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authorYuchen Pei <me@ypei.me>2019-03-15 20:35:55 +0100
committerYuchen Pei <me@ypei.me>2019-03-15 20:35:55 +0100
commitda4476318290453a247521b1244ba18feb34cf33 (patch)
treed0ab1eec5500fad0332889aa7e136da815c277e0
parent97d05feace09e730d42dc99adf8b76e06d15f4e0 (diff)
minor changes
-rw-r--r--posts/2019-03-14-great-but-manageable-expectations.md9
1 files changed, 5 insertions, 4 deletions
diff --git a/posts/2019-03-14-great-but-manageable-expectations.md b/posts/2019-03-14-great-but-manageable-expectations.md
index d812043..889f674 100644
--- a/posts/2019-03-14-great-but-manageable-expectations.md
+++ b/posts/2019-03-14-great-but-manageable-expectations.md
@@ -375,6 +375,7 @@ I will break the proof into two parts:
$c_1$, $c_2$ and the function $C(c_1, c_2)$ are important to the
practicality and usefulness of Conjecture 0.
+Part 1 can be derived using Conjecture 1.
We use the notations $p_I$ and $q_I$ to be $q$ and $p$ conditioned on
the subsampling index $I$, just like in the proof of the subsampling theorems (Claim 19 and 24).
Then
@@ -385,16 +386,16 @@ $$D_\lambda(q_I || p_I) = D_\lambda(p_I || q_I)
= D_\lambda(\mu_0 || \mu_0) = D_\lambda(\mu_1 || \mu_1) = 0 & I \in \mathcal I_\notin
\end{cases}$$
-and that $p = |\mathcal I|^{-1} \sum_{I \in \mathcal I} p_I$ and
+Since $p = |\mathcal I|^{-1} \sum_{I \in \mathcal I} p_I$ and
$q = |\mathcal I|^{-1} \sum_{I \in \mathcal I} q_I$ and
-$|\mathcal I_\in| = r |\mathcal I|$.
+$|\mathcal I_\in| = r |\mathcal I|$, by Conjecture 1, we have Part 1.
**Remark in the proof**. As we can see here, instead of trying to prove Conjecture 1,
it suffices to prove a weaker version of it, by specialising on mixture of Gaussians,
-to have a Claim 26 without any conjectural assumptions.
+in order to have a Claim 26 without any conjectural assumptions.
I have in fact posted the Conjecture on [Stackexchange](https://math.stackexchange.com/questions/3147963/an-inequality-related-to-the-renyi-divergence).
-Now let us prove Item 2.
+Now let us verify Part 2.
Using Claim 27 and Example 1, we have