joint continuous density function

Let $X_1,X_2,\mathrm{\dots },X_n$ be $n$ random variables all defined on the same probability space. The joint continuous density function of $X_1,X_2,\mathrm{\dots },X_n$, denoted by $f_{X_1,X_2,\mathrm{\dots },X_n}(x_1,x_2,\mathrm{\dots },x_n)$, is the function$f_{X_1,X_2,\mathrm{\dots },X_n}:{ℝ}^n\to ℝ$ such that for any domain $D\subset {ℝ}^n$, we have

${\displaystyle {\int }_D}{f}_{X_1,X_2,\mathrm{\dots },X_n}(u_1,u_2,\mathrm{\dots },u_n)d{u}_{1}d{u}_2\mathrm{\dots }d{u}_n=\text{Prob}(X_1,X_2,\mathrm{\dots },X_n\in D)$

As in the case where $n=1$, this function satisfies:

1. $f_{X_1,X_2,\mathrm{\dots },X_n}(x_1,\mathrm{\dots },x_n)\ge 0$ $\forall (x_1,\mathrm{\dots },x_n)$

2. ${\int }_{x_1,\mathrm{\dots },x_n}{f}_{X_1,X_2,\mathrm{\dots },X_n}(u_1,u_2,\mathrm{\dots },u_n)𝑑u_1𝑑u_2\mathrm{\dots }𝑑u_n=1$

As in the single variable case, $f_{X_1,X_2,\mathrm{\dots },X_n}$ does not represent the probability that each of the random variables takes on each of the values.