# Variance  (Page 2/4)

 Page 2 / 4

Remarks

• If the pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ is independent, it is uncorrelated. The converse is not true, as examples in the next section show.
• If the ${a}_{i}=±1$ and all pairs are uncorrelated, then
$\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[\sum _{k=1}^{n},{a}_{i},{X}_{i}\right]=\sum _{k=1}^{n}\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[{X}_{i}\right]$
The variance add even if the coefficients are negative.

We calculate variances for some common distributions. Some details are omitted—usually details of algebraic manipulation or the straightforward evaluation of integrals. In somecases we use well known sums of infinite series or values of definite integrals. A number of pertinent facts are summarized in Appendix B . Some Mathematical Aids. The results below are included in the table in Appendix C .

Variances of some discrete distributions

1. Indicator function $X={I}_{E}P\left(E\right)=p,q=1-p$ $E\left[X\right]=p$
$E\left[{X}^{2}\right]-{E}^{2}\left[X\right]=E\left[{I}_{E}^{2}\right]-{p}^{2}=E\left[{I}_{E}\right]-{p}^{2}=p-{p}^{2}=p\left(1-p\right)=pq$
2. Simple random variable $X=\sum _{i=1}^{n}{t}_{i}{I}_{{A}_{i}}$ (primitive form) $P\left({A}_{i}\right)={p}_{i}$ .
$\mathrm{Var}\left[X\right]=\sum _{i=1}^{n}{t}_{i}^{2}{p}_{i}{q}_{i}-2\sum _{i
3. Binomial $\left(n,p\right)$ . $X=\sum _{i=1}^{n}{I}_{{E}_{i}}\text{with}\left\{{I}_{{E}_{i}}:1\le i\le n\right\}\text{iid}P\left({E}_{i}\right)=p$
$\mathrm{Var}\left[X\right]=\sum _{i=1}^{n}\mathrm{Var}\left[{I}_{{E}_{i}}\right]=\sum _{i=1}^{n}pq=npq$
4. Geometric $\left(p\right)$ . $P\left(X=k\right)=p{q}^{k}\forall k\ge 0$ $E\left[X\right]=q/p$
We use a trick: $E\left[{X}^{2}\right]=E\left[X\left(X-1\right)\right]+E\left[X\right]$
$E\left[{X}^{2}\right]=p\sum _{k=0}^{\infty }k\left(k-1\right){q}^{k}+q/p=p{q}^{2}\sum _{k=2}^{\infty }k\left(k-1\right){q}^{k-2}+q/p=p{q}^{2}\frac{2}{{\left(1-q\right)}^{3}}+q/p=2\frac{{q}^{2}}{{p}^{2}}+q/p$
$\mathrm{Var}\left[X\right]=2\frac{{q}^{2}}{{p}^{2}}+q/p-{\left(q/p\right)}^{2}=q/{p}^{2}$
5. Poisson $\left(\mu \right)$ $P\left(X=k\right)={e}^{-\mu }\phantom{\rule{0.166667em}{0ex}}\frac{{\mu }^{k}}{k!}\forall k\ge 0$
Using $E\left[{X}^{2}\right]=E\left[X\left(X-1\right)\right]+E\left[X\right]$ , we have
$E\left[{X}^{2}\right]={e}^{-\mu }\sum _{k=2}^{\infty }k\left(k-1\right)\frac{{\mu }^{k}}{k!}+\mu ={e}^{-\mu }{\mu }^{2}\sum _{k=2}^{\infty }\frac{{\mu }^{k-2}}{\left(k-2\right)!}+\mu ={\mu }^{2}+\mu$
Thus, $\mathrm{Var}\left[X\right]={\mu }^{2}+\mu -{\mu }^{2}=\mu$ . Note that both the mean and the variance have common value μ .

