Joint Probability Distribution

In some experiment, we might want to study simultaneous outcomes of several random variables.
If and are two discrete random variables, the probability distribution for their simultaneous occurrence can be represented by a function with values
Definition 3.8:
$\fbox{\parbox{5in}{ The function $f(x ,y)$ is a \textbf{joint probability distr... ...-plane, \begin{displaymath} P[(X,Y)\in A]=\sum \sum_A f(x,y) \end{displaymath}}}$

Example 3.14: Two refills for a ballpoint pen are selected at random from a box that contains 3 blue refills, 2 red refills, and 3 green refills. If is the number of blue refills and is the number of red refills selected, find
the joint probability function

$\displaystyle f(x,y)=\frac{ \left( \begin{array}{l} 3 \\ x \\ \end{array}... ...{array}\right) }{ \left( \begin{array}{l} 8 \\ 2 \\ \end{array}\right) }$
$P[(X,Y) \in A]$ , where is the region $\{(x,y)\vert x+y \leq 1\}$ .

$\displaystyle P[(X,Y) \in A]=P(X+Y \leq 1)$

$\displaystyle =f(0,0)+f(0,1)+f(1,0)$

$\displaystyle =\frac{3}{28}+\frac{3}{14}+\frac{9}{28}=\frac{9}{14}$

			x		Row
		0	1	2	Totals
	0	$\frac{3}{28}$	$\frac{9}{28}$	$\frac{3}{28}$	$\frac{15}{28}$

y	1	$\frac{3}{14}$	$\frac{3}{14}$		$\frac{3}{7}$

	2	$\frac{1}{28}$			$\frac{1}{28}$

Column		$\frac{5}{14}$	$\frac{15}{28}$	$\frac{3}{28}$	1
Totals

Definition 3.9:
$\fbox{\parbox{5in}{ The function $f(x ,y)$\ is a \textbf{joint density function}... ...nt \int_A f(x,y) dx dy$ \end{enumerate}For any region $A$\ in the $xy$-plane, }}$
Example 3.15: A candy company distributes boxes of chocolates with a mixture of creams, toffees, and nuts coated in both light and dark chocolate.
For randomly selected box, let and , respectively, be the proportions of the light and dark chocolates that are creams.
The joint density function is as follows:

$\displaystyle f(x,y)= \left\lbrace \begin{array}{l} \frac{2}{5}(2x+3y),~0\leq x \leq 1,~ 0 \leq y \leq 1 \\ 0, ~elsewhere\\ \end{array}\right\rbrace$

Verify $\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) dx dy =1$
$\fbox{\parbox{5in}{ \begin{small} \begin{displaymath} \int_{-\infty}^\infty \int... ...th}\begin{displaymath} =\frac{2}{5}+\frac{3}{5}=1 \end{displaymath}\end{small}}}$
$P[(X,Y) \in A]$ , where is the region ${(x,y) \vert 0 < x < \frac{1}{2}, \frac{1}{4} < y < \frac{1}{2}}$ ,
$\fbox{\parbox{5in}{ \begin{small} \begin{displaymath} P[(X,Y) \in A]=P(0 < X < \... ...extbar_{\frac{1}{4}}^{\frac{1}{2}}=\frac{13}{160} \end{displaymath}\end{small}}}$

Definition 3.10:
$\fbox{\parbox{5in}{ The \textbf{marginal distributions} of $X$ alone and of $Y$... ...h(y)=\int_{-\infty}^\infty f(x,y)dx \end{displaymath}for the continuous case }}$

Example 3.16: Show that the column and row totals of the following table give the marginal distribution of alone and of alone.

			x		Row
		0	1	2	Totals
	0	$\frac{3}{28}$	$\frac{9}{28}$	$\frac{3}{28}$	$\frac{15}{28}$

y	1	$\frac{3}{14}$	$\frac{3}{14}$		$\frac{3}{7}$

	2	$\frac{1}{28}$			$\frac{1}{28}$

Column		$\frac{5}{14}$	$\frac{15}{28}$	$\frac{3}{28}$	1
Totals

Solution:

