24 Variances

In this chapter, we introduce the concept of the variance of a random variable. Like the expectation, the variance serves as a single scalar characteristic that summarizes a probability distribution and is often used as a central characteristic quantity of random variables. The variance serves as a measure of the “variability” of a random variable. Ultimately, the variance of a random variable is a special covariance, namely the covariance of a random variable with itself, as discussed in the chapter on covariances. Closely related to the concept of variance is the concept of standard deviation, which we introduce in parallel. We supplement these concepts in the present chapter with their descriptive-statistical equivalents, the so-called sample variance and sample standard deviation, as well as with their occurrence in conditional distributions.

24.1 Definition

Definition 24.1 (Variance and standard deviation) Let \(\xi\) be a random variable with finite expectation \(\mathbb{E}(\xi)\). Then the variance of \(\xi\) is defined as \[\begin{equation} \mathbb{V}(\xi) := \mathbb{E}\left((\xi - \mathbb{E}(\xi))^2\right) \end{equation}\] and the standard deviation of \(\xi\) is defined as \[\begin{equation} \mathbb{S}(\xi) := \sqrt{\mathbb{V}(\xi)}. \end{equation}\]

Thus, by Definition 23.3, the variance of a random variable \(\xi\) with outcome space \(\mathcal{X}\) is the expectation of the function \[\begin{equation} f: \mathcal{X} \to \mathcal{Z}, x \mapsto f(x) := (x - \mathbb{E}(\xi))^2 \end{equation}\] and therefore the expected squared deviation of a random variable from its expectation. Squaring the deviation of the random variable from its expectation is necessary because otherwise, by Theorem 23.1, it would always hold that \[\begin{equation} \mathbb{E}(\xi-\mathbb{E}(\xi)) = \mathbb{E}(\xi) - \mathbb{E}(\xi) = 0. \end{equation}\] Intuitively, the variance thus measures the dispersion or variability of a random variable. In particular, the Chebyshev inequality, discussed in more detail in the chapter on inequalities, states that \[\begin{equation} \mathbb{P}(|\xi - \mathbb{E}(\xi)| \ge x) \le \frac{\mathbb{V}(\xi)}{x^2}. \end{equation}\] With the Chebyshev inequality, for example, \[\begin{equation} \mathbb{P}\left(|\xi - \mathbb{E}(\xi)| \ge 2 \sqrt{\mathbb{V}(\xi)}\right) \le \frac{\mathbb{V}(\xi)}{\left(2 \sqrt{\mathbb{V}(\xi)}\right)^2} = \frac{1}{4} \end{equation}\] and \[\begin{equation} \mathbb{P}\left(|\xi - \mathbb{E}(\xi)| \ge 3 \sqrt{\mathbb{V}(\xi)}\right) \le \frac{\mathbb{V}(\xi)}{\left(3 \sqrt{\mathbb{V}(\xi)}\right)^2} = \frac{1}{9}. \end{equation}\] For a random variable, it therefore always holds that the probability of an absolute deviation from twice its standard deviation is at most \(1/4\), so, from a frequentist perspective, this applies to only about a quarter of its realizations, and the probability of an absolute deviation from three times its standard deviation is at most \(1/9\), so, from a frequentist perspective, this applies to only about one tenth of its realizations, each independently of the exact form of the distribution of the random variable. Besides the variance, however, there are many other measures of the variability of random variables. Here we mention the expected absolute deviation of a random variable from its expectation, \[\begin{equation} \mathbb{A}(\xi) := \mathbb{E}(|\xi - \mathbb{E}(\xi)|), \end{equation}\] and the so-called entropy of a random variable, \[\begin{equation} \mathbb{H}(\xi) := -\mathbb{E}(\ln p(x)). \end{equation}\] We first illustrate the definition of variance in Definition 24.1 using a few examples.

