Home page for accesible maths 2.6 Expectation and Related Summaries 2.6.2 Variance 2.6.4 Standardisation

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2.6.3 Properties of Summary Measures

Both summing and integration are linear operators:

$\sum_{i=-\infty}^{\infty}ca_{i}=c\sum_{i=-\infty}^{\infty}a_{i}$
$\sum_{i=-\infty}^{\infty}(a_{i}+b_{i})=\sum_{i=-\infty}^{\infty}a_{i}+\sum_{i=% -\infty}^{\infty}b_{i}$
$\int_{-\infty}^{\infty}ca(s)\,\mathrm{d}s=c\int_{-\infty}^{\infty}a(s)\,% \mathrm{d}s$
$\int_{-\infty}^{\infty}(a(s)+b(s))\,\mathrm{d}s=\int_{-\infty}^{\infty}a(s)\,% \mathrm{d}s+\int_{-\infty}^{\infty}b(s)\,\mathrm{d}s$ .

So expectation, whether for a continuous (set $a(x)=g(x)f_{X}(x)$ and $b(x)=h(x)f_{X}(x)$ ) or a discrete (set $a_{r}=g(r)p_{R}(r)$ and $b_{r}=h(r)p_{R}(r)$ ) random variable, obeys the two rules of linearity which follow directly from the definition. For arbitrary functions $g$ and $h$ , and a constant $c$ :

$\operatorname{\mathsf{E}}\left[{g(R)+h(R)}\right]=\operatorname{\mathsf{E}}% \left[{g(R)}\right]+\operatorname{\mathsf{E}}\left[{h(R)}\right]$ ,
$\operatorname{\mathsf{E}}\left[{cg(R)}\right]=c\operatorname{\mathsf{E}}\left[% {g(R)}\right]$ .

The expectation, variance and standard deviation respectively of the linear function $aR+b$ of the random variable $R$ for constants $a$ and $b$ are:

$\operatorname{\mathsf{E}}\left[{aR+b}\right]=a\operatorname{\mathsf{E}}\left[{% R}\right]+b$ ,
${\operatorname{\mathsf{Var}}}\left[{aR+b}\right]=a^{2}{\operatorname{\mathsf{% Var}}}\left[{R}\right]$ ,
${\operatorname{\mathsf{StdDev}}}\left[{aR+b}\right]=|a|{\operatorname{\mathsf{% StdDev}}}\left[{R}\right]$ .

The first formula is a direct consequence of the two properties of linearity. The second formula arises because $\operatorname{\mathsf{E}}\left[{(aR+b)^{2}}\right]=\operatorname{\mathsf{E}}% \left[{a^{2}R^{2}+2abR+b^{2}}\right]=a^{2}\operatorname{\mathsf{E}}\left[{R^{2% }}\right]+2ab\operatorname{\mathsf{E}}\left[{R}\right]+b^{2}$ , so

\displaystyle{\operatorname{\mathsf{Var}}}\left[{aR+b}\right]=\operatorname{% \mathsf{E}}\left[{(aR+b)^{2}}\right]-\operatorname{\mathsf{E}}\left[{aR+b}% \right]^{2}=a^{2}\left(\operatorname{\mathsf{E}}\left[{R^{2}}\right]-% \operatorname{\mathsf{E}}\left[{R}\right]^{2}\right).

Finally, recall that the standard deviation is the positive square root of the variance.

Example 2.6.1.

What are the expectation and variance of the discrete probability distribution given below?

$r$	$-1$	$0$	$1$
$p(r)$	$1/3$	$1/6$	$1/2$

Solution.

$\operatorname{\mathsf{E}}\left[{R}\right]=\sum_{i=-\infty}^{\infty}ip(i)={% \color[rgb]{0.76,0.01,0}-1(1/3)}{\color[rgb]{0.76,0.01,0}+0(1/6)}{\color[rgb]{% 0.76,0.01,0}+1(1/2)}{\color[rgb]{0.76,0.01,0}=1/6.}$
$\operatorname{\mathsf{E}}\left[{R^{2}}\right]=\sum_{i=-\infty}^{\infty}i^{2}p(% i)={\color[rgb]{0.76,0.01,0}(-1)^{2}(1/3)}{\color[rgb]{0.76,0.01,0}+0^{2}(1/6)% }{\color[rgb]{0.76,0.01,0}+1^{2}(1/2)}{\color[rgb]{0.76,0.01,0}=5/6.}$
${{\operatorname{\mathsf{Var}}}\left[{R}\right]={\color[rgb]{0.76,0.01,0}% \operatorname{\mathsf{E}}\left[{R^{2}}\right]-\operatorname{\mathsf{E}}\left[{% R}\right]^{2}=5/6-1/36=29/36}.}$

Example 2.6.2.

A triangular pdf: the random variable $X$ in Example 2.4.2 has pdf

\displaystyle f_{X}(x)=\left\{\begin{array}[]{ll}1+x&\quad-1<x\leq 0\\ 1-x&\quad 0<x\leq 1\\ 0&\quad\text{otherwise}\end{array}\right.

