I have a distribution $P$ and I have all the distributions $Q$s that are at most d distance from $P$. Let's assume the convex combination of all those $Q$s are $S$. Then what would be the maximum distance between $P$ and $S$? Here, by distance I assumed Total Variation Distance. My intuition tells me the maximum distance should be d, but I can't get a formal proof.


1 Answer 1


Fix $\epsilon > 0$ then there exists some $A\in\mathcal{F}$ so that

$$\begin{align} d(P,\sum_i \lambda_i Q_i) &< \left|\; P(A) - \sum_i \lambda_i Q_i(A) \;\right| + \epsilon\\\\ &= \left|\; \sum_i \lambda_i (P(A) - Q_i(A)) \;\right| + \epsilon\\\\ &\leq \sum_i \lambda_i |P(A) - Q_i(A)| + \epsilon\\\\ &\leq \sum_i \lambda_i d(P, Q_i) + \epsilon\\\\ &\leq \sum_i \lambda_i d + \epsilon = d + \epsilon \end{align}$$

As this holds for any $\epsilon > 0$, let $\epsilon \to 0$, so

$$d(P,\sum_i \lambda_i Q_i) \leq d$$


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