# 3.5: Variance of Discrete Random Variables

- Page ID
- 12766

We now look at our second numerical characteristic associated to random variables.

### Definition \(\PageIndex{1}\)

The * variance* of a random variable \(X\) is given by

$$\sigma^2 = \text{Var}(X) = \text{E}[(X-\mu)^2],\notag$$

where \(\mu\) denotes the expected value of \(X\). The

*of \(X\) is given by*

**standard deviation**$$\sigma = \text{SD}(X) = \sqrt{\text{Var}(X)}.\notag$$

In words, the variance of a random variable is the average of the squared deviations of the random variable from its mean (expected value). Notice that the variance of a random variable will result in a number with units squared, but the standard deviation will have the same units as the random variable. Thus, the standard deviation is easier to interpret, which is why we make a point to define it.

The variance and standard deviation give us a ** measure of spread **for random variables. The standard deviation is interpreted as a measure of how "spread out'' the possible values of \(X\) are with respect to the mean of \(X\), \(\mu = \text{E}[X]\).

### Example \(\PageIndex{1}\)

Consider the two random variables \(X_1\) and \(X_2\), whose probability mass functions are given by the histograms in Figure 1 below. Note that \(X_1\) and \(X_2\) have the same mean. However, in looking at the histograms, we see that the possible values of \(X_2\) are more "spread out" from the mean, indicating that the variance (and standard deviation) of \(X_2\) is larger.

Figure 1: Histograms for random variables \(X_1\) and \(X_2\), both with same expected value different variance.

Theorem 3.4.1 actually tells us how to compute variance, since it is given by finding the expected value of a *function* applied to the random variable. First, if \(X\) is a discrete random variable with possible values \(x_1, x_2, \ldots, x_i, \ldots\), and probability mass function \(p(x)\), then the variance of \(X\) is given by

$$\text{Var}(X) = \sum_{i} (x_i - \mu)^2\cdot p(x_i).\notag$$

The above formula follows directly from Definition 3.5.1. However, there is an alternate formula for calculating variance, given by the following theorem, that is often easier to use.

### Theorem \(\PageIndex{1}\)

Let \(X\) be any random variable, with mean \(\mu\). Then the variance of \(X\) is

$$\text{Var}(X) = \text{E}[X^2] - \mu^2.$$

**Proof**-
By the definition of

*variance*(Definition 3.5.1) and the linearity of expectation, we have the following:

\begin{align*}

\text{Var}(X)&= \text{E}[(X-\mu)^2]\\

&= \text{E}[X^2+\mu^2-2X\mu]\\

&= \text{E}[X^2]+\text{E}[\mu^2]-\text{E}[2X\mu]\\

&= \text{E}[X^2] + \mu^2-2\mu \text{E}[X] \quad (\text{Note: since}\ \mu\ \text{is constant, we can take it out from the expected value})\\

&= \text{E}[X^2] + \mu^2-2\mu^2\\

&= \text{E}[X^2] -\mu^2

\end{align*}

### Example \(\PageIndex{2}\)

Continuing in the context of Example 3.4.1, we calculate the variance and standard deviation of the random variable \(X\) denoting the number of heads obtained in two tosses of a fair coin. Using the alternate formula for variance, we need to first calculate \(E[X^2]\), for which we use Theorem 3.4.1:

$$E[X^2] = 0^2\cdot p(0) + 1^2\cdot p(1) + 2^2\cdot p(2) = 0 + 0.5 + 1 = 1.5.\notag$$

In Example 3.4.1, we found that \(\mu = E[X] = 1\). Thus, we find

\begin{align*}

\text{Var}(X) &= E[X^2] - \mu^2 = 1.5 - 1 = 0.5 \\

\Rightarrow\ \text{SD}(X) &= \sqrt{\text{Var}(X)} = \sqrt{0.5} \approx 0.707

\end{align*}

### Exercise \(\PageIndex{1}\)

Consider the context of Example 3.4.2, where we defined the random variable \(X\) to be our winnings on a single play of game involving flipping a fair coin three times. We found that \(\text{E}[X] = 1.25\). Now find the variance and standard deviation of \(X\).

**Answer**-
First, find \(\text{E}[X^2]\):

\begin{align*}

\text{E}[X^2] &= \sum_i x_i^2\cdot p(x_i) \\

&= (-1)^2\cdot\frac{1}{8} + 1^2\cdot\frac{1}{2} + 2^2\cdot\frac{1}{4} + 3^2\cdot\frac{1}{8} = \frac{11}{4} = 2.75

\end{align*} Now, we use the alternate formula for calculating variance:

\begin{align*}

\text{Var}(X) &= \text{E}[X^2] - \text{E}[X]^2 = 2.75 - 1.25^2 = 1.1875 \\

\Rightarrow \text{SD}(X) &= \sqrt{1.1875} \approx 1.0897

\end{align*}

Given that the variance of a random variable is defined to be the expected value of* squared* deviations from the mean, variance is not linear as expected value is. We do have the following useful property of variance though.

### Theorem \(\PageIndex{2}\)

Let \(X\) be a random variable, and \(a, b\) be constants. Then the following holds:

$$\text{Var}(aX + b) = a^2\text{Var}(X).\notag$$

### Exercise \(\PageIndex{2}\)

Prove Theorem 3.5.2.

**Answer**-
First, let \(\mu = \text{E}[X]\) and note that by the linearity of expectation we have

$$\text{E}[aX + b] = a\text{E}[X] + b = a\mu + b. \notag$$

Now, we use the alternate formula for variance given in Theorem 3.5.1 to prove the result:

\begin{align*}

\text{Var}(aX + b) &= \text{E}\left[(aX+b)^2\right] - \left(\text{E}[aX + b]\right)^2 \\

&= \text{E}[a^2X^2 +2abX + b^2] - \left(a\mu + b\right)^2\\

&= a^2\text{E}[X^2] + 2ab\text{E}[X] + b^2 - a^2\mu^2 - 2ab\mu - b^2 \\

&= a^2\text{E}[X^2] - a^2\mu^2 = a^2(\text{E}[X^2] - \mu^2) = a^2\text{Var}(X)

\end{align*}

Theorem 3.5.2 easily follows from a little algebraic modification. Note that the "\(+\ b\)'' disappears in the formula. There is an intuitive reason for this. Namely, the "\(+\ b\)'' corresponds to a *horizontal shift* of the probability mass function for the random variable. Such a transformation to this function is not going to affect the *spread*, i.e., the variance will not change.

As with expected values, for many of the common probability distributions, the variance is given by a parameter or a function of the parameters for the distribution.

Distribution |
Expected Value |

Bernoulli(\(p\)) | \(p(1-p)\) |

binomial(\(n, p\)) | \(np(1-p)\) |

hypergeometric(\(N, n, m\)) | \(\frac{n(m/N)(1 - m/N)(N-n)}{N-1}\) |

geometric(\(p\)) | \(\frac{1-p}{p^2}\) |

negative binomial(\(r, p\)) | \(\frac{r(1-p)}{p^2}\) |

Poisson(\(\lambda\)) | \(\lambda\) |