Find

(a)

$\operatorname{\mathsf{E}}\left[{X}\right]$ ,
(b)

${\operatorname{\mathsf{Var}}}\left[{X}\right]$ ,
(c)

$\operatorname{\mathsf{E}}\left[{(2X+1)^{2}}\right]$ , and
(d)

$\operatorname{\mathsf{E}}\left[{|X|}\right]$ .

Solution. We need only consider the intervals where $f_{X}(x)>0$ .

(a)

$\displaystyle\operatorname{\mathsf{E}}\left[{X}\right]$ $\displaystyle={\color[rgb]{0.76,0.01,0}\int_{-1}^{0}t(1+t)\,\mathrm{d}t+\int_{% 0}^{1}t(1-t)\,\mathrm{d}t}$

$\displaystyle={\color[rgb]{0.76,0.01,0}[t^{2}/2+t^{3}/3]_{-1}^{0}+[t^{2}/2-t^{% 3}/3]_{0}^{1}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}-(1/2-1/3)+(1/2-1/3)=0,}$

which we could also write down directly by symmetry.
(b)

$\displaystyle\operatorname{\mathsf{E}}\left[{X^{2}}\right]$ $\displaystyle={\color[rgb]{0.76,0.01,0}\int_{-1}^{0}t^{2}(1+t)\,\mathrm{d}t+% \int_{0}^{1}t^{2}(1-t)\,\mathrm{d}t}$

$\displaystyle={\color[rgb]{0.76,0.01,0}[t^{3}/3+t^{4}/4]_{-1}^{0}+[t^{3}/3-t^{% 4}/4]_{0}^{1}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}-(-1/3+1/4)+(1/3-1/4)=1/6.}$

So ${\operatorname{\mathsf{Var}}}\left[{X}\right]=1/6$ .
(c)

By linearity, ${\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{E}}\left[{(2X+1)^{2}}\right]=4% \operatorname{\mathsf{E}}\left[{X^{2}}\right]+4\operatorname{\mathsf{E}}\left[% {X}\right]+1=4/6+1=5/3}$ .
(d)

$|t|=-t$ when $t\leq 0$ and $|t|=t$ when $t\geq 0$ , so

$\displaystyle\operatorname{\mathsf{E}}\left[{|X|}\right]$ $\displaystyle={\color[rgb]{0.76,0.01,0}\int_{-1}^{0}-t(1+t)\,\mathrm{d}t+\int_% {0}^{1}t(1-t)\,\mathrm{d}x}$

$\displaystyle={\color[rgb]{0.76,0.01,0}-[t^{2}/2+t^{3}/3]_{-1}^{0}+[t^{2}/2-t^% {3}/3]_{0}^{1}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}(1/2-1/3)+(1/2-1/3)=1/3.}$

Example 2.6.3.

A random variable $X$ has a pdf of

\displaystyle f_{X}(x)=\left\{\begin{array}[]{ll}1/x^{2}&\quad 1<x<\infty\\ 0&\quad\text{otherwise}\end{array}\right.

Find $\operatorname{\mathsf{E}}\left[{X}\right]$ .

Solution.

	$\displaystyle\operatorname{\mathsf{E}}\left[{X}\right]$	$\displaystyle=\int_{1}^{\infty}t\times\frac{1}{t^{2}}\,\mathrm{d}t=\lim_{r% \rightarrow\infty}{\color[rgb]{0.76,0.01,0}\int_{1}^{r}\frac{1}{t}\,\mathrm{d}t}$
		$\displaystyle=\lim_{r\rightarrow\infty}{\color[rgb]{0.76,0.01,0}[\log t]_{1}^{% r}=\infty.}$

The expectation is infinity! An example where the expectation is not even well defined is given in Chapter 3.

	$\displaystyle\operatorname{\mathsf{E}}\left[{X}\right]$	$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{-1}^{0}t(1+t)\,\mathrm{d}t+\int_{% 0}^{1}t(1-t)\,\mathrm{d}t}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}[t^{2}/2+t^{3}/3]_{-1}^{0}+[t^{2}/2-t^{% 3}/3]_{0}^{1}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}-(1/2-1/3)+(1/2-1/3)=0,}$

	$\displaystyle\operatorname{\mathsf{E}}\left[{X^{2}}\right]$	$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{-1}^{0}t^{2}(1+t)\,\mathrm{d}t+% \int_{0}^{1}t^{2}(1-t)\,\mathrm{d}t}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}[t^{3}/3+t^{4}/4]_{-1}^{0}+[t^{3}/3-t^{% 4}/4]_{0}^{1}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}-(-1/3+1/4)+(1/3-1/4)=1/6.}$

	$\displaystyle\operatorname{\mathsf{E}}\left[{\|X\|}\right]$	$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{-1}^{0}-t(1+t)\,\mathrm{d}t+\int_% {0}^{1}t(1-t)\,\mathrm{d}x}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}-[t^{2}/2+t^{3}/3]_{-1}^{0}+[t^{2}/2-t^% {3}/3]_{0}^{1}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}(1/2-1/3)+(1/2-1/3)=1/3.}$