Some absolutely continuous distributions

1. Uniform on $\left(a,b\right)$ ${f}_{X}\left(t\right)=\frac{1}{b-a}a $E\left[X\right]=\frac{a+b}{2}$
$E\left[{X}^{2}\right]=\frac{1}{b-a}{\int }_{a}^{b}{t}^{2}dt=\frac{{b}^{3}-{a}^{3}}{3\left(b-a\right)}\text{so}\mathrm{Var}\left[X\right]=\frac{{b}^{3}-{a}^{3}}{3\left(b-a\right)}-\frac{{\left(a+b\right)}^{2}}{4}=\frac{{\left(b-a\right)}^{2}}{12}$
2. Symmetric triangular $\left(a,b\right)$ Because of the shift property (V2) , we may center the distribution at the origin. Then the distribution is symmetric triangular $\left(-c,c\right)$ , where $c=\left(b-a\right)/2$ . Because of the symmetry
$\mathrm{Var}\left[X\right]=E\left[{X}^{2}\right]={\int }_{-c}^{c}{t}^{2}{f}_{X}\left(t\right)dt=2{\int }_{0}^{c}{t}^{2}{f}_{X}\left(t\right)\phantom{\rule{0.166667em}{0ex}}dt$
Now, in this case,
${f}_{X}\left(t\right)=\frac{c-t}{{c}^{2}}0\le t\le c\text{so}\phantom{\rule{4.pt}{0ex}}\text{that}E\left[{X}^{2}\right]=\frac{2}{{c}^{2}}{\int }_{0}^{c}\left(c{t}^{2}-{t}^{3}\right)dt=\frac{{c}^{2}}{6}=\frac{{\left(b-a\right)}^{2}}{24}$
3. Exponential $\left(\lambda \right)$ ${f}_{X}\left(t\right)=\lambda {e}^{-\lambda t},t\ge 0E\left[X\right]=1/\lambda$
$E\left[{X}^{2}\right]={\int }_{0}^{\infty }\lambda {t}^{2}{e}^{-\lambda t}dt=\frac{2}{{\lambda }^{2}}\text{so}\text{that}\mathrm{Var}\left[X\right]=1/{\lambda }^{2}$
4. Gamma $\left(\alpha ,\lambda \right)$ ${f}_{X}\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\lambda }^{\alpha }{t}^{\alpha -1}{e}^{-\lambda t}t\ge 0E\left[X\right]=\frac{\alpha }{\lambda }$
$E\left[{X}^{2}\right]=\frac{1}{\Gamma \left(\alpha \right)}{\int }_{0}^{\infty }{\lambda }^{\alpha }{t}^{\alpha +1}{e}^{-\lambda t}dt=\frac{\Gamma \left(\alpha +2\right)}{{\lambda }^{2}\Gamma \left(\alpha \right)}=\frac{\alpha \left(\alpha +1\right)}{{\lambda }^{2}}$
Hence $\mathrm{Var}\left[X\right]=\alpha /{\lambda }^{2}$ .
5. Normal $\left(\mu ,{\sigma }^{2}\right)$ $E\left[X\right]=\mu$
Consider $Y\sim N\left(0,1\right),E\left[Y\right]=0,\mathrm{Var}\left[Y\right]=\frac{2}{\sqrt{2\pi }}{\int }_{0}^{\infty }{t}^{2}{e}^{-{t}^{2}/2}dt=1$ .
$X=\sigma Y+\mu \text{implies}\mathrm{Var}\left[X\right]={\sigma }^{2}\mathrm{Var}\left[Y\right]={\sigma }^{2}$

Extensions of some previous examples

In the unit on expectations, we calculate the mean for a variety of cases. We revisit some of those examples and calculate the variances.

## Expected winnings ( Example 8 From "mathematical expectation: simple random variables")

A bettor places three bets at $2.00 each. The first pays$10.00 with probability 0.15, the second $8.00 with probability 0.20, and the third$20.00 with probability 0.10.

SOLUTION

The net gain may be expressed

$X=10{I}_{A}+8{I}_{B}+20{I}_{C}-6,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{with}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(A\right)=0.15,\phantom{\rule{0.277778em}{0ex}}P\left(B\right)=0.20,\phantom{\rule{0.277778em}{0ex}}P\left(C\right)=0.10$

We may reasonbly suppose the class $\left\{A,\phantom{\rule{0.166667em}{0ex}}B,\phantom{\rule{0.166667em}{0ex}}C\right\}$ is independent (this assumption is not necessary in computing the mean). Then

$\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[X\right]={10}^{2}P\left(A\right)\left[1-P\left(A\right)\right]+{8}^{2}P\left(B\right)\left[1-P\left(B\right)\right]+{20}^{2}P\left(C\right)\left[1-P\left(C\right)\right]$

Calculation is straightforward. We may use MATLAB to perform the arithmetic.

c = [10 8 20];p = 0.01*[15 20 10];q = 1 - p; VX = sum(c.^2.*p.*q)VX = 58.9900

## A function of X ( Example 9 From "mathematical expectation: simple random variables")

Suppose X in a primitive form is

$X=-3{I}_{{C}_{1}}-{I}_{{C}_{2}}+2{I}_{{C}_{3}}-3{I}_{{C}_{4}}+4{I}_{{C}_{5}}-{I}_{{C}_{6}}+{I}_{{C}_{7}}+2{I}_{{C}_{8}}+3{I}_{{C}_{9}}+2{I}_{{C}_{10}}$

with probabilities $P\left({C}_{i}\right)=0.08,\phantom{\rule{0.277778em}{0ex}}0.11,\phantom{\rule{0.277778em}{0ex}}0.06,\phantom{\rule{0.277778em}{0ex}}0.13,\phantom{\rule{0.277778em}{0ex}}0.05,\phantom{\rule{0.277778em}{0ex}}0.08,\phantom{\rule{0.277778em}{0ex}}0.12,\phantom{\rule{0.277778em}{0ex}}0.07,\phantom{\rule{0.277778em}{0ex}}0.14,\phantom{\rule{0.277778em}{0ex}}0.16$ .

Let $g\left(t\right)={t}^{2}+2t$ . Determine $E\left[g\left(X\right)\right]$ and $\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[g\left(X\right)\right]$

c = [-3 -1 2 -3 4 -1 1 2 3 2]; % Original coefficientspc = 0.01*[8 11 6 13 5 8 12 7 14 16]; % Probabilities for C_jG = c.^2 + 2*c % g(c_j) EG = G*pc' % Direct calculation E[g(X)]EG = 6.4200 VG = (G.^2)*pc' - EG^2 % Direct calculation Var[g(X)]VG = 40.8036 [Z,PZ]= csort(G,pc); % Distribution for Z = g(X) EZ = Z*PZ' % E[Z]EZ = 6.4200 VZ = (Z.^2)*PZ' - EZ^2 % Var[Z]VZ = 40.8036

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