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Revision difference : joint discrete density function |
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| 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:\\ |
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:\\ |
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| $f_{X_1, X_2, ..., X_n}: R^n \to R$\\ |
$f_{X_1, X_2, ..., X_n}: R^n \to R$\\ |
| $f_{X_1, X_2, ..., X_n}(x_1,x_2,...,x_n) = P[X_1 = x_1, X_2 = x_2, ... , X_n = x_n]$\\ |
$f_{X_1, X_2, ..., X_n}(x_1,x_2,...,x_n) = P[X_1 = x_1, X_2 = x_2, ... , X_n = x_n]$\\ |
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| 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.\\ |
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.\\ |
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| Also, as in the case where $n=1$, this function satisfies:\\ |
Also, as in the case where $n=1$, this function satisfies:\\ |
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| \begin{enumerate} |
\begin{enumerate} |
| \item $f_{X_1, X_2, ..., X_n}(x_1,...,x_n) \geq 0$ $\forall (x_1,...,x_n)$ |
\item $f_{X_1, X_2, ..., X_n}(x_1,...,x_n) \geq 0$ $\forall (x_1,...,x_n)$ |
| \item $\sum_{x_1, ... ,x_n}^{} { f_{X_1, X_2, ..., X_n}(x_1,...,x_n) }= 1$ |
\item $\sum_{x_1, ... ,x_n}^{} { f_{X_1, X_2, ..., X_n}(x_1,...,x_n) }= 1$ |
| \end{enumerate} |
\end{enumerate} |
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| In this case, $f_{X_1, X_2, ..., X_n}(x_1,...,x_n) = P[ X_1 = x_1, X_2 = x_2, ... , X_n = x_n ]$. |
In this case, $f_{X_1, X_2, ..., X_n}(x_1,...,x_n) = P[ X_1 = x_1, X_2 = x_2, ... , X_n = x_n ]$. |
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