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Lemma A.2 (Hanson-Wright Inequality Ver10 Let [MATH] and [MATH] Then for some absolute constant [MATH] , for every [MATH] [EQUATION] |
Throughout this proof, we will use [MATH] to denote a set of labeled samples [MATH] , where [MATH] and [MATH] , where [MATH] . We will use [MATH] to denote the cardinality of [MATH] |
The following lemma proves property (i) of Definition 2.2 Lemma A.3 Let [MATH] and [MATH] If the multiset [MATH] consists of [MATH] pairs [MATH] where [MATH] and [MATH] it satisfies [MATH] [MATH] [MATH] with probability at least [MATH] |
Proof. Let [MATH] be the size of [MATH] Consider the unlabeled set of samples [MATH] drawn from [MATH] .To establish (i), we note that the probability that a coordinate of a sample [MATH] has absolute value at least [MATH] |
is at most [MATH] by Fact A.1 . By a union bound over [MATH] coordinates, the probability that all coordinates of all samples have absolute value smaller than [MATH] |
is at least [MATH] . In this case, [MATH] assuming [MATH] Also note that [MATH] By Fact A.1 , the probability that [MATH] is at most [MATH] By a union bound, the probability that all [MATH] are smaller than [MATH] is at least [MATH] Hence, [MATH] holds for all the samples with probability at least [MATH] This completes... |
We will require the following technical claim: Claim A.4 Let [MATH] be of size at least a sufficiently large multiple of [MATH] With probability at least [MATH] , we have that for any unit vector [MATH] and [MATH] it holds |
[EQUATION] Proof. We start with the following claim: Claim A.5 Let [MATH] be a set of [MATH] independent samples from [MATH] For [MATH] , we have with probability at least [MATH] , for all [MATH] |
[MATH] and all [MATH] have [MATH] Proof. To prove the claimed tail for all [MATH] , we first show that for any [MATH] , where [MATH] the claimed tail bound divided by [MATH] (i.e., [MATH] holds with probability [MATH] Then, by a simple union bound, we get the desired upper bound for all [MATH] |
Given [MATH] , we have that [MATH] where rfc is the complementary error function. We define the function [MATH] Observe that [MATH] is a sum of [MATH] independent Bernoulli random variables with mean [MATH] Since [MATH] , by the Chernoff bound, [MATH] |
with probability at most [MATH] . Let [MATH] be such that [MATH] Since [MATH] for all [MATH] we have that [MATH] and hence [MATH] Thus, we have that for all [MATH] [MATH] and this bound suffices. |
When [MATH] , note that [MATH] Here we need to use a more explicit version of the Chernoff bound. That gives that [MATH] with probability at most [MATH] where [MATH] is the KL-divergence between Bernoulli’s with probabilities [MATH] and [MATH] . When [MATH] [MATH] [MATH] , we obtain |
[EQUATION] Thus, we have that [MATH] with probability at most [MATH] in this case. Note that [MATH] for [MATH] for [MATH] By a union bound, we have that for [MATH] for integers [MATH] |
that [MATH] for all such [MATH] is [MATH] Note that the largest [MATH] has [MATH] and since [MATH] and [MATH] , this means that [MATH] |
If [MATH] [MATH] and so the result is trivial. If [MATH] , then [MATH] . Otherwise, there is a [MATH] with [MATH] and thus [MATH] This completes the proof of Claim A.5 |
Let [MATH] be a [MATH] -cover of the set of unit vectors including all coordinate directions of size [MATH] Then, by a union bound, Claim A.5 holds for [MATH] for all [MATH] |
except with probability [MATH] where [MATH] . This is smaller than [MATH] when [MATH] which holds when [MATH] is a sufficiently large multiple of [MATH] . The latter statement in turn holds when [MATH] is a sufficiently large multiple of [MATH] We assume this holds in the following. |
Now consider a unit vector [MATH] and [MATH] Let [MATH] and [MATH] . Let [MATH] have [MATH] Then let [MATH] If [MATH] has [MATH] , using the triangle inequality, it follows that either [MATH] or else [MATH] So we let [MATH] and we have |
