Mean and Variance of the Common Distributions

The mean (or expectation, or first moment) $\mu=\mathbb{E}[X]$ is the probability-weighted average of the values of $X$: a sum $\sum_k k\,\mathbb{P}(X=k)$ in the discrete case (Definition 13.1) and an integral $\int_{-\infty}^{\infty} x f(x)\,dx$ in the continuous case (Definition 13.2). It locates the centre of the distribution. The variance $\sigma^2=\mathrm{Var}(X)=\mathbb{E}[(X-\mu)^2]$ (Definition 13.4) is the mean squared distance from that centre, so it measures spread; being an average of squares it is never negative, and its square root $\sigma=\mathrm{SD}(X)=\sqrt{\mathrm{Var}(X)}$ — the standard deviation — restores the original units. In practice you rarely expand $\mathbb{E}[(X-\mu)^2]$ directly. Instead you use the computational formula (Proposition 13.3) $\mathrm{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2,$ which trades the deviation for the second moment $\mathbb{E}[X^2]$ (the mean square, Definition 13.3) minus the square of the mean. Finally, both summaries behave predictably under an affine change of variable (Proposition 13.4): $\mathbb{E}(aX+b)=a\mathbb{E}(X)+b$, while $\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)$ — shifting by $b$ moves the centre but not the spread, and scaling by $a$ multiplies the variance by $a^2$.

Three named discrete distributions cover most discrete models, and their moments are derived in §13.1-§13.2. A Bernoulli variable $X\sim\mathrm{Ber}(p)$ is a single trial scoring $1$ with probability $p$ and $0$ otherwise; its mean is $\mathbb{E}[X]=1\cdot p+0\cdot(1-p)=p$ (Example 13.2) and its variance is $\mathrm{Var}(X)=p(1-p)$ (Example 13.9), largest at $p=\tfrac12$ (maximum uncertainty) and zero at $p=0$ or $p=1$ (a sure outcome). A binomial variable $X\sim\mathrm{Bin}(n,p)$ counts the successes in $n$ such independent trials, so both moments simply scale by $n$: $\mathbb{E}[X]=np$ (Example 13.3) and $\mathrm{Var}(X)=np(1-p)$ (Example 13.10). Notice $\mathrm{Var}(X)=\mathbb{E}[X]\cdot(1-p)$, so for a binomial the variance is always the mean times $1-p$. A geometric variable $X\sim\mathrm{Geom}(p)$ counts the number of independent trials up to and including the first success, with p.m.f. $\mathbb{P}(X=k)=(1-p)^{k-1}p$ for $k\ge 1$; summing the series gives $\mathbb{E}[X]=\frac{1}{p}$ (Example 13.4), and with $\mathbb{E}[X^2]=\frac{1+q}{p^2}$ (where $q=1-p$) the computational formula yields $\mathrm{Var}(X)=\frac{q}{p^2}=\frac{1-p}{p^2}$ (Example 13.12). A rare success (small $p$) means a long, highly variable wait, which is why both $\frac{1}{p}$ and $\frac{1-p}{p^2}$ blow up as $p\to 0$.

The continuous uniform distribution $X\sim\mathrm{Unif}[a,b]$ spreads probability evenly over the interval $[a,b]$, with constant density $f(x)=\frac{1}{b-a}$ there and $0$ outside. Its mean is found by integration (Example 13.7): $\mathbb{E}[X]=\int_a^b x\cdot\frac{1}{b-a}\,dx=\frac{b^2-a^2}{2(b-a)}=\frac{a+b}{2},$ exactly the midpoint of the interval, as symmetry demands. For the variance (Example 13.11) first compute the second moment $\mathbb{E}[X^2]=\int_a^b x^2\cdot\frac{1}{b-a}\,dx=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3},$ and then the computational formula gives $\mathrm{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2=\frac{a^2+ab+b^2}{3}-\left(\frac{a+b}{2}\right)^2=\frac{(b-a)^2}{12}.$ The variance depends only on the width $b-a$, never on where the interval sits: sliding $[a,b]$ along the line (replacing $X$ by $X+c$) leaves the spread unchanged, in line with $\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)$. The standard deviation is $\sigma=\frac{b-a}{\sqrt{12}}$.

The four results together form a compact lookup table. Here $p\in(0,1)$ is a success probability, $n$ a number of trials, and $a<b$ the endpoints of an interval.

| Distribution | Mean $\mathbb{E}[X]$ | Variance $\mathrm{Var}(X)$ | | --- | --- | --- | | $\mathrm{Ber}(p)$ | $p$ | $p(1-p)$ | | $\mathrm{Bin}(n,p)$ | $np$ | $np(1-p)$ | | $\mathrm{Geom}(p)$ | $\dfrac{1}{p}$ | $\dfrac{1-p}{p^2}$ | | $\mathrm{Unif}[a,b]$ | $\dfrac{a+b}{2}$ | $\dfrac{(b-a)^2}{12}$ |

Applying it is a three-step recipe. First, name the family and read off its parameters from the wording (a single yes/no trial is Bernoulli; a count of successes in a fixed number of trials is binomial; a count of trials until the first success is geometric; an evenly-spread continuous quantity on an interval is uniform). Second, substitute the parameters into the mean and variance formulas — no summation or integration needed, since the derivations are already done. Third, if a spread in the original units is wanted, take $\sigma=\sqrt{\mathrm{Var}(X)}$. For instance, $\mathrm{Bin}(10,0.3)$ has mean $10(0.3)=3$ and variance $10(0.3)(0.7)=2.1$; $\mathrm{Geom}(0.25)$ has mean $\frac{1}{0.25}=4$ and variance $\frac{0.75}{0.25^2}=12$; and $\mathrm{Unif}[2,8]$ has mean $\frac{2+8}{2}=5$ and variance $\frac{(8-2)^2}{12}=3$.