Variance and Standard Deviation

Definition of the Variance

Mathematical Definition of the Variance

Computation Formula for the Variance

Discrete random variable: $D(X) = \sum_{i=1}^{\infty} (x_i - \mu)^2 p_i$

Continuous random variable: $D(X) = \int_{-\infty}^{+\infty} (x - \mu)^2 f(x) dx$

Simplified Computation Formula for the Variance

Proof: $D(X) = E[(X - \mu)^2] = E(X^2 - 2\mu X + \mu^2) = E(X^2) - 2\mu E(X) + \mu^2 = E(X^2) - \mu^2 = E(X^2) - [E(X)]^2$

Intuitive Understanding of the Variance

Understanding: the variance is the average of the squared deviations of the values of a random variable from its expectation; it reflects the degree of dispersion of the values taken by a random variable.

Examples of the Variance

Example 1: rolling a fair die, with $X$ denoting the number that appears

• $E(X) = 3.5$
• $E(X^2) = \sum_{i=1}^6 i^2 \cdot \frac{1}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$
• $D(X) = E(X^2) - [E(X)]^2 = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{35}{12}$

Example 2: a random variable $X$ uniformly distributed on the interval $[0, 1]$

• $E(X) = \frac{1}{2}$
• $E(X^2) = \int_0^1 x^2 \cdot 1 dx = \frac{x^3}{3}\big|_0^1 = \frac{1}{3}$
• $D(X) = E(X^2) - [E(X)]^2 = \frac{1}{3} - (\frac{1}{2})^2 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$

Definition of the Standard Deviation

Mathematical Definition of the Standard Deviation

Intuitive Understanding of the Standard Deviation

Understanding: the standard deviation has the same dimension as the random variable, and it directly reflects the degree of dispersion of the values taken by a random variable.

Examples of the Standard Deviation

Example 3: continuing Example 1, the standard deviation of the die roll

• $\sigma = \sqrt{D(X)} = \sqrt{\frac{35}{12}} \approx 1.71$

Example 4: continuing Example 2, the standard deviation of the uniform distribution

• $\sigma = \sqrt{D(X)} = \sqrt{\frac{1}{12}} \approx 0.289$

Properties of the Variance

Non-negativity

Property 1: $D(X) \geq 0$

Proof: $D(X) = E[(X - \mu)^2] \geq 0$, because the squared term is non-negative.

Variance of a Constant

Property 2: $D(c) = 0$, where $c$ is a constant.

Proof: $D(c) = E[(c - c)^2] = E(0) = 0$

Variance under a Linear Transformation

Property 3: $D(aX + b) = a^2 D(X)$, where $a, b$ are constants.

Proof: $D(aX + b) = E[(aX + b - E(aX + b))^2] = E[(aX + b - aE(X) - b)^2] = E[(aX - aE(X))^2] = a^2 E[(X - E(X))^2] = a^2 D(X)$

Variance of Independent Random Variables

Property 4: if $X$ and $Y$ are independent, then $D(X + Y) = D(X) + D(Y)$

Proof: $D(X + Y) = E[(X + Y - E(X + Y))^2] = E[(X - E(X) + Y - E(Y))^2]$ $= E[(X - E(X))^2] + E[(Y - E(Y))^2] + 2E[(X - E(X))(Y - E(Y))]$ Since $X$ and $Y$ are independent, $E[(X - E(X))(Y - E(Y))] = E(X - E(X))E(Y - E(Y)) = 0$, so: $D(X + Y) = D(X) + D(Y)$

Additivity of the Variance

Property 5: if $X_1, X_2, \dots, X_n$ are pairwise independent, then: $D(X_1 + X_2 + \dots + X_n) = D(X_1) + D(X_2) + \dots + D(X_n)$

Variance of Common Distributions

Two-Point Distribution

Proof:

• $E(X) = p$
• $E(X^2) = 0^2 \cdot (1-p) + 1^2 \cdot p = p$
• $D(X) = E(X^2) - [E(X)]^2 = p - p^2 = p(1-p)$

Binomial Distribution

Proof:

• $E(X) = np$
• $E(X^2) = \sum_{k=0}^n k^2 C_n^k p^k(1-p)^{n-k}$
• Computation gives $E(X^2) = np + n(n-1)p^2$
• $D(X) = E(X^2) - [E(X)]^2 = np + n(n-1)p^2 - n^2p^2 = np(1-p)$

Poisson Distribution

Proof:

• $E(X) = \lambda$
• $E(X^2) = \sum_{k=0}^{\infty} k^2 \frac{\lambda^k}{k!} e^{-\lambda} = \lambda + \lambda^2$
• $D(X) = E(X^2) - [E(X)]^2 = \lambda + \lambda^2 - \lambda^2 = \lambda$

Uniform Distribution

Proof:

• $E(X) = \frac{a+b}{2}$
• $E(X^2) = \int_a^b x^2 \cdot \frac{1}{b-a} dx = \frac{b^3-a^3}{3(b-a)} = \frac{a^2+ab+b^2}{3}$
• $D(X) = E(X^2) - [E(X)]^2 = \frac{a^2+ab+b^2}{3} - (\frac{a+b}{2})^2 = \frac{(b-a)^2}{12}$

Exponential Distribution

Proof:

• $E(X) = \frac{1}{\lambda}$
• $E(X^2) = \int_0^{\infty} x^2 \cdot \lambda e^{-\lambda x} dx = \frac{2}{\lambda^2}$
• $D(X) = E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - (\frac{1}{\lambda})^2 = \frac{1}{\lambda^2}$

Normal Distribution

Proof:

• $E(X) = \mu$
• Computation by integration gives $E(X^2) = \mu^2 + \sigma^2$
• $D(X) = E(X^2) - [E(X)]^2 = \mu^2 + \sigma^2 - \mu^2 = \sigma^2$

Chebyshev’s Inequality

Statement of Chebyshev’s Inequality

Applications of Chebyshev’s Inequality

Application 1: estimating the probability that a random variable deviates from its expectation Application 2: proving the law of large numbers Application 3: applications in statistical inference

Applications of the Variance

Applications in Statistics

Application 1: the sample variance The sample variance is an unbiased estimator of the population variance.

Application 2: parameter estimation The variance is an important indicator for measuring the quality of an estimator.

Applications in Finance

Application 3: risk assessment The variance is an important indicator for measuring investment risk.

Application 4: portfolio theory Portfolios are optimized through the variance.

Exercises

Exercise 1

Given $E(X)=2, D(X)=3$, find $D(3X-1)$.

Exercise 2

Given that the random variable $X$ follows an exponential distribution with parameter $\lambda$, find $D(X)$.

Exercise 3

Given that the random variable $X$ follows a binomial distribution with parameters $n, p$, find $D(X)$.

Exercise 4

Given the distribution law $P(X=1)=0.2, P(X=2)=0.5, P(X=3)=0.3$ of $X$, find $D(X)$.

Exercise 5

Given that the random variable $X$ follows a Poisson distribution with parameter $\lambda$, find $D(X)$.

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Summary

Symbols Used in This Article

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