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Thanks for your excellent work! It inspires me a lot.
In Eq.(22), you mentioned that by the Markov's inequality, $P(Z(i) > T/M') \ge \frac{\Delta(n,m) - T/M'}{1 - T/M'}$.
However, here is my derivation: $P(Z(i) > T/M') = P(1 - Z(i) < 1 - T/M')$. By Markov's inequality, $P(1 - Z(i) < 1 - T/M') \ge 1 - \frac{1 - E(Z_i)}{1 - T/M'} = \frac{E(Z_i) - T/M'}{1 - T/M'} $. By step 3, $E(Z_i) \ge \Delta(n, m) / M'$, so it seems that $\Delta(n,m)$ in Eq.(22) is missing a $M'$.
Is there anything wrong in my derivation? Thanks!!!!
The text was updated successfully, but these errors were encountered:
Thanks for your excellent work! It inspires me a lot.
In Eq.(22), you mentioned that by the Markov's inequality,$P(Z(i) > T/M') \ge \frac{\Delta(n,m) - T/M'}{1 - T/M'}$ .
However, here is my derivation:$P(Z(i) > T/M') = P(1 - Z(i) < 1 - T/M')$ . By Markov's inequality, $P(1 - Z(i) < 1 - T/M') \ge 1 - \frac{1 - E(Z_i)}{1 - T/M'} = \frac{E(Z_i) - T/M'}{1 - T/M'} $ . By step 3, $E(Z_i) \ge \Delta(n, m) / M'$ , so it seems that $\Delta(n,m)$ in Eq.(22) is missing a $M'$ .
Is there anything wrong in my derivation? Thanks!!!!
The text was updated successfully, but these errors were encountered: