Q.6.12 - A First Course in Probability

Question

Show that the jointly continuous (discrete) random variables X1, ... , Xn are independent if and only if their joint probability density (mass) function f(x1, ... , xn) can be written as f(x1, ... , xn) = n i=1 gi(xi) for nonnegative functions gi(x), i = 1, ... , n 

Step-by-Step Solution

Answer

X₁,.....X_n are independent only if $f(x_1,......x_n)= \prod _{i=1}^n{g}_i\left(x_i\right)$

Step 1 : Content Introduction

We need to proof X₁,.....X_n are independent only if $f(x_1,......x_n)= \prod _{i=1}^n{g}_i\left(x_i\right)$

Step 2: Content Explanation

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

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

Let us choose $B_1=(-\infty ,x_1),B_2(-\infty ,....x_2),.....,B_n=(-\infty ,x_n)$

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

The derivative with respect to each variable is

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

Letting $g_i\left(x_i\right)=f{x}_i\left(x_i\right)$ we then obtained  $f(x_1,....x_n)=\prod _{i=1}^n{g}_i\left(x_i\right)$

Step 3: Conclusion

Let $f(x_1,....x_n)=\prod _{i=1}^n{g}_i\left(x_i\right)$

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

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

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