Question

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

Short answer

\(E[X^2]\geq E[X]^2\) has been proved by showing that variance of a random variable is always nonnegative.

Step-by-step solution

Understand the Definition of Variance

The variance of a random variable \(X\), denoted by \(Var(X)\) or \(\sigma^2\), is defined as the expectation of the squared deviation of \(X\) from its mean \(E[X]\). In mathematical terms, \(Var(X) = E[(X-E[X])^2]\).

Expand the Definition of Variance

If we expand the term within the expectation in the variance definition, we will get \(Var(X) = E[X^2 - 2XE[X] + E[X]^2]\). By the linearity property of expectation, this can be further simplified to \(Var(X) = E[X^2] - 2E[X]E[X] + E[X]^2 = E[X^2] - E[X]^2\).

Apply Non-Negativity of Variance

Since the variance of a random variable is always nonnegative, we have that \(Var(X) = E[X^2] - E[X]^2\geq 0\).

Rearrange the inequality

If we rearrange the inequality \(0 \leq Var(X) = E[X^2] - E[X]^2\), we obtain the inequality \(E[X^2]\geq E[X]^2\), which completes the proof.

Key concepts

Expectation of Random Variables

The expectation, or expected value, represents the average or mean value a random variable is likely to obtain if an experiment were repeated many times. In simpler terms, it's what you’d typically expect the outcome of an experiment to be. It's denoted as \(E[X]\) for a random variable \(X\). To calculate the expectation, you multiply each possible value the random variable can take by the probability of that value and then sum all those products.

For instance, if you roll a fair six-sided die, the expected value of the roll is \(E[X] = \frac{1}{6}(1) + \frac{1}{6}(2) + \frac{1}{6}(3) + \frac{1}{6}(4) + \frac{1}{6}(5) + \frac{1}{6}(6) = 3.5\), even though you can never actually roll a 3.5 with a fair die.

Linearity Property of Expectation

The linearity property of expectation states that for any two random variables, \(X\) and \(Y\), and any two real numbers, \(a\) and \(b\), the expectation of their linear combination is the same as the linear combination of their expectations. Mathematically, this is expressed as:\[E[aX + bY] = aE[X] + bE[Y]\]

Applying Linearity to Simplify Calculations

Consider a random variable \(X\), and you want to find the expected value of \(2X + 3\). According to the linearity property, this is simply \(2E[X] + 3\), which is much simpler than calculating the expectation from the definition each time. This property is particularly useful when dealing with the expectations of sums or differences of random variables.

Non-Negativity of Variance

Variance measures the spread or dispersion of values that a random variable can take. It tells us how much the values of the variable are likely to differ from the mean (the expectation). Importantly, variance can never be negative. Why? Because it's calculated based on the square of differences between the random variable's outcomes and the mean. Since any number squared is positive or zero, the variance, being an average of these squares, must also be non-negative.

This concept becomes critical when we are comparing the mean and the expected squared value of the random variable. As seen from the solution steps, since the variance \(Var(X) = E[X^2] - E[X]^2\) is always greater than or equal to zero, rearranging the terms yields the inequality \(E[X^2] \textcolor{seagreen}{\text{\textgreater=}} \) E[X]^2\. This inequality is a foundational aspect of probability theory known as the variance inequality or Chebyshev's inequality.