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<record version="5" id="573">
 <title>joint discrete density function</title>
 <name>JointDiscreteDensityFunction</name>
 <created>2001-10-26 19:32:52</created>
 <modified>2004-03-03 04:40:34</modified>
 <type>Definition</type>
 <creator id="2727" name="mathcam"/>
 <author id="2760" name="yark"/>
 <author id="23" name="Riemann"/>
 <classification>
	<category scheme="msc" code="60E05"/>
 </classification>
 <synonyms>
	<synonym concept="joint discrete density function" alias="joint probability function"/>
	<synonym concept="joint discrete density function" alias="joint distribution"/>
 </synonyms>
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 <content>\PMlinkescapeword{continuous}
\PMlinkescapeword{difference}
\PMlinkescapeword{satisfies}

Let $X_1, X_2, ..., X_n$ be $n$ random variables all defined on the same probability space. The \textbf{joint discrete density function} of $X_1, X_2, ..., X_n$, denoted by $f_{X_1, X_2, ..., X_n}(x_1,x_2,...,x_n)$, is the following function:\\
\par
$f_{X_1, X_2, ..., X_n}: R^n \to R$\\
$f_{X_1, X_2, ..., X_n}(x_1,x_2,...,x_n) = P[X_1 = x_1, X_2 = x_2, ... , X_n = x_n]$\\
\par
As in the single variable case, sometimes it's expressed as $p_{X_1, X_2, ..., X_n}(x_1,x_2,...,x_n)$ to mark the difference between this function and the continuous joint density function.\\
\par
Also, as in the case where $n=1$, this function satisfies:\\
\par
\begin{enumerate}
\item $f_{X_1, X_2, ..., X_n}(x_1,...,x_n) \geq  0$    $\forall (x_1,...,x_n)$
\item $\sum_{x_1, ... ,x_n}^{} {  f_{X_1, X_2, ..., X_n}(x_1,...,x_n) }= 1$
\end{enumerate}
\par
In this case, $f_{X_1, X_2, ..., X_n}(x_1,...,x_n) = P[ X_1 = x_1, X_2 = x_2, ... , X_n = x_n ]$.</content>
</record>