Example 24.1 (Variance of a discrete random variable) Let \(\xi\) be a random variable with outcome space \(\mathcal{X} := \{-1,0,1\}\) and PMF \[\begin{equation} p(-1) = \frac{1}{4}, \quad p(0) = \frac{1}{2}, \quad p(1) = \frac{1}{4}. \end{equation}\] Then \[\begin{equation} \mathbb{V}(\xi) = \frac{1}{2}. \end{equation}\]

Proof. By Definition 24.1, using the expectation of the same random variable determined in Example 23.1, we obtain \[\begin{align} \begin{split} \mathbb{V}(\xi) & = \mathbb{E}\left((\xi - \mathbb{E}(\xi))^2\right) \\ & = \mathbb{E}\left((\xi - 0)^2\right) \\ & = \mathbb{E}\left(\xi^2\right) \\ & = \sum_{x \in \mathcal{X}} x^2 \,p(x) \\ & = (-1)^2 \cdot p(-1) + 0^2 \cdot p(0) + 1^2 \cdot p(1) \\ & = 1 \cdot \frac{1}{4} + 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} \\ & = \frac{1}{2}. \\ \end{split} \end{align}\]

Example 24.2 (Variance of a Bernoulli random variable) Let \(\xi \sim \mbox{Bern}(\mu)\). Then the variance of \(\xi\) is given by \[\begin{equation} \mathbb{V}(\xi) = \mu(1-\mu). \end{equation}\]

Proof. \(\xi\) is a discrete random variable and \(\mathbb{E}(\xi) = \mu\). Thus, \[\begin{align} \begin{split} \mathbb{V}(\xi) & = \mathbb{E}\left((\xi - \mu)^2\right) \\ & = \sum_{x \in \{0,1\}} (x - \mu)^2 \mbox{Bern}(x;\mu) \\ & = (0 - \mu)^2 \mu^0(1-\mu)^{1-0} + (1 - \mu)^2\mu^1(1-\mu)^{1-1} \\ & = \mu^2 (1-\mu) + (1 - \mu)^2\mu \\ & = \left(\mu^2 + (1 - \mu)\mu\right)(1-\mu) \\ & = \left(\mu^2 + \mu - \mu^2\right)(1 - \mu) \\ & = \mu(1-\mu). \end{split} \end{align}\]

Example 24.3 (Variance of a normally distributed random variable) Let \(\xi \sim N(\mu,\sigma^2)\). Then the variance of \(\xi\) is given by \[\begin{equation} \mathbb{V}(\xi) = \sigma^2. \end{equation}\]

Proof. We omit the proof.

Analogously to the concept of the sample mean, one defines the descriptive-statistical equivalents of Definition 24.1 as follows.

Definition 24.2 (Sample variance and sample standard deviation) Let \(\xi_1,...,\xi_n\) be random variables and let \(\bar{\xi}\) be their sample mean. The sample variance of \(\xi_1,...,\xi_n\) is defined as \[\begin{equation} S^2 := \frac{1}{n-1}\sum_{i=1}^n (\xi_i - \bar{\xi})^2 \end{equation}\] and the sample standard deviation of \(\xi_1,...,\xi_n\) is defined as \[\begin{equation} S := \sqrt{S^2}. \end{equation}\]

As in the context of the sample mean, it is central to recognize that \(\mathbb{V}(\xi)\) and \(\mathbb{S}(\xi)\) are characteristics of a random variable \(\xi\), whereas \(S^2\) and \(S\) are characteristics of a sample \(\xi_1,...,\xi_n\). Since \(S^2\) and \(S\) are functions of random variables, \(S^2\) and \(S\) are also themselves random variables. We denote realizations of \(S^2\) and \(S\) by \(s^2\) and \(s\), respectively.

Example 24.4 (Sample variance and sample standard deviation) As an example of the realizations of a sample variance and a sample standard deviation, we again consider the realizations of the random variables \(\xi_1,...,\xi_{10} \sim N(1,2)\) shown in Table 24.1, with sample mean realization \(\bar{x} = 0.68\).

Table 24.1: Sample realization

\(x_1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)	\(x_7\)	\(x_8\)	\(x_9\)	\(x_{10}\)
0.54	1.01	-3.28	0.35	2.75	-0.51	2.32	1.49	0.96	1.25

The sample variance realization is then \[\begin{equation} s^2 = \frac{1}{9}\sum_{i=1}^{10} (x_i - \bar{x})^2 = \frac{1}{9}\sum_{i=1}^{10} (x_i - 0.68)^2 = \frac{25.37}{9} = 2.82 \end{equation}\] and the sample standard deviation realization is correspondingly \[\begin{equation} s = \sqrt{s^2} = \sqrt{2.82} = 1.68. \end{equation}\]

The concepts of variance and standard deviation are evidently equivalent to one another by Definition 24.1, since one is obtained directly from the other by taking the square root or by squaring. Historically, variance is closely interwoven both with the exponential argument of the normal distribution and with the sums of squares of analysis of variance, and is therefore often easier to handle than standard deviation in model building.