[EQUATION] If we iterate this procedure, for large enough [MATH] , we will have [MATH] and so [MATH] and hence we have [MATH] By Claim A.5 , we have [MATH] and so |
[EQUATION] So we have shown that, for all [MATH] and [MATH] [MATH] Since this is trivial for [MATH] , we are done. The following lemma proves property (ii) of Definition 2.2 |
Lemma A.6 For all [MATH] , we have that [MATH] Proof. The proof of the lemma will make essential use of the following elementary fact: |
Fact A.7 For any pair of real random variables [MATH] , integers [MATH] , and [MATH] , it holds: [EQUATION] Notice that when [MATH] , the RHS is greater than [MATH] and hence the inequality is trivial. When [MATH] , by condition (i) of Definition 2.2 , the LHS is [MATH] which makes the inequality trivial as well. For t... |
[EQUATION] By condition (i) of Definition 2.2 , we have that [MATH] Thus, we can set the integer parameter [MATH] to be [MATH] By Claim A.4 , the term [MATH] is at most |
[MATH] Due to the simple fact [EQUATION] we have [EQUATION] We first establish an upper bound for [MATH] in which case [MATH] . Each term in the summation above satisfies |
[EQUATION] Note that there exists a sufficiently large universal constant [MATH] such that for [MATH] , we have that [MATH] which implies that the exponential term is negligible for this range of [MATH] There are at most [MATH] such terms in the summation, which yields an upper bound of the summation as [MATH] for suff... |
We now prove an upper bound for the case that [MATH] . For [MATH] , the first term in the min function satisfies [MATH] Similarly, the second term in the min function satisfies |
[MATH] The first term monotonically decreases as [MATH] gets larger. The second term monotonically increases as [MATH] gets larger. We can thus conclude that |
[EQUATION] There are at most [MATH] such terms, which yields an upper bound of [EQUATION] Now note that for [MATH] , the above is smaller than [MATH] because we can assume [MATH] . Hence, we obtain an upper bound of [MATH] which completes the proof. |
The following lemma proves property (iii) of Definition 2.2 Lemma A.8 For [MATH] we have that [MATH] with probability at least [MATH] |
Proof. The proof requires two simple technical claims. Our first claim gives an explicit formula for the distribution of the projections of [MATH] in any direction: |
Claim A.9 For any [MATH] with [MATH] we have that [EQUATION] where [MATH] and [MATH] are independent. Proof. Let [MATH] Note that [MATH] and are independent. Define [MATH] It is now easy to verify that the claim statement holds. |
Our second claim gives a tight concentration inequality for [MATH] Claim A.10 For any [MATH] with [MATH] and [MATH] , we have that for some absolute constant [MATH] |
[EQUATION] where [MATH] and [MATH] Proof. Let [MATH] be a [MATH] -dimensional vector, where each [MATH] is independently drawn from [MATH] Let [MATH] be a diagonal matrix whose diagonal entries [MATH] |
equal [MATH] for [MATH] , and [MATH] for [MATH] Using Claim A.9 [MATH] can be written as [MATH] . Applying the Hanson-Wright inequality (Lemma A.2 ), we get |
[EQUATION] where [MATH] and [MATH] This completes the proof. We now have the necessary ingredients to prove Lemma A.8 Let [MATH] Since [MATH] and |
[MATH] we have [MATH] Given that [MATH] and Claim A.10 we have [MATH] Taking a union bound over [MATH] unit basis vectors of [MATH] we get that with probability at least [MATH] no coordinate of [MATH] has magnitude larger than |
[MATH] , and hence [MATH] The following lemma proves property (iv) of Definition 2.2 Lemma A.11 For [MATH] , we have that [MATH] with probability at least [MATH] |
Proof. The proof idea is the following: By standard results, it is straightforward to handle the concentration of the empirical covariance of a distribution with bounded support. Hence, we split the distribution into two parts. One part contains most of the probability mass and has almost identical covariance as the or... |
