Consider a problem as follows: We have a set of points in a high-dimensional Euclidean space . We want to project the points onto a space of low dimension in such a way that pairwise distances of the points are approximately the same as before.
Formally, we are looking for a map such that for any pair of original points , distorts little from , where is the Euclidean norm, i.e. is the distance between and in Euclidean space.
This problem has various important applications in both theory and practice. In many tasks, the data points are drawn from a high dimensional space, however, computations on high-dimensional data are usually hard due to the infamous "curse of dimensionality". The computational tasks can be greatly eased if we can project the data points onto a space of low dimension while the pairwise relations between the points are approximately preserved.
The Johnson-Lindenstrauss Theorem states that it is possible to project points in a space of arbitrarily high dimension onto an -dimensional space, such that the pairwise distances between the points are approximately preserved.
- For any and any positive integer , let be a positive integer such that
- Then for any set of points in , there is a map such that for all ,
- Furthermore, this map can be found in expected polynomial time.
The random projections
The map is done by random projection. There are several ways of applying the random projection. We adopt the one in the original Johnson-Lindenstrauss paper.
|The projection (due to Johnson-Lindenstrauss)
- Let be a random matrix that projects onto a uniform random k-dimensional subspace.
- Multiply by a fixed scalar . For every , is mapped to .
The projected point is a vector in .
The purpose of multiplying the scalar is to guarantee that .
Besides the uniform random subspace, there are other choices of random projections known to have good performances, including:
- A matrix whose entries follow i.i.d. normal distributions. (Due to Indyk-Motwani)
- A matrix whose entries are i.i.d. . (Due to Achlioptas)
In both cases, the matrix is also multiplied by a fixed scalar for normalization.
A proof of the Theorem
We present a proof due to Dasgupta-Gupta, which is much simpler than the original proof of Johnson-Lindenstrauss. The proof is for the projection onto uniform random subspace. The idea of the proof is outlined as follows:
- To bound the distortions to pairwise distances, it is sufficient to bound the distortions to the length of unit vectors.
- A uniform random subspace of a fixed unit vector is identically distributed as a fixed subspace of a uniform random unit vector. We can fix the subspace as the first k coordinates of the vector, thus it is sufficient to bound the length (norm) of the first k coordinates of a uniform random unit vector.
- Prove that for a uniform random unit vector, the length of its first k coordinates is concentrated to the expectation.
From pairwise distances to norms of unit vectors
Let be a vector in the original space, the random matrix projects onto a uniformly random k-dimensional subspace of . We only need to show that
Think of as a for some . Then by applying the union bound to all pairs of the points in , the random projection violates the distortion requirement with probability at most
so has the desirable low-distortion with probability at least . Thus, the low-distortion embedding can be found by trying for expected times (recalling the analysis fo geometric distribution).
We can further simplify the problem by normalizing the . Note that for nonzero 's, the statement that
is equivalent to that
Thus, we only need to bound the distortions for the unit vectors, i.e. the vectors that . The rest of the proof is to prove the following lemma for the unit vector in .
- For any unit vector , it holds that
As we argued above, this lemma implies the Johnson-Lindenstrauss Theorem.
Random projection of fixed unit vector fixed projection of random unit vector
Let be a fixed unit vector in . Let be a random matrix which projects the points in onto a uniformly random -dimensional subspace of .
Let be a uniformly random unit vector in . Let be such a fixed matrix which extracts the first coordinates of the vectors in , i.e. for any , .
In other words, is a random projection of a fixed unit vector; and is a fixed projection of a uniformly random unit vector.
A key observation is that:
- The distribution of is the same as the distribution of .
The proof of this observation is omitted here.
With this observation, it is sufficient to work on the subspace of the first coordinates of the uniformly random unit vector . Our task is now reduced to the following lemma.
- Let be a uniformly random unit vector in . Let be the projection of to the subspace of the first -coordinates of .
Due to the above observation, Lemma 3.2 implies Lemma 3.1 and thus proves the Johnson-Lindenstrauss theorem.
Note that . Due to the linearity of expectations,
Since is a uniform random unit vector, it holds that . And due to the symmetry, all 's are equal. Thus, for all . Therefore,
Lemma 3.2 actually states that is well-concentrated to its expectation.
Concentration of the norm of the first entries of uniform random unit vector
We now prove Lemma 3.2. Specifically, we will prove the direction:
The direction is proved with the same argument.
Due to the discussion in the last section, this can be interpreted as a concentration bound for , which is a sum of . This hints us to use Chernoff-like bounds. However, for uniformly random unit vector , 's are not independent (because of the constraint that ). We overcome this by generating uniform unit vectors from independent normal distributions.
The following is a very useful fact regarding the generation of uniform unit vectors.
|Generating uniform unit vector
- Let be i.i.d. random variables, each drawn from the normal distribution . Let . Then
- is a uniformly random unit vector.
Then for ,
To avoid writing a lot of 's. We write . The first inequality (the lower tail) of Lemma 3.2 can be written as:
The probability is a tail probability of the sum of independent variables. The 's are not 0-1 variables, thus we cannot directly apply the Chernoff bounds. However, the following two key ingredients of the Chernoff bounds are satisfiable for the above sum:
- The 's are independent.
- Because 's are normal, it is known that the moment generating functions for 's can be computed as follows:
- If follows the normal distribution , then , for
Therefore, we can re-apply the technique of the Chernoff bound (applying Markov's inequality to the moment generating function and optimizing the parameter ) to bound the probability :
The last term is minimized when
which is is for the choice of k in the Johnson-Lindenstrauss theorem that
So we have proved that
With the same argument, the other direction can be proved so that
which is also for .
Lemma 3.2 is proved. As we discussed in the previous sections, Lemma 3.2 implies Lemma 3.1, which implies the Johnson-Lindenstrauss theorem.