Both variance and sample variance measure the variability of a random variable in squared units. An example of this is the calculation of the sample variance in Example 24.4, if one imagines the realizations in Table 24.1 as length measurements in meters (m). According to Definition 24.2, the unit is then squared when the sample variance is calculated, \[\begin{equation} s^2 = \frac{1}{9} \sum_{i=1}^{10} \left(x_i \mbox{ m} - 0.68 \mbox { m}\right)^2 = \frac{1}{9} \cdot 25.37 \mbox{ m}^2 = 2.82 \mbox{ m}^2. \end{equation}\] The resulting measure of variability would therefore have the unit of an area (\(\mbox{m}^2\)), even though the original data represent lengths in meters (\(\mbox{m}\)). Taking the square root corrects the units back to the unit of the data and, as the standard deviation, then provides a measure of variability in the original unit, \[\begin{equation} s = \sqrt{2.82 \mbox{ m}^2} = \sqrt{2.82} \sqrt{\mbox{ m}^2} = 1.68 \mbox{ m}. \end{equation}\] However, in most application contexts one is more interested in the ratios of variabilities of different samples, for which both sample variance and sample standard deviation are equally suitable.

24.2 Properties

The variance has the property of not taking negative values.

Theorem 24.1 (Non-negativity of variance) Let \(\xi\) be a random variable. Then \[\begin{equation} \mathbb{V}(\xi) \ge 0. \end{equation}\]

Proof. We consider the case of a discrete random variable. Then first \[\begin{equation} (x - \mathbb{E}(\xi))^2 \ge 0 \mbox{ for all } x \in \mathcal{X}. \end{equation}\] Furthermore, for the PMF \(p\) of \(\xi\), \[\begin{equation} p(x) \ge 0 \mbox{ for all } x \in \mathcal{X}. \end{equation}\] Thus, \[\begin{equation} p(x)(x - \mathbb{E}(\xi))^2 \ge 0 \mbox{ for all } x \in \mathcal{X}. \end{equation}\] But then \[\begin{equation} \mathbb{V}(\xi) = \sum_{x \in \mathcal{X}} p(x)(x - \mathbb{E}(\xi))^2 \ge 0. \end{equation}\] The non-negativity of the variance for continuous random variables is shown analogously.

Calculating variances is often facilitated by the following theorem, the so-called variance shift theorem, especially when the expectation of the squared random variable is easy to determine or known.

Theorem 24.2 (Variance shift theorem) Let \(\xi\) be a random variable. Then \[\begin{equation} \mathbb{V}(\xi) = \mathbb{E}\left(\xi^2 \right) - \mathbb{E}(\xi)^2. \end{equation}\]

Proof. With the definition of variance and the linearity of expectation, we have \[\begin{align} \begin{split} \mathbb{V}(\xi) & = \mathbb{E}\left((\xi - \mathbb{E}(\xi))^2\right) \\ & = \mathbb{E}\left(\xi^2 - 2\xi\mathbb{E}(\xi) + \mathbb{E}(\xi)^2 \right) \\ & = \mathbb{E}(\xi^2) - 2\mathbb{E}(\xi)\mathbb{E}(\xi) + \mathbb{E}\left(\mathbb{E}(\xi)^2\right) \\ & = \mathbb{E}(\xi^2) - 2\mathbb{E}(\xi)^2 + \mathbb{E}(\xi)^2 \\ & = \mathbb{E}(\xi^2) - \mathbb{E}(\xi)^2. \end{split} \end{align}\]

In analogy to Theorem 23.1, the following result holds for variance.

Theorem 24.3 (Variance and standard deviation under linear-affine transformation of a random variable) Let \(\xi\) be a random variable and let \[\begin{equation} f: \mathcal{X} \to \mathcal{Z}, x \mapsto f(x) := ax + b \mbox{ for } a,b \in \mathbb{R} \end{equation}\] be a linear-affine function. Then \[\begin{equation} \mathbb{V}(f(\xi)) = \mathbb{V}(a\xi + b) = a^2 \mathbb{V}(\xi) \end{equation}\] and \[\begin{equation} \mathbb{S}(f(\xi)) = \mathbb{S}(a\xi + b) = |a| \mathbb{S}(\xi). \end{equation}\]

