W08 - Examples

Functions on two random variables

PMF of XY squared from chart

PMF of XY squared from chart

Suppose the joint PMF of $X$ and $Y$ is given by this chart:

$Y\downarrow X\to$	$1$	$2$
$-1$	0.2	0.2
$0$	0.35	0.1
$1$	0.05	0.1

Define $W=X{Y}^2$.

(a) Find the PMF $P_W(w)$.

(b) Find the expectation $E[W]$.

Link to original

Max and Min from joint PDF

Max and Min from joint PDF

Suppose the joint PDF of $X$ and $Y$ is given by:

$$f_{X,Y}(x,y)=\{\begin{array}{ll}\frac{3}{2}(x^2+y^2)& x,y\in [0,1]\\ 0& \text{otherwise}\end{array}$$

Find the PDF of (a) $W=\mathrm{Max}(X,Y)$, and of (b) $W=\mathrm{Min}(X,Y)$.

Solution

(a)

(1) Compute CDF of $W$:

Convert to event form:

$$\begin{array}{c}{F}_W(w)=P[\mathrm{Max}(X,Y)\le w]\\ \\ \gg \gg P[X\le w\text{and}Y\le w]\end{array}$$

Integrate PDF over the region, assuming $w\in [0,1]$:

$$\begin{array}{c}{\int }_{-\infty }^w{\int }_{-\infty }^w{f}_{X,Y}(x,y)dxdy\\ \\ \gg \gg {\int }_0^w{\int }_0^w\frac{3}{2}(x^2+y^2)dxdy\gg \gg w^4\end{array}$$

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(2) Differentiate to find $f_W(w)$:

$f_W=\frac{d}{dw}{F}_W(w)$:

$$f_W(w)=\{\begin{array}{ll}4w^3& w\in [0,1]\\ 0& \text{otherwise}\end{array}$$

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(b)

(1) Compute CDF of $W$:

Convert to event form:

$$\begin{array}{c}{F}_W(w)=P[\mathrm{Min}(X,Y)\le w]\\ \\ \gg \gg 1-P[\mathrm{Min}(X,Y)>w]\\ \\ \gg \gg 1-P[X>w\text{and}Y>w]\end{array}$$

Integrate PDF over the region:

$$\begin{array}{c}P[X>w\text{and}Y>w]\gg \gg {\int }_w^1{\int }_w^1\frac{3}{2}(x^2+y^2)dxdy\\ \\ \gg \gg w^4-w^3-w+1\end{array}$$

Therefore:

$$F_W(w)=-w^4+w^3+w$$

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(2) Differentiate to find $f_W(w)$:

$f_W=\frac{d}{dw}{F}_W(w)$:

$$f_W(w)=\{\begin{array}{ll}-4w^3+3w^2+1& w\in [0,1]\\ 0& \text{otherwise}\end{array}$$ Link to original

PDF of a quotient

PDF of a quotient

Suppose the joint PDF of $X$ and $Y$ is given by:

$$f_{X,Y}(x,y)=\{\begin{array}{ll}\lambda \mu e^{-(\lambda x+\mu y)}& x,y\ge 0\\ 0& \text{otherwise}\end{array}$$

Find the PDF of $W=g(X,Y)$ for $g(X,Y)=Y/X$.

Solution

(1) Find the CDF using logic:

$$\begin{array}{c}{F}_W(w)=P[Y/X\le w]\\ \\ \gg \gg P[Y\le wX]\end{array}$$

Integrate over this region:

$$\begin{array}{rl}P[Y\le wX]& ={\int }_0^{\infty }{\int }_0^{wx}{f}_{X,Y}(x,y)dydx\\ & \\ & \gg \gg {\int }_0^{\infty }\lambda e^{-\lambda x}{\int }_0^{wx}\mu e^{-\mu y}dydx\\ & \\ & \gg \gg {\int }_0^{\infty }\lambda e^{-\lambda x}\left(-e^{-\mu wx}+1\right)dx\\ & \\ & \gg \gg 1-\frac{\lambda }{\lambda +\mu w}\end{array}$$

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(2) Differentiate to find PDF:

Compute $\frac{d}{dw}{F}_W(w)$:

$$f_W(w)=\{\begin{array}{ll}\frac{\lambda \mu }{(\lambda +\mu w{)}^2}& w\ge 0\\ & \\ 0& \text{otherwise}\end{array}$$Link to original