$\displaystyle P(X=0)=g(0)=\sum_{y=0}^2 f(0,y)=f(0,0)+f(0,1)+f(0,2)$

$\displaystyle =\frac{3}{28}+\frac{3}{14}+\frac{1}{28}=\frac{5}{14}$

$\displaystyle P(X=1)=g(1)=\sum_{y=0}^2 f(1,y)=f(1,0)+f(1,1)+f(1,2)$

$\displaystyle =\frac{9}{28}+\frac{3}{14}+0=\frac{15}{28}$

$\displaystyle P(X=2)=g(2)=\sum_{y=0}^2 f(2,y)=f(2,0)+f(2,1)+f(2,2)$

$\displaystyle =\frac{3}{28}+0+0=\frac{3}{28}$

x	0	1	2
g(x)	$\frac{5}{14}$	$\frac{15}{28}$	$\frac{3}{28}$

Example 3.17: Find and for the following joint density function.

$\displaystyle f(x,y)= \left\lbrace \begin{array}{l} \frac{2}{5}(2x+3y),~0\leq x \leq 1,~ 0 \leq y \leq 1 \\ 0, ~elsewhere\\ \end{array}\right\rbrace$
$\fbox{\parbox{5in}{ \begin{small}\begin{displaymath} =\int_{-\infty}^\infty f(x,... ...th}for $~0\leq x \leq 1,~ 0 \leq y \leq 1$\ and $g(x)=0$, elsewhere\end{small}}}$
$\fbox{\parbox{5in}{ \begin{small}\begin{displaymath} =\int_{-\infty}^\infty f(x,... ...th}for $~0\leq x \leq 1,~ 0 \leq y \leq 1$\ and $h(y)=0$, elsewhere\end{small}}}$

Definition 3.11:
$\fbox{\parbox{5in}{ Let $X$\ and $Y$\ be two random variables, \underline{discre... ... \begin{displaymath} f(x\vert y)=\frac{f(x,y)}{h(y)}, h(y)>0 \end{displaymath}}}$
Evaluate the probability that falls between and given that is known.

$\displaystyle P(a<X<b\vert Y=y)=\sum_x f(x\vert y), for the discrete case$

$\displaystyle P(a<X<b\vert Y=y)=\int_a^b f(x\vert y),~for~the~continuous~case$

Example 3.18: Referring to Example 3.14, find the conditional distribution of , given that , and use it to determine $P(X = 0\vert Y = 1)$ .
Solution:

$\displaystyle h(y=1)=\sum_{x=0}^2 f(x,1)=\frac{3}{14}+\frac{3}{14}+0=\frac{3}{7}$

$\displaystyle f(x\vert 1)=\frac{f(x,1)}{h(1)}=\frac{7}{3}f(x,1),x=0,1,2$

$\displaystyle f(0\vert 1)=\frac{7}{3}f(0,1)=\frac{7}{3}*\frac{3}{14}=\frac{1}{2}$

$\displaystyle f(1\vert 1)=\frac{7}{3}f(1,1)=\frac{7}{3}*\frac{3}{14}=\frac{1}{2}$

$\displaystyle f(2\vert 1)=\frac{7}{3}f(2,1)=\frac{7}{3}*0=0$

$\displaystyle \Longrightarrow P(X=0\vert Y=1)=f(0\vert 1)=\frac{1}{2}$

x 0 1 2

f(x|1) $\frac{1}{2}$ $\frac{1}{2}$ 0

Example 3.19: The joint density for the random variables , where is the unit temperature change and is the proportion of spectrum shift that a certain atomic particle produces is

$\displaystyle f(x,y)= \left\lbrace \begin{array}{l} 10xy^2, 0< x<y<1 \\ 0, elsewhere\\ \end{array}\right\rbrace$
Find the marginal densities , and the conditional density $f(y\vert x)$ .
$\fbox{\parbox{5in}{ \begin{displaymath} g(x)=\int_{-\infty}^\infty f(x,y)dy=\int... ...x)}=\frac{10xy^2}{\frac{10x(1-x^3)}{3}}=\frac{3y^2}{(1-x^3)} \end{displaymath}}}$
Find the probability that the spectrum shifts more than half of the total observations, given the temperature is increased to 0.25 unit.
$\fbox{\parbox{5in}{ \begin{displaymath} P(Y>\frac{1}{2}\vert X=0.25)=\int_{1/2}^... ....25)dy=\int_{1/2}^1 \frac{3y^2}{(1-0.25^3)}dy=\frac{8}{9} \end{displaymath}\ }}$

Definition 3.12:
$\fbox{\parbox{5in}{ Let $X$ and $Y$ be two random variables, discrete or conti... ...playmath} f(x,y)=g(x)h(y), for all (x,y) within their range \end{displaymath}}}$
Example 3.21: Show that the random variables of Example 3.14 are not statistically independent.