Proof. We obtain \[\begin{align} \begin{split} \mathbb{V}(f(\xi)) & = \mathbb{V}(a\xi + b) \\ & = \mathbb{E}\left((a\xi+b-\mathbb{E}(a\xi + b))^2\right) \\ & = \mathbb{E}\left((a\xi+b-a\mathbb{E}(\xi)-b)^2\right) \\ & = \mathbb{E}\left((a\xi-a\mathbb{E}(\xi))^2\right) \\ & = \mathbb{E}\left(a(\xi - \mathbb{E}(\xi))^2\right) \\ & = \mathbb{E}\left(a^2(\xi - \mathbb{E}(\xi))^2\right) \\ & = a^2\mathbb{E}\left((\xi - \mathbb{E}(\xi))^2\right) \\ & = a^2\mathbb{V}(\xi). \\ \end{split} \end{align}\] Taking square roots while noting that \[\begin{equation} a \ge 0 \Rightarrow a^2 \ge 0 \Rightarrow \sqrt{a^2} = a \ge 0 \Leftrightarrow \sqrt{a^2} = |a| \mbox{ for } a \ge 0 \end{equation}\] and \[\begin{equation} a < 0 \Rightarrow a^2 > 0 \Rightarrow \sqrt{a^2} = -a > 0 \Leftrightarrow \sqrt{a^2} = |a| \mbox{ for } a < 0 \end{equation}\] then shows the result for the standard deviation.

The variance of the linear combination of two independent random variables can be written as a modified linear combination of the variances of the random variables.

Theorem 24.4 (Variance under linear combination of two independent random variables) Let \(\xi_1\) and \(\xi_2\) be two independent random variables and let \(a_1,a_2 \in \mathbb{R}\). Then \[\begin{equation} \mathbb{V}\left(a_1\xi_1 + a_2\xi_2\right) = a_1^2\mathbb{V}(\xi_1) + a_2^2\mathbb{V}(\xi_1). \end{equation}\]

Proof. First, we record that, by Theorem 23.2, \[\begin{equation} \mathbb{E}\left(a_1\xi_1 + a_2\xi_2\right) = a_1\mathbb{E}(\xi_1) + a_2\mathbb{E}(\xi_2). \end{equation}\] Furthermore, by Theorem 23.3, we have \[\begin{align} \begin{split} \mathbb{E}\left((\xi_1 - \mathbb{E}(\xi_1))(\xi_2 - \mathbb{E}(\xi_2))\right) & = \mathbb{E}\left(\xi_1\xi_2 - \mathbb{E}(\xi_1)\xi_2- \xi_1\mathbb{E}(\xi_2) + \mathbb{E}(\xi_1)\mathbb{E}(\xi_2)\right) \\ & = \mathbb{E}(\xi_1\xi_2) - \mathbb{E}(\xi_1)\mathbb{E}(\xi_2) - \mathbb{E}(\xi_1)\mathbb{E}(\xi_2) + \mathbb{E}(\xi_1)\mathbb{E}(\xi_2) \\ & = \mathbb{E}(\xi_1)\mathbb{E}(\xi_2) - \mathbb{E}(\xi_1)\mathbb{E}(\xi_2) \\ & = 0. \end{split} \end{align}\] It follows that \[\begin{align} \begin{split} \mathbb{V}(a_1\xi_1 + a_2\xi_2) & = \mathbb{E}\left((a_1\xi_1 + a_2\xi_2 - \mathbb{E}\left(a_1\xi_1 + a_2\xi_2\right))^2\right) \\ & = \mathbb{E}\left((a_1\xi_1 + a_2\xi_2 - a_1\mathbb{E}(\xi_1) - a_2\mathbb{E}(\xi_2))^2\right) \\ & = \mathbb{E}\left((a_1\xi_1 - a_1\mathbb{E}(\xi_1) + a_2\xi_2 - a_2\mathbb{E}(\xi_2))^2\right) \\ & = \mathbb{E}\left(a_1(\xi_1 - \mathbb{E}(\xi_1)) + (a_2(\xi_2 - \mathbb{E}(\xi_2)))^2\right) \\ & = \mathbb{E}\left(a_1(\xi_1 - \mathbb{E}(\xi_1)))^2 + 2a_1(\xi_1 - \mathbb{E}(\xi_1))(a_2(\xi_2 - \mathbb{E}(\xi_2)) + (a_2(\xi_2 - \mathbb{E}(\xi_2)))^2\right) \\ & = \mathbb{E}\left(a_1^2(\xi_1 - \mathbb{E}(\xi_1))^2 + 2a_1a_2(\xi_1 - \mathbb{E}(\xi_1))(\xi_2 - \mathbb{E}(\xi_2)) + a_2^2(\xi_2 - \mathbb{E}(\xi_2))^2\right) \\ & = a_1^2\mathbb{E}\left((\xi_1 - \mathbb{E}(\xi_1))^2\right) + 2a_1a_2\mathbb{E}\left((\xi_1 - \mathbb{E}(\xi_1))(\xi_2 - \mathbb{E}(\xi_2))\right) + a_2^2\mathbb{E}\left((\xi_2 - \mathbb{E}(\xi_2))^2\right) \\ & = a_1^2\mathbb{V}(\xi_1) + 2a_1a_2\mathbb{E}\left((\xi_1 - \mathbb{E}(\xi_1))(\xi_2 - \mathbb{E}(\xi_2))\right) + a_2^2\mathbb{V}(\xi_2) \\ & = a_1^2\mathbb{V}(\xi_1) + 2a_1a_2\cdot 0 + a_2^2\mathbb{V}(\xi_2) \\ & = a_1^2\mathbb{V}(\xi_1) + a_2^2\mathbb{V}(\xi_2). \\ \end{split} \end{align}\]

