note that by definition, variance is always non-negative.
02

Expand the expression inside expectation

The expression inside the expected value can be expanded as follows: \(Var(X) = E[X^2 - 2X E[X] + (E[X])^2] = E[X^2] - 2(E[X])^2 + (E[X])^2\), which simplifies to \(Var(X) = E[X^2] - (E[X])^2\). This can be rearranged to \(E[X^2] = Var(X) + (E[X])^2\).
03

Use the properties of the variance

By definition, variance is always non-negative. Therefore, \(Var(X) \geq 0\). So for any random variable \(X\) with existing variance, the expression \(Var(X) + (E[X])^2\) is always greater than or equal to \((E[X])^2\). Substituting this into the earlier equation gives \(E[X^2] \geq (E[X])^2\).

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