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authorThibaut Horel <thibaut.horel@gmail.com>2015-02-06 15:48:24 -0500
committerThibaut Horel <thibaut.horel@gmail.com>2015-02-06 15:48:24 -0500
commit0ff14f56819acfc7be77f9237e18417d465b2266 (patch)
tree32576d399ce36de031188e1ffff5b8e3f56b4336 /paper/sections/results.tex
parent724dae4487559d7e52c5ac56b9059d124b664a13 (diff)
downloadcascades-0ff14f56819acfc7be77f9237e18417d465b2266.tar.gz
Compression
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@@ -203,7 +203,7 @@ with probability $1-\frac{1}{m}$. If $\theta_i = 0$ and $\hat \theta > \eta$,
then $\|\hat \theta - \theta\|_2 \geq |\hat \theta_i-\theta_j| > \eta$, which
is a contradiction. Therefore we get no false positives. If $\theta_i \leq \eta
+ \epsilon$, then $|\hat{\theta}_i- \theta_i| < \epsilon \implies \theta_j
-> \eta$. Therefore, we get all strong parents.
+> \eta$ and we get all strong parents.
\end{proof}
Assuming we know a lower bound $\alpha$ on $\Theta_{i,j}$,