W05 - Examples

Function on a random variable

Expectation of function on RV given by chart

Expectation of function on RV given by chart

Suppose that $g:\mathbb{R}\to \mathbb{R}$ 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:

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

Variance of uniform random variable

Variance of uniform random variable

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

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

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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:

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

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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:

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

PDF of derived from CDF

PDF of derived from CDF

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

(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:

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

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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:

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

Probabilities via CDF

Probabilities via CDF

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

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

Link to original

Continuous wait time

Earthquake wait time

Earthquake wait time

Suppose the San Andreas fault produces major earthquakes modeled by a Poisson process, with an average of 1 major earthquake every 100 years.

(a) What is the probability that there will not be a major earthquake in the next 20 years?

(b) What is the probability that three earthquakes will strike within the next 20 years?

Solution

(a)

Since the average wait time is 100 years, we set $\lambda =0.01$ earthquakes per year. Set $X\sim \mathrm{Exp}(0.01)$ and compute:

$$P[X>20]=e^{-\lambda \cdot 20}\gg \gg e^{-0.01\cdot 20}\gg \gg \approx 0.82$$

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(b)

The same Poisson process has the same $\lambda =0.01$ earthquakes per year. Set $X\sim \mathrm{Erlang}(3,0.01)$, so:

$$\begin{array}{c}{f}_X(t)=\frac{{\lambda }^{\ell }}{(\ell -1)!}{t}^{\ell -1}{e}^{-\lambda t}\\ \\ \gg \gg \frac{(0.01{)}^3}{(3-1)!}{t}^{3-1}{e}^{-0.01\cdot t}\gg \gg \frac{10^{-6}}{2}{t}^2{e}^{-0.01\cdot t}\end{array}$$

Now compute:

$$\begin{array}{c}P[X\le 20]={\int }_0^{20}{f}_X(x)dx\\ \\ \gg \gg {\int }_0^{20}\frac{10^{-6}}{2}{t}^2{e}^{-0.01\cdot t}dt\gg \gg \approx 0.00115\end{array}$$Link to original