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We need to proofX₁,.....X_nare independent only if$f(x_1,......x_n)=\underset{i=1}{\overset{n}{鈭�}}{g}_i\left(x_i\right)$

Let X₁,.....X_n be independent variables.

by the definition of independence $P(X_1鈭�B_1).......P(X_n......B_n)$ for all subsets B₁,....B_n of the real numbers.

Let us choose localid="1647605707823" $B_1=(-鈭�,x_1),B_2(-鈭�,....x_2),.....,B_n=(-鈭�,x_n)$

However, the cumulative distribution is defined as $F_x\left(x\right)=P(X鈮�x)$

The derivative with respect to each variable is

$$\begin{gathered} f(x_1,....x_n)=\frac{{未}^n}{未x_1....未x_n}F(x_1,....x_n) \\ =\frac{{未}^n}{未x_1}F{x}_1\left(x_1\right).......\frac{{未}^n}{未x_n}F{x}_n\left(x_n\right) \end{gathered}$$

Letting$g_i\left(x_i\right)=f{x}_i\left(x_i\right)$we then obtainedrole="math" localid="1647606023027" $f(x_1,....x_n)=\underset{i=1}{\overset{n}{鈭�}}{g}_i\left(x_i\right)$

Let $f(x_1,....x_n)=\underset{i=1}{\overset{n}{鈭�}}{g}_i\left(x_i\right)$

Let B₁, B₂, ......B_n be subsets of real numbers.

$$\begin{gathered} P(X_1鈭�B_1,........X_n鈭�B_n) \\ ={鈭�}_{B_1}.....{鈭�}_{B_n}f(x_1,....x_n)d{x}_n...d{x}_1 \\ ={鈭�}_{B_1}...{鈭�}_{B_n}\underset{i=1}{\overset{n}{鈭�}}{g}_i\left(x_i\right)d{x}_n....d{x}_1 \\ ={鈭�}_{B_1}{g}_1\left(x_1\right)d{x}_1........{鈭�}_{B_n}{g}_n\left(x_n\right)d{x}_n \\ =P(X_1鈭�B_1)....P(X_n鈭�B_n) \end{gathered}$$

We have prove that $X_1,.....X_n$ are independent variables.