Trace distance between two probability distributions - prove

Click For Summary

Homework Help Overview

The discussion revolves around proving a relationship between two probability distributions, specifically focusing on the trace distance defined as \(D(p_x,q_x) = \frac{1}{2} \sum_x |p_x - q_x|\). Participants are tasked with demonstrating that this trace distance can be expressed as the maximum difference over all subsets of the distributions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the triangle inequality and the properties of probability distributions summing to one. There are attempts to manipulate the expressions involving the differences between the distributions.

Discussion Status

The discussion includes various attempts to understand the relationship between the trace distance and the maximum difference over subsets. Some participants have provided insights into the structure of the problem, while others are still exploring the implications of their findings. There is no explicit consensus yet, but the conversation appears to be productive.

Contextual Notes

Participants are working under the constraints of the definitions of probability distributions and the properties that arise from them, such as the sum of probabilities equating to one. There is an ongoing examination of how these properties influence the proof being sought.

Emil_M
Messages
45
Reaction score
2

Homework Statement


Let ##\{p_x\}## and ##\{q_x\}## be two probability distributions over the same index set ##\{x\}={1,2,...,N}##. Then, the trace distance between them is given by ##D(p_x,q_x):=\frac{1}{2} \sum_x |p_x-q_x|##.

Prove that ##D(p_x,q,_x)=max_S |p(S)-q(S)|=max_S | \sum_{x \in S} p_x - \sum_{x \in S} q_x|##, where the maximization is over all subsets ##S## of the index set ##\{x\}##.

Homework Equations


See above

The Attempt at a Solution


<br /> \begin{align*}<br /> \frac{1}{2} \sum_x |p_x-q_x| &amp;\geq | \sum_{x \in S} p_x - \sum_{x \in S} q_x|\\<br /> &amp;= |\sum_{x \in S} p_x-q_x|<br /> \end{align*}<br />

Then, I have tried playing around with the triangle inequality, but that didn't go anywhere...

Thanks for you help!
 
Last edited:
Physics news on Phys.org
The key piece of information here is that all the probabilities will sum to 1.
If p(1) - q(1) = z, then the sum of q(2) to q(N) minus the sum of p(2) to p(N) will also be z.
 
Emil_M said:

Homework Statement


Let ##\{p_x\}## and ##\{q_x\}## be two probability distributions over the same index set ##\{x\}={1,2,...,N}##. Then, the trace distance between them is given by ##D(p_x,q_x):=\frac{1}{2} \sum_x |p_x-q_x|##.

Prove that ##D(p_x,q,_x)=max_S |p(S)-q(S)|=max_S | \sum_{x \in S} p_x - \sum_{x \in S} q_x|##, where the maximization is over all subsets ##S## of the index set ##\{x\}##.

Homework Equations


See above

The Attempt at a Solution


<br /> \begin{align*}<br /> \frac{1}{2} \sum_x |p_x-q_x| &amp;\geq | \sum_{x \in S} p_x - \sum_{x \in S} q_x|\\<br /> &amp;= |\sum_{x \in S} p_x-q_x|<br /> \end{align*}<br />

Then, I have tried playing around with the triangle inequality, but that didn't go anywhere...

Thanks for you help!

If you let ##p_x - q_x = r_x##, the ##r_x## sum to zero. We can re-write ##D(p_x,q_x)## as ##(1/2) \sum_x |r_x| ## and ##p(S) - q(S) = \sum_{x \in S} r_x##.

We can write
\sum_{x \in S} r_x = \underbrace{\sum_{x \in S, r_x &gt; 0} r_x }_ {\geq 0} +<br /> \underbrace{\sum_{x \in S, r_x &lt; 0} r_x }_ {\leq 0}
Think about what properties ##S## must have in order that the absolute value of the above sum be maximal, say at the subset ##S = S_0##.
 
Last edited:
Thanks for your help! I get it now
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
6
Views
4K
Replies
4
Views
3K