Let [MATH] denote the distribution of [MATH] , and [MATH] be the distribution of [MATH] conditional on the event [MATH] . In this part of the proof, we argue that the covariance of the [MATH] is similar to the [MATH] Denote the matrix [MATH] as [MATH] and the matrix |
[MATH] as [MATH] For any unit vector [MATH] , we have that [EQUATION] By the argument in the proof of Lemma A.3 , we have that [MATH] , the term [MATH] |
is less than the second moment of the [MATH] -tail of [MATH] of which we are going to obtain an upper bound. Let [MATH] satisfy [MATH] The second moment of the [MATH] -tail is by definition |
[EQUATION] Applying Claim A.10 with [MATH] we get [MATH] for some absolute constant [MATH] . Notice that [MATH] which is equivalent to [MATH] Pick appropriate [MATH] such that |
[MATH] Notice that [MATH] and we can rewrite Equation ( as [EQUATION] By Equation ( ), [MATH] and hence [MATH] Now we argue that, with high probability, the empirical covariance concentrates to [MATH] The sampling procedure from [MATH] can be thought as the following: draw [MATH] i.i.d samples from [MATH] , if [MATH] |
is satisfied for all samples, return the samples, otherwise declare failure. Hence, with probability [MATH] , the samples in [MATH] can be seen as distributed as [MATH] . If we set [MATH] in Corollary 5.52 of Ver10 , with probability at least [MATH] |
[MATH] Finally, we have [EQUATION] Given [MATH] we have [MATH] and hence with probability at least [MATH] [MATH] This completes the proof of Proposition 2.3 |
A.1 Handling Approximate Identity Covariance In this subsection, we discuss the case where the covariance matrix of [MATH] is not exactly identity, but instead satisfies [MATH] We will prove that a slightly modified deterministic regularity condition (i.e., Definition 2.2 ) still holds for this setting. The same algori... |
Proposition A.12 Let [MATH] where [MATH] and [MATH] If the multiset [MATH] consists of [MATH] labeled samples [MATH] , where [MATH] and [MATH] , where [MATH] then [MATH] is [MATH] -good with respect to [MATH] with probability at least [MATH] |
Proof. We prove the all the conditions in Definition 2.2 will be satisfied by reducing to the identity covariance case and applying Proposition A.12 . Let [MATH] , we have that pair [MATH] follows the setting of Proposition 2.3 , where [MATH] and [MATH] Let [MATH] be the set that contains pairs [MATH] |
1. For condition (i), we can apply Proposition A.12 and get [MATH] and [MATH] 2. For condition (ii), we have [EQUATION] 3. For condition (iii), we have [MATH] |
4. For condition (iv), notice that by Proposition 2.3 [MATH] . Multipling [MATH] on both sides yields [MATH] [EQUATION] Given the above proposition, we can first robustly estimate the covariance matrix to appropriate accuracy, using the algorithm of DKK 16 , and then apply our robust LR algorithm for the known covarian... |
Appendix B Proof of Proposition 2.1 We start by proving concentration bounds for [MATH] In particular, we show: Lemma B.1 We have that [MATH] |
Proof. Since [MATH] , for any [MATH] , we have that [EQUATION] Since [MATH] is a symmetric matrix, we have [MATH] So, to bound [MATH] it suffices to bound [MATH] for unit vectors [MATH] |
By definition of [MATH] for any [MATH] we have that [EQUATION] For unit vectors [MATH] , the RHS is bounded from above as follows: |
[EQUATION] where the third line follows from the fact that [MATH] , the fourth line holds since the probability must be less than [MATH] [MATH] satisfies condition (i) of Definition 2.2 and Equation ; and the fifth line follows from condition (iii) of Definition 2.2 |
Corollary B.2 We have that [MATH] where the [MATH] term denotes a matrix of spectral norm [MATH] Proof. By definition, we have that [MATH] . Thus, we can write |
[EQUATION] where the second line uses the fact that [MATH] , the goodness of [MATH] (condition (iv) in Definition 2.2 ), and Lemma B.1 . Specifically, Lemma B.1 implies that [MATH] Therefore, we have that |
[MATH] Lemma B.3 We have that [MATH] where the [MATH] term denotes a vector with [MATH] -norm at most [MATH] Proof. By definition, we have that [MATH] . Since [MATH] is a good set, by condition (iii) of Definition 2.2 , we have [MATH] . Since [MATH] , it follows that [MATH] . Using the valid inequality [MATH] and Lemma... |
Corollary B.4 We have [MATH] , where [MATH] [MATH] Proof. By definition, we can write [EQUATION] where the first line follows by definition, second line follows from Corollary B.2 the third line follows from the fact that [MATH] the fourth line follows from Lemma B.3 This completes the proof. |
Case of Small Spectral Norm. We are now ready to analyze the case that the vector [MATH] is returned by the algorithm in Step In this case, we have that |
[MATH] which implies for any unit vector [MATH] , we have that [MATH] Since [MATH] , Lemma B.3 gives that [EQUATION] If [MATH] , the proof will be complete. Otherwise Corollary B.4 becomes [MATH] which implies |
[EQUATION] and hence [MATH] This proves part (i) of Proposition 2.1 Case of Large Spectral Norm. We next show the correctness of the algorithm when it returns a filter in Step 10 |
We start by proving that if [MATH] for a sufficiently large universal constant [MATH] then a value [MATH] satisfying the condition in Step 10 exists. |
We need to prove the following two facts of about [MATH] (1) [MATH] , and (2) [MATH] If [MATH] , we must have [MATH] which yields a contradiction. If [MATH] , we have [MATH] in Corollary B.4 which will be assumed for the rest of the proof. For sufficiently large [MATH] , we have |
[EQUATION] where [MATH] is the matrix defined in Corollary B.4 Hence, we have [MATH] The proof of the first fact is complete. By the definition of [MATH] |
[EQUATION] which is equivalent to [EQUATION] for a non-negative constant [MATH] The proof of the second fact is complete. Moreover, using the inequality [MATH] |
and Lemma B.3 as above, we get that [EQUATION] where we used the fact that [MATH] Suppose for the sake of contradiction that for all [MATH] we have that |
[EQUATION] which implies [EQUATION] Using ( 11 ), and assume that [MATH] , we obtain that for all [MATH] we have that [EQUATION] |
We now have the following sequence of inequalities: [EQUATION] Rearranging the above, we get that [EQUATION] Combined with ( 10 ), we obtain the following: there exists constants [MATH] such that |
[EQUATION] Notice that because [MATH] , we have [MATH] . By the above equation, for sufficiently small [MATH] , we have [MATH] , which is a contradiction if [MATH] is sufficiently large. Therefore, it must be the case that for some value of [MATH] the condition in Step 10 is satisfied. |
The following claim completes the proof: Claim B.5 We have that [MATH] Proof. Recall that [MATH] with [MATH] and [MATH] disjoint multisets such that [MATH] |
We can similarly write [MATH] with [MATH] and [MATH] Since [EQUATION] it suffices to show that [MATH] Note that [MATH] is the number of points rejected by the filter that lie in [MATH] |
Note that the fraction of elements of [MATH] that are removed to produce [MATH] (i.e., satisfy [MATH] ) is at most [MATH] This follows from property (ii) of Definition 2.2 |
Hence, it holds that [MATH] On the other hand, Step 10 of the algorithm ensures that the fraction of elements of [MATH] that are rejected by the filter is at least [MATH] . Note that |
[MATH] is the number of points rejected by the filter that lie in [MATH] Therefore, we can write: [EQUATION] where the second line uses the fact that [MATH] |
and the last line uses the fact that [MATH] This completes the proof of the claim. Appendix C Deterministic Regularity Conditions for Algorithm |
We start by formally defining the set of deterministic regularity conditions under which our main algorithm succeeds: Definition C.1 |
Let [MATH] [MATH] , and [MATH] We say that a multiset [MATH] of elements in [MATH] is [MATH] -representative (with respect to [MATH] if the following conditions are satisfied: |
1. (i) For all [MATH] [MATH] (ii) For any unit vector [MATH] and [MATH] , we have [EQUATION] (iii) [MATH] (iv) [MATH] 2. For all [MATH] , let [MATH] Then the following hold: |
