Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mu_i,\nu_i$ be probability measure on a finite space $\Omega_i,i=1,2,\dots,n$. Define $\mu=\prod\limits_{i=1}^{n}\mu_i$ and $\nu=\prod\limits_{i=1}^{n}\nu_i$ on $\Omega=\prod\limits_{i=1}^{n}\Omega_i$, show that  $$\|\mu-\nu\| \le \sum\limits_{i=1}^{n}\|\mu_i-\nu_i\|$$ where $\|\mu-\nu\|$ denote the total variation distance between $\mu$ and $\nu$.  

I know how to do this using coupling, is there a way to do it without coupling?

I try to write $$\|\mu-\nu\|={1 \over 2}\sum\limits_{x=(x_1,x_2,\dots,x_n) \in \Omega}|\prod\limits_{i=1}^{n}\mu_i(x_i)-\prod\limits_{i=1}^{n}\nu_i(x_i)|$$

and use the fact that $\prod\limits_{i=1}^{n}\mu_i \le \sum\limits_{i=1}^{n}\mu_i$ and $\prod\limits_{i=1}^{n}\nu_i \le \sum\limits_{i=1}^{n}\nu_i$, but I didn't succeed.

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

This is a direct consequence of the fact that for every nonnegative $a_i$ and $b_i$, $$|(a_1\cdots a_n)-(b_1\cdots b_n)|\leqslant\sum\limits_{i=1}^n|a_i-b_i|\,(a_1\cdots a_{i-1})(b_{i+1}\cdots b_n). $$ Hence, $$ 2\|\mu-\nu\|=\sum\limits_x\left|\mu_1(x_1)\cdots\mu_n(x_n)-\nu_1(x_1)\cdots\nu_n(x_n)\right|\leqslant\sum\limits_{i=1}^n\Delta_i, $$ with $$ \Delta_i=\sum\limits_{x_i}|\mu_i(x_i)-\nu_i(x_i)|\,\sum\limits_{\widehat x_i}\mu_1(x_1)\cdots\mu_{i-1}(x_{i-1})\nu_{i+1}(x_{i+1})\cdots\nu_n(x_n), $$ where $\widehat x_i=(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)$. Each sum over $\widehat x_i$ is a product of masses of probability measures hence $$ \Delta_i=\sum\limits_{x_i}|\mu_i(x_i)-\nu_i(x_i)|=2\|\mu_i-\nu_i\|, $$ and you are done.

Edit The first inequality is a consequence of the triangular inequality between the numbers $(c_i)_{0\leqslant i\leqslant n}$ defined by $c_i=(a_1\cdots a_{i})(b_{i+1}\cdots b_n)$ for $1\leqslant i\leqslant n-1$, $c_0=b_1\cdots b_n$ and $c_n=a_1\cdots a_n$ since, for every $1\leqslant i\leqslant n$, $$ c_{i}-c_{i-1}=(a_{i}-b_{i})(a_1\cdots a_{i-1})(b_{i+1}\cdots b_n). $$

share|improve this answer
    
how can I proove the first inequality? By expanding a few case, say $n=2,n=3$, I see that it holds, but I can't prove it formally. –  Nicolas Essis-Breton Oct 13 '11 at 18:38
    
See edit. $ $ $ $ –  Did Oct 13 '11 at 20:45
    
I see, clever. Thanks for the clarification. –  Nicolas Essis-Breton Oct 14 '11 at 19:50
add comment

First, notice that the difference of two probability measures is a signed measure of total variance at most $2$.

Hence your problem reduces to showing that $\|\mu\|\leq\sum\limits_{i=1}^{n}\|\mu_i\|$ for signed measures $\mu_i$ on finite spaces $\Omega_i$ with $\|\mu_i\|\leq 2$.

I will show you how to do the case $n=2$.

Then we have to show that $$ \sum\limits_{x\in\Omega_1,y\in\Omega_2} \lvert\mu_1(x)\mu_2(y)\rvert\leq \sum\limits_{x\in\Omega_1} \lvert\mu_1(x)\rvert +\sum\limits_{y\in\Omega_2} \lvert\mu_2(y)\rvert. $$ But this inequality is equivalent to $$\|\mu_1\|\cdot\|\mu_2\|\leq\|\mu_1\|+\|\mu_2\|,$$ which is easily verified using $\|\mu_i\|\leq 2$:

Edit: The last inequality can be seen as follows: $$ \|\mu_1\|\cdot\|\mu_2\|\leq\|\mu_1\|^2/2+\|\mu_2\|^2/2\leq\|\mu_1\|+\|\mu_2\|. $$

share|improve this answer
    
thanks for this elegant argument. –  Nicolas Essis-Breton Oct 13 '11 at 17:45
    
why does the last inequality holds? I don't see how $\|\mu_i\| \le 2$ helps to prove it. –  Nicolas Essis-Breton Oct 13 '11 at 21:22
    
Please see my edit. It is a general fact that $ab\leq\frac{a^2+b^2}{2}$. This follows from the binomial formula. –  Rasmus Oct 13 '11 at 22:57
    
Thanks for the clarification. –  Nicolas Essis-Breton Oct 14 '11 at 0:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.