### add a macro for Var, and use it

parent defbcae6
 ... ... @@ -189,11 +189,11 @@ is $2$-independent). Hence $\E[X_{r, j}] = 1 / 2^r$. By linearity of expectation, $\E[Y_{r}] = d / 2^r$. We will also use the variance of these variables -- note that $${\rm Var}[X_{r, j}] \leq \E[X_{r, j}^2] = \E[X_{r, j}] = 1/2^r$$ $$\Var[X_{r, j}] \leq \E[X_{r, j}^2] = \E[X_{r, j}] = 1/2^r$$ And because $h$ is $2$-independent, the variables $X_{r, j}$ and $X_{r, j'}$ are independent for $j \neq j'$, and hence: $${{\rm Var}}[Y_{r}] = \sum_{j : f_j > 0} {\rm Var}[X_{r, j}] \leq d / 2^r$$ $$\Var[Y_{r}] = \sum_{j : f_j > 0} \Var[X_{r, j}] \leq d / 2^r$$ Now, let $a$ be the smallest integer such that $2^{a + 1/2} \geq 3d$. Then we have: ... ... @@ -208,7 +208,7 @@ $$\Pr[\hat{d} \leq d / 3] = \Pr[ Y_{b + 1} = 0] \leq \Pr[ \vert Y_{b + 1} - \E[Y_{b + 1}] \vert \geq d / 2^{b + 1} ]$$ Using Chebyshev's inequality, we get: $$\Pr[\hat{d} < d / 3] \leq {{\rm Var}[Y_b] \over (d / 2^{b + 1})^2} \leq$$ \Pr[\hat{d} < d / 3] \leq {\Var[Y_b] \over (d / 2^{b + 1})^2} \leq {2^{b + 1} \over d} \leq {\sqrt{2} \over 3} \qed ... ...
 ... ... @@ -170,6 +170,7 @@ \def\E{{\bb E}} \def\Pr{{\rm Pr}\mkern0.5mu} \def\Prsub#1{{\rm Pr}_{#1}} \def\Var{{\rm Var}\mkern0.5mu} % Vektory \def\t{{\bf t}} ... ...
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