More general statements about the variance of the sum of non-independent random variables can be made only with the help of the concept of covariance introduced in the chapter on covariances.

24.3 Conditional variance

Definition 24.3 (Conditional variance) Given a random vector \(\xi := (\xi_1,\xi_2)\) with outcome space \(\mathcal{X} := \mathcal{X}_1 \times \mathcal{X}_2\). Then the conditional variance of \(\xi_1\) given \(\xi_2 = x_2\) is defined as \[\begin{equation} \mathbb{V}(\xi_1|\xi_2 = x_2) = \mathbb{E}\left(\left(\xi_1 - \mathbb{E}(\xi_1|\xi_2 = x_2)\right)^2|\xi_2 = x_2\right) \end{equation}\] and the conditional standard deviation of \(\xi_1\) given \(\xi_2 = x_2\) is defined as \[\begin{equation} \mathbb{S}(\xi_1|\xi_2 = x_2) = \sqrt{\mathbb{V}(\xi_1|\xi_2 = x_2)}. \end{equation}\]

Conditional variance is defined in the sense of conditional expectation; like a conditional expectation, a conditional variance is therefore a random variable. The shift theorem also holds for conditional variance, as stated in the following theorem.

Theorem 24.5 (Shift theorem of conditional variance) Given a random vector \(\xi := (\xi_1,\xi_2)\), we have \[\begin{equation} \mathbb{V}(\xi_1|\xi_2 = x_2) = \mathbb{E}\left(\xi_1^2|\xi_2 = x_2\right) - \mathbb{E}(\xi_1|\xi_2 = x_2)^2 \end{equation}\]

Proof. We restrict ourselves to the proof for a discrete random vector. The proof for a continuous random vector follows analogously. With the definition of conditional expectation (Definition 23.6), we have \[\begin{align} \begin{split} \mathbb{V}(\xi_1|\xi_2 = x_2) & = \mathbb{E}\left(\left(\xi_1 - \mathbb{E}(\xi_1|\xi_2 = x_2)\right)^2|\xi_2 = x_2\right) \\ & = \mathbb{E}\left(\xi_1^2 - 2\xi_1\mathbb{E}(\xi_1|\xi_2 = x_2) + \mathbb{E}(\xi_1|\xi_2 = x_2)^2|\xi_2 = x_2\right) \\ & = \sum_{x_1 \in \mathcal{X}_1}p(x_1|x_2)\left(x_1^2 - 2x_1\mathbb{E}(\xi_1|\xi_2 = x_2) + \mathbb{E}(\xi_1|\xi_2 = x_2)^2\right) \\ & = \sum_{x_1 \in \mathcal{X}_1}x_1^2p(x_1|x_2) - \sum_{x_1 \in \mathcal{X}_1}2x_1\mathbb{E}(\xi_1|\xi_2 = x_2)p(x_1|x_2) + \sum_{x_1 \in \mathcal{X}_1}\mathbb{E}(\xi_1|\xi_2 = x_2)^2p(x_1|x_2) \\ & = \mathbb{E}(\xi_1^2|\xi_2 = x_2) - 2\mathbb{E}(\xi_1|\xi_2 = x_2)\sum_{x_1 \in \mathcal{X}_1}x_1p(x_1|x_2) + \mathbb{E}(\xi_1|\xi_2 = x_2)^2\sum_{x_1 \in \mathcal{X}_1}p(x_1|x_2) \\ & = \mathbb{E}(\xi_1^2|\xi_2 = x_2) - 2\mathbb{E}(\xi_1|\xi_2 = x_2)\mathbb{E}(\xi_1|\xi_2 = x_2) + \mathbb{E}(\xi_1|\xi_2 = x_2)^2 \cdot 1 \\ & = \mathbb{E}(\xi_1^2|\xi_2 = x_2) - 2\mathbb{E}(\xi_1|\xi_2 = x_2)^2 + \mathbb{E}(\xi_1|\xi_2 = x_2)^2 \\ & = \mathbb{E}(\xi_1^2|\xi_2 = x_2) - \mathbb{E}(\xi_1|\xi_2 = x_2)^2 \end{split} \end{align}\]

The following theorem, the so-called law of iterated variance, establishes a connection between the variance of a random variable, its conditional variance, and the variance of its conditional expectation.

Theorem 24.6 (Law of iterated variance) Given a random vector \(\xi := (\xi_1,\xi_2)\), we have \[\begin{equation} \mathbb{V}(\xi_1) = \mathbb{E}(\mathbb{V}(\xi_1|\xi_2)) + \mathbb{V}(\mathbb{E}(\xi_1|\xi_2)). \end{equation}\]

Proof. We restrict ourselves to the proof for a discrete random vector. The proof for a continuous random vector follows analogously. With the variance shift theorem (Theorem 24.2), the law of iterated expectation (Theorem 23.4), and the shift theorem of conditional variance (Theorem 24.5), we have \[\begin{align} \begin{split} \mathbb{V}(\xi_1) & = \mathbb{E}(\xi_1^2) - \mathbb{E}(\xi_1)^2 \\ & = \mathbb{E}\left(\mathbb{E}(\xi_1^2|\xi_2)\right) - \mathbb{E}\left(\mathbb{E}(\xi_1|\xi_2)\right)^2 \\ & = \sum_{x_2 \in \mathcal{X}_2}p(x_2)\mathbb{E}(\xi_1^2|\xi_2 = x_2) - \mathbb{E}\left(\mathbb{E}(\xi_1|\xi_2)\right)^2 \\ & = \sum_{x_2 \in \mathcal{X}_2}p(x_2)\left(\mathbb{V}(\xi_1|\xi_2 = x_2) + \mathbb{E}(\xi_1|\xi_2 = x_2)^2 \right) - \mathbb{E}\left(\mathbb{E}(\xi_1|\xi_2)\right)^2 \\ & = \sum_{x_2 \in \mathcal{X}_2}p(x_2)\mathbb{V}(\xi_1|\xi_2 = x_2) + \sum_{x_2 \in \mathcal{X}_2}p(x_2)\mathbb{E}(\xi_1|\xi_2 = x_2)^2 - \mathbb{E}\left(\mathbb{E}(\xi_1|\xi_2)\right)^2 \\ & = \mathbb{E}(\mathbb{V}(\xi_1|\xi_2)) + \left(\mathbb{E}\left(\mathbb{E}(\xi_1|\xi_2)^2\right) - \mathbb{E}\left(\mathbb{E}(\xi_1|\xi_2)\right)^2\right) \\ & = \mathbb{E}(\mathbb{V}(\xi_1|\xi_2)) + \mathbb{V}\left(\mathbb{E}(\xi_1|\xi_2)\right) \end{split} \end{align}\]

Study questions

State the definition of the variance and standard deviation of a random variable.
Explain the definition of the variance and standard deviation of a random variable.
Compute the variance of a Bernoulli random variable.
State the definition of sample variance and sample standard deviation.
Explain the definitions of sample variance and sample standard deviation.
State the theorem on the non-negativity of variance.
State the theorem on the variance shift theorem.
State the theorem on variance and standard deviation under linear-affine transformation of a random variable.
State the theorem on variance under linear combination of two independent random variables.

Study question answers

See Definition 24.1.
See the explanations of Definition 24.1.
See Example 24.2.
See Definition 24.2.
See the explanations of Definition 24.2 and Example 24.4.
See Theorem 24.1.
See Theorem 24.2.
See Theorem 24.3.
See Theorem 24.4.