$\displaystyle f(0,1)=\frac{3}{14},g(0)=\sum_{y=0}^2 f(0,y)=\frac{5}{14}, h(1)=\sum_{x=0}^2 f(x,1)=\frac{3}{7}$

$\displaystyle \Longrightarrow f(0,1) \neq g(0)*h(1)$
therefore and are not statistically independent.

Example: In a binary communications channel, let denote the bit sent by the transmitter and let denote the bit received at the other end of the channel. Due to noise in the channel we do not always have . A joint probability distribution is given as

x

0 1

y 0 0.45 0.03 0.48

1 0.05 0.47 0.52

0.5 0.5

x

0 1

y 0 f(0,0) f(1,0) h(0)

1 f(0,1) f(1,1) h(1)

g(0) g(1)
and are not independent because

$\displaystyle f(0,0) \neq g(0)h(0) \Longrightarrow 0.45 \neq 0.5*0.48$
$P(X=x,Y=y)=P[(X=x) \cap (Y=y)]$ : it is the probability that and simultaneously.
$f(0,0)=P(X=0,Y=0)=P[(X=0) \cap (Y=0)]$
So

$\displaystyle =P[(X=0) \cap (Y=0)]+ P[(X=0) \cap (Y=1)]=f(0,0)+f(0,1)$
$\Longrightarrow P[Y=0 \vert X=0]=\frac{P[(X=0) \cap (Y=0)]}{P[X=0]}=\frac{f(0,0)}{g(0)}$

Sent 0 & Received 0: NO error.

$\displaystyle P[Y=0 \vert X=0]=\frac{f(0,0)}{g(0)}=\frac{0.45}{0.9}=0.9$
Sent 1 & Received 0: ERROR

$\displaystyle P[Y=0 \vert X=1]=\frac{f(1,0)}{g(1)}=\frac{0.03}{0.5}=0.06$
Sent 0 & Received 1: ERROR

$\displaystyle P[Y=1 \vert X=0]=\frac{f(0,1)}{g(0)}=\frac{0.05}{0.5}=0.1$
Sent 1 & Received 1: NO error.

$\displaystyle P[Y=1 \vert X=1]=\frac{f(1,1)}{g(1)}=\frac{0.47}{0.5}=0.94$
Notice that

$\displaystyle P[Y=0 \vert X=0]+P[Y=1 \vert X=0]=1$

$\displaystyle P[Y=0 \vert X=1]+P[Y=1 \vert X=1]=1$

Definition 3.13:
$\fbox{\parbox{5in}{ Let $X_1, X_2,\ldots ,X_n$\ be $n$\ random variables, discre... ...(x_n) \end{displaymath}for all $(x_1, x_2 ,\ldots ,x_n)$\ within their range. }}$

Example 3.22: Suppose that the shelf life, in years, of a certain perishable food product packaged in cardboard containers is a random variable whose probability density function is given by

$\displaystyle f(x,y)= \left\lbrace \begin{array}{l} e^{-x},~x>0 \\ 0, ~elsewhere\\ \end{array}\right\rbrace$
Let $X_1, X_2,\ldots ,X_n$ represent the shelf lives for three of these containers selected independently and find
Solution:

$\displaystyle f(x_1,x_2,x_3)=f(x_1)f(x_2)f(x_3)=e^{-x_1-x_2-x_3}$
for and elsewhere

$\displaystyle P(X_1<2, 1< X_2<3, X_3>2)=\int_2^\infty \int_1^3 \int_0^2 e^{-x_1-x_2-x_3}dx_1dx_2dx_3$

$\displaystyle =(1-e^{-2})(e^{-1}-e^{-3})e^{-2}=0.0372$

Cem Ozdogan 2012-02-15

x	0	1	2
f(x\|1)	$\frac{1}{2}$	$\frac{1}{2}$	0

		x
		0	1
y	0	0.45	0.03	0.48
	1	0.05	0.47	0.52
		0.5	0.5

		x
		0	1
y	0	f(0,0)	f(1,0)	h(0)
	1	f(0,1)	f(1,1)	h(1)
		g(0)	g(1)