(i) For all [MATH] [MATH] (ii) For any [MATH] [EQUATION] (iii) [MATH] (iv) [MATH] 3. For all [MATH] , let [MATH] and for [MATH] let [MATH] . Then we have |
(i) For all [MATH] [MATH] (ii) For every [MATH] with [MATH] we have that [EQUATION] (iii) We have that [MATH] (iv) We have that [MATH] , where [MATH] |
4. [MATH] We now establish that a sufficiently large set of uncorrupted samples will satisfy the above conditions with high probability: |
Proposition C.2 If [MATH] is a set of uncorrupted samples with size [MATH] larger than [MATH] , then except with probability [MATH] [MATH] is [MATH] -representative. |
The rest of this section will focus on establishing the above proposition. Condition 1(i) and (ii) follow from Claim A.4 and (iii) and (iv) follow from Lemma C.3 . Condition 2 (i) and (ii) follow from Lemma C.4 , (iii) and (iv) from Lemma C.5 . Condition 3 (i) follows from Claim A.4 and the bound on [MATH] in [MATH] (i... |
It follows from standard results on estimating the mean and covariance matrix of a Gaussian that: Lemma C.3 For [MATH] , with probability at least [MATH] we have that [MATH] and [MATH] |
In addition to Claim A.4 , we want that: Lemma C.4 If Claim A.4 holds then, with probability at least [MATH] , we have that, for any [MATH] and any [MATH] |
[MATH] where [MATH] Proof. By the triangle inequality, [MATH] By Claim A.4 , we have that for all [MATH] and [MATH] that [EQUATION] |
Since [MATH] , we can apply Claim A.5 to it to get that, except with probability [MATH] that, for all [MATH] [EQUATION] Since [MATH] , we have: |
[EQUATION] Lemma C.5 For [MATH] , with probability at least [MATH] , for any [MATH] we have that [MATH] [MATH] , where [MATH] Proof. |
It follows from standard results on estimating the mean and covariance matrix of a Gaussian that [MATH] [MATH] . By Lemma C.3 , we have |
[MATH] By Proposition C.6 and Lemma C.3 , we have [EQUATION] Proposition C.6 For [MATH] , with probability at least [MATH] [MATH] |
Proof. The proof is very similar to that of Lemma A.8 Note that [MATH] . As long as [MATH] and [MATH] we will have: Corollary C.7 |
[MATH] We can combine this corollary with Claim A.4 in a similar way to Lemma A.6 to obtain that: Lemma C.8 If Claim A.4 and the previous Lemma C.4 hold, we have that, for any [MATH] , unit vector [MATH] and any [MATH] |
[MATH] , where [MATH] and [MATH] is [MATH] with the elements with [MATH] , where [MATH] Proof. By Claim A.4 [MATH] when [MATH] Thus, the maximum value of [MATH] for which [MATH] is non-zero is [MATH] . When [MATH] , the lemma is trivial. So we need to show it for [MATH] |
We apply Fact A.7 to obtain that [EQUATION] where the last inequality holds due to Corollary C.7 and Claim A.4 When [MATH] , each term has [MATH] . In this case, we have [MATH] for the exponential term and [MATH] . Hence, the second term is always smaller. We have |
[MATH] and so the sum of these terms is at most [MATH] Each of the [MATH] terms is bounded by [MATH] and the sum is over [MATH] terms and so in this case we have |
[EQUATION] We now consider the case when [MATH] . When [MATH] , we have [EQUATION] When [MATH] , we have [EQUATION] Since there are [MATH] terms and [MATH] for [MATH] and [MATH] , we have that |
[EQUATION] This completes the proof. Next we show that, with high probability, the expectations in 3 (iii) and (iv) are close under [MATH] and [MATH] when we condition them similarly: |
Lemma C.9 For any [MATH] let [MATH] be [MATH] conditioned on [MATH] . Then, except with probability [MATH] , for all [MATH] and all [MATH] and all unit vectors [MATH] , we have that |
[EQUATION] and [EQUATION] Proof. We let [MATH] and note that [MATH] in [MATH] dimensions. We note that [MATH] for some unit vector [MATH] . We can consider [MATH] to be the distribution of [MATH] conditioned on [MATH] and [MATH] as consisting of at least [MATH] independent samples from [MATH] It suffices to show that, ... |
[EQUATION] and [EQUATION] Now consider a fixed [MATH] . For any [MATH] [EQUATION] Thus, the central moments of [MATH] satisfy [EQUATION] |
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