- #1
Emil_M
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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
[itex]
\begin{align*}
\frac{1}{2} \sum_x |p_x-q_x| &\geq | \sum_{x \in S} p_x - \sum_{x \in S} q_x|\\
&= |\sum_{x \in S} p_x-q_x|
\end{align*}
[/itex]
Then, I have tried playing around with the triangle inequality, but that didn't go anywhere...
Thanks for you help!
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