Examples

Expectation of function on RV given by chart

Suppose that $g:ℝ\to ℝ$ in such a way that $g:1\mapsto 4$ and $g:2\mapsto 1$ and $g:3\mapsto 87$ and no other values are mapped to $4,1,87$. Define $Y=g(X)$.

$X:$	1	2	3
$P_X(k):$	$1/7$	$2/7$	$4/7$
$Y:$	4	1	87

Then:

$$E[X]=1\cdot \frac{1}{7}+2\cdot \frac{2}{7}+3\cdot \frac{4}{7}\gg \gg \frac{17}{7}$$

And:

$$E[Y]=4\cdot \frac{1}{7}+1\cdot \frac{2}{7}+87\cdot \frac{4}{7}\gg \gg \frac{354}{7}$$

Therefore:

$$E[5X+2Y+3]\gg \gg 5\cdot \frac{17}{7}+2\cdot \frac{354}{7}+3\gg \gg \frac{814}{7}$$

PMF through many-to-one function

Suppose the PMF of $X$ is given by:

$$P_X(x)=\{\begin{array}{cc}0.2& x=-1\\ 0.4& x=0\\ 0.4& x=+1\\ 0& \text{else}\end{array}$$

Now suppose $Y=|X|$. That is to say, $Y=g(X)$ and $g(X)=|X|$.

What is the PMF of $Y$?

Solution

Notice that $g(+1)=1$ and $g(-1)=1$. So $P_X(+1)$ and $P_X(-1)$ are combined into $P_Y(1)$:

$$P_Y(y)=\{\begin{array}{cc}0.4& y=0\\ 0.6& y=1\\ 0& \text{else}\end{array}$$

Variance of uniform random variable

The uniform random variable $X$ on $[a,b]$ has distribution given by $P[c\le X\le d]={\displaystyle \frac{d-c}{b-a}}$ when $a\le c\le d\le b$.

(a) Find $\mathrm{Var}[X]$ using the shorter formula.

(b) Find $\mathrm{Var}[3X]$ using “squaring the scale factor.”

(c) Find $\mathrm{Var}[3X]$ directly.

Solution (a)

(1) Compute density.

The density for $X$ is:

$$f_X(x)=\{\begin{array}{cc}{\displaystyle \frac{1}{b-a}}& \text{for }x\in [a,b]\\ 0& \text{otherwise}\end{array}$$

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(2) Compute $E[X]$ and $E[X^2]$ directly using integral formulas.

Compute $E[X]$:

$$E[X]={\int }_a^b\frac{x}{b-a}dx\gg \gg \frac{b+a}{2}$$

Now compute $E[X^2]$:

$$E[X^2]={\int }_a^b\frac{x^2}{b-a}dx\gg \gg \frac{1}{3}(b^2+ba+a^2)$$

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(3) Find variance using short formula.

Plug in:

$$\begin{array}{c}\mathrm{Var}[X]=E[X^2]-E{[X]}^2\\ \\ \gg \gg \frac{1}{3}(b^2+ab+a^2)-{\left(\frac{b+a}{2}\right)}^2\\ \\ \gg \gg \frac{{(b-a)}^2}{12}\end{array}$$

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(b)

(1) “Squaring the scale factor” formula:

$$\mathrm{Var}[aX+b]=a^2\mathrm{Var}[X]$$

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(2) Plugging in:

$$\mathrm{Var}[3X]\gg \gg 9\mathrm{Var}[X]\gg \gg \frac{9}{12}{(b-a)}^2$$

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(c)

(1) Density.

The variable $3X$ will have $1/3$ the density spread over the interval $[3a,3b]$.

Density is then:

$$f_{3X}(x)=\{\begin{array}{cc}{\displaystyle \frac{1}{3b-3a}}& \text{on }[3a,3b]\\ 0& \text{otherwise}\end{array}$$

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(2) Plug into prior variance formula.

Use $a⇝3a$ and $b⇝3b$.

Get variance:

$$\mathrm{Var}[3X]=\frac{{(3b-3a)}^2}{12}$$

Simplify:

$$\gg \gg \frac{{(3(b-a))}^2}{12}\gg \gg \frac{9}{12}{(b-a)}^2$$

Average pay raise

Suppose the average salary at Company A is $52,000. Each employee is given a 3% raise and a $2000 bonus. What is the average salary now?

Solution

Let $X$ be a random variable indicating the starting salary of each employee.

Then $Y=g(X)=(1.03)X+2000$ is a random variable giving the new salary of each employee.

We can calculate the expectation by linearity:

$$\begin{array}{c}E[Y]=E[g(X)]\gg \gg E[(1.03)X+2000]\\ \\ \gg \gg (1.03)E[X]+2000\gg \gg (1.03)(52000)+2000\gg \gg 55560\end{array}$$

PDF of derived from CDF

Suppose that $F_X(x)={\displaystyle \frac{1}{1+e^{-x}}}$.

(a) Find the PDF of $X$. $$ (b) Find the PDF of $e^X$.

Solution

(a)

Formula:

$$F_X(x)={\int }_{-\infty }^x{f}_X(t)dt⟹f_X(x)=\frac{d}{dx}{F}_X(x)$$

Plug in:

$$\begin{array}{c}{f}_X(x)=\frac{d}{dx}(1+e^{-x}{)}^{-1}\gg \gg -(1+e^{-x}{)}^{-2}\cdot (-e^{-x})\\ \\ \gg \gg \frac{e^{-x}}{{(1+e^{-x})}^2}\end{array}$$

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(b)

By definition:

$$F_{e^X}(x)=P[e^X\le x]$$

Since $e^X$ is increasing, we know:

$$e^X\le a⟺X\le \ln a$$

Therefore:

$$\begin{array}{c}{F}_{e^X}(x)=F_X(\ln x)\\ \\ \gg \gg \frac{1}{1+e^{-\ln x}}\gg \gg \frac{1}{1+x^{-1}}\end{array}$$

Then using differentiation:

$$\begin{array}{c}{f}_{e^X}(x)=\frac{d}{dx}\left(\frac{1}{1+x^{-1}}\right)\\ \\ \gg \gg -{(1+x^{-1})}^{-2}\cdot (-x^{-2})\gg \gg \frac{1}{{(x+1)}^2}\end{array}$$

Probabilities via CDF

Suppose the CDF of $X$ is given by $F_X(x)={\displaystyle \frac{1}{1+e^{-x}}}$. Compute:

(a) $P[X\le 1]$ $$ (b) $P[X<1]$ $$ (c) $P[-0.5\le X\le 0.2]$ $$ (d) $P[-2\le X]$

Solution

(a)

$$F_X(1)=\frac{1}{1+e^{-1}}\gg \gg \approx 0.731$$

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(b) Same as (a) because $P[X=1]=0$ (single point in a continuous distribution).

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(c)

$$F_X(0.2)-F_X(-0.5)\gg \gg \frac{1}{1+e^{-0.2}}-\frac{1}{1+e^{0.5}}\gg \gg \approx 0.172$$

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(d)

$$1-F_X(-2)\gg \gg 1-\frac{1}{1+e^2}\gg \gg \approx 0.881$$