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5.4 Continuous Random Variables

If $X$ and $Y$ are both continuous random variables their joint probability density function (pdf) is defined from

\displaystyle F_{XY}(x,y)=\int_{-\infty}^{y}\int_{-\infty}^{x}f_{XY}(s,t)\,% \mathrm{d}s\,\mathrm{d}t=\int_{-\infty}^{x}\int_{-\infty}^{y}f_{XY}(s,t)\,% \mathrm{d}t\,\mathrm{d}s.

Equivalently

\displaystyle f_{XY}(x,y)=\frac{\partial^{2}F_{XY}(x,y)}{\partial x\partial y}% =\frac{\partial^{2}F_{XY}(x,y)}{\partial y\partial x}.

For simplicity, we usually only state $F_{XY}(x,y)$ for values of $(x,y)$ such that $f_{XY}(x,y)>0$ . So if $F_{XY}$ is not defined for a particular $(x,y)$ pair, then $f_{XY}(x,y)=0$ at that point.

Properties of $f_{XY}(x,y)$ :

1.

Positivity: $f_{XY}(x,y)\geq 0$ for all $(x,y)$ ,
2.

Summability: $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(s,t)\,\mathrm{d}s\,% \mathrm{d}t=1$ .
3.

Just as for a univariate random variable, we can find the probability of event $A$ , i.e. $\operatorname{\mathsf{P}}\left({(X,Y)\in A}\right)$ , by integrating the pdf over the event $A$ :

$\displaystyle\operatorname{\mathsf{P}}\left({(X,Y)\in A}\right)=\int\int_{A}f_% {XY}(s,t)\,\mathrm{d}s\,\mathrm{d}t.$

For a univariate rv $X$ it is sometimes helpful to think the probability that it is in a region $A$ as the area under the density curve.

For a bivariate rv $(X,Y)$ it is sometimes helpful to think the probability that it is in a region $A$ as the volume under the density surface.

In particular

\displaystyle\operatorname{\mathsf{P}}\left({X\in[x,x+\delta x],Y\in[y,y+% \delta y]}\right)\approx f_{XY}(x,y)\delta x\delta y.

Example 5.4.1.

The random variables $(X,Y)$ have joint distribution function

\displaystyle F_{XY}(x,y)=\frac{x^{2}y+y^{2}x}{16}

for $0<x<2$ , $0<y<2$ . Recall that, for simplicity of presentation, we only specify the cdf where the density is non-zero. Obtain:

(a)

the joint pdf,
(b)

$\operatorname{\mathsf{P}}\left({X<1,Y<1}\right)$ ,
(c)

$\operatorname{\mathsf{P}}\left({X<1}\right)$ ,
(d)

$\operatorname{\mathsf{P}}\left({X^{2}+Y^{2}\leq 1}\right)$ ,
(e)

$\operatorname{\mathsf{P}}\left({X>Y}\right)$ ,

and explain how you could have obtained the answer to (e) without any calculation.

Solution.

(a)

The joint pdf is

$\displaystyle f_{XY}(x,y)$ $\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\partial^{2}}{\partial x\partial y% }F_{XY}(x,y)}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{16}\frac{\partial^{2}}{% \partial x\partial y}\left(x^{2}y+y^{2}x\right)}$

for $0<x<2$ , $0<y<2$ . So

$\displaystyle f_{XY}(x,y)=\left\{\begin{array}[]{ll}{\color[rgb]{0.76,0.01,0}(% x+y)/8}&\quad 0<x<2,\ 0<y<2\\ {\color[rgb]{0.76,0.01,0}0}&\quad\text{otherwise}\end{array}\right.$

(b)

$\operatorname{\mathsf{P}}\left({X<1,Y<1}\right)$ , two approaches: cdf and pdf.

cdf: $\operatorname{\mathsf{P}}\left({X<1,Y<1}\right)={\color[rgb]{0.76,0.01,0}F_{XY% }(1,1)=(1^{2}\times 1+1^{2}\times 1)/16=1/8.}$

pdf: (draw picture)

	$\displaystyle\operatorname{\mathsf{P}}\left({X<1,Y<1}\right)$	$\displaystyle=\int_{t=-\infty}^{1}\int_{s=-\infty}^{1}f_{XY}(s,t)\,\mathrm{d}s% \,\mathrm{d}t$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{t=0}^{1}\int_{s=0}^{1}% {s+t}\,\mathrm{d}s\,\mathrm{d}t}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{t=0}^{1}\left[\frac{1}% {2}s^{2}+st\right]_{s=0}^{1}\,\mathrm{d}t}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{t=0}^{1}\frac{1}{2}+t% \,\mathrm{d}t}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\left[\frac{1}{2}(t+t^{2})% \right]_{t=0}^{1}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{8}}$

(c)

$\operatorname{\mathsf{P}}\left({X<1}\right)$ , two approaches: cdf and pdf.

cdf: because $Y<2$ ,

$\displaystyle\operatorname{\mathsf{P}}\left({X<1}\right)$ $\displaystyle=F_{X,Y}(1,\infty)=F_{X,Y}(1,2)$

$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1^{2}\times 2+2^{2}\times 1}{16}=% \frac{3}{8}}$

pdf: (draw picture) Using calculations from part (b),

$\displaystyle\operatorname{\mathsf{P}}\left({X<1}\right)$ $\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{t=0}^{2}\int_{s=0}^{1}% s+t\,\mathrm{d}s\,\mathrm{d}t}$

$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{t=0}^{2}\frac{1}{2}+t% \,\mathrm{d}t}$

$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\left[\frac{1}{2}(t+t^{2})% \right]_{t=0}^{2}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{3}{8}.}$

(d)

$\operatorname{\mathsf{P}}\left({X^{2}+Y^{2}<1}\right)$ , pdf:

Unnumbered Figure: Link

	$\displaystyle\operatorname{\mathsf{P}}\left({X^{2}+Y^{2}<1}\right)$	$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{s=0}^{1}\int_{t=0}^{% \sqrt{1-s^{2}}}s+t\,\mathrm{d}t\,\mathrm{d}s}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{0}^{1}\left[st+\frac{1% }{2}t^{2}\right]_{t=0}^{\sqrt{1-s^{2}}}\,\mathrm{d}s}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{0}^{1}s\sqrt{1-s^{2}}+% \frac{1}{2}-\frac{1}{2}s^{2}\,\mathrm{d}s}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\left[-\frac{1}{3}(1-s^{2})^% {3/2}+\frac{1}{2}s-\frac{1}{6}s^{3}\right]_{0}^{1}}{\color[rgb]{0.76,0.01,0}% \text{by inspection}}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\left(-0+\frac{1}{2}-\frac{1% }{6}\right)-\left(-\frac{1}{3}+0-0\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{12}.}$

(e)

$\operatorname{\mathsf{P}}\left({X>Y}\right)$ , pdf:

$\displaystyle\operatorname{\mathsf{P}}\left({X>Y}\right)$ $\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{s=0}^{2}\int_{t=0}^{s}% s+t\,\mathrm{d}t\,\mathrm{d}s}$

$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{s=0}^{2}\left[st+\frac% {1}{2}t^{2}\right]_{t=0}^{s}\,\mathrm{d}x}$

$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{s=0}^{2}\frac{3}{2}s^{% 2}\,\mathrm{d}s}$

$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\left[\frac{1}{2}s^{3}\right% ]_{s=0}^{2}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}1/2.}$

Without integration? $\operatorname{\mathsf{P}}\left({Y=X=0}\right)$ , and by symmetry of ${\color[rgb]{0.76,0.01,0}f_{XY}(x,y)}$ about the ${\color[rgb]{0.76,0.01,0}x=y}$ line there is equal chance of ${\color[rgb]{0.76,0.01,0}X>Y}$ and ${\color[rgb]{0.76,0.01,0}Y>X}$ .

Example 5.4.2.

The joint distribution (in years) for the lifetimes $X$ and $Y$ of two computer components has joint pdf

\displaystyle f_{XY}(x,y)=\left\{\begin{array}[]{ll}\beta\exp(-\beta x)\exp(-y% )&\quad 0<x<\infty,\ 0<y<\infty\\ 0&\quad\text{ otherwise}\end{array}\right.

Find the probabilities of the following events:

(a)

Both components have lifetimes exceeding one year.
(b)

Component $Y$ has a longer lifetime than component $X$ .

Solution.

(a)

	$\displaystyle\operatorname{\mathsf{P}}\left({X>1,Y>1}\right)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{t=1}^{\infty}\int_{s=1}^{\infty}% \beta\exp(-\beta s)\exp(-t)\,\mathrm{d}s\,\mathrm{d}t}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{1}^{\infty}\left[-\exp(-\beta s)% \exp(-t)\right]_{s=1}^{\infty}\,\mathrm{d}t}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{1}^{\infty}\exp(-\beta)\exp(-t)\,% \mathrm{d}t}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\left[-\exp(-\beta)\exp(-t)\right]_{1}^% {\infty}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\exp(-\beta)\exp(-1)=\exp(-[\beta+1]).}$

Unnumbered Figure: Link

(b)

	$\displaystyle\operatorname{\mathsf{P}}\left({Y>X}\right)$	$\displaystyle=\int_{s=0}^{\infty}\int_{t=s}^{\infty}\beta\exp(-\beta s)\exp(-t% )\,\mathrm{d}t\,\mathrm{d}s$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{0}^{\infty}\left[-\beta\exp(-% \beta s)\exp(-t)\right]_{t=s}^{\infty}\,\mathrm{d}s}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{0}^{\infty}\beta\exp(-\beta s)% \exp(-s)\,\mathrm{d}s}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{0}^{\infty}\beta\exp(-[\beta+1]s)% \,\mathrm{d}s}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\beta}{\beta+1}.}$

	$\displaystyle f_{XY}(x,y)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\partial^{2}}{\partial x\partial y% }F_{XY}(x,y)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{16}\frac{\partial^{2}}{% \partial x\partial y}\left(x^{2}y+y^{2}x\right)}$

	$\displaystyle\operatorname{\mathsf{P}}\left({X<1}\right)$	$\displaystyle=F_{X,Y}(1,\infty)=F_{X,Y}(1,2)$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1^{2}\times 2+2^{2}\times 1}{16}=% \frac{3}{8}}$

	$\displaystyle\operatorname{\mathsf{P}}\left({X<1}\right)$	$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{t=0}^{2}\int_{s=0}^{1}% s+t\,\mathrm{d}s\,\mathrm{d}t}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{t=0}^{2}\frac{1}{2}+t% \,\mathrm{d}t}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\left[\frac{1}{2}(t+t^{2})% \right]_{t=0}^{2}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{3}{8}.}$

	$\displaystyle\operatorname{\mathsf{P}}\left({X>Y}\right)$	$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{s=0}^{2}\int_{t=0}^{s}% s+t\,\mathrm{d}t\,\mathrm{d}s}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{s=0}^{2}\left[st+\frac% {1}{2}t^{2}\right]_{t=0}^{s}\,\mathrm{d}x}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\int_{s=0}^{2}\frac{3}{2}s^{% 2}\,\mathrm{d}s}$
		$\displaystyle=\frac{1}{8}{\color[rgb]{0.76,0.01,0}\left[\frac{1}{2}s^{3}\right% ]_{s=0}^{2}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1/2.}$