Databases Reference
In-Depth Information
Then the dot product
u
i
·
v
j
is the
predicted preference
of user
i
for item
j
,
and you want that to be as close as possible to the
actual preference
x
i
,
j
.
So, you want to find the best choices of
U
and
V
that minimize the
squared differences between prediction and observation on every‐
thing you actually know, and the idea is that if it's really good on stuff
you know, it will also be good on stuff you're guessing. This should
sound familiar to you—it's mean squared error, like we used for linear
regression.
Now you get to choose a parameter, namely the number
d
defined as
how may latent features you want to use
. The matrix
U
will have a row
for each user and a column for each latent feature, and the matrix
V
will have a row for each item and a column for each latent feature.
How do you choose
d
?
It's typically about 100, because it's more than
20 (as we told you, through the course of developing the product, we
found that we had a pretty good grasp on someone if we ask them 20
questions) and it's as much as you care to add before it's computa‐
tionally too much work.
The resulting latent features are the basis of a well-defined
subspace of the total
n
-dimensional space of potential latent
variables. There's no reason to think this solution is unique
if there are a bunch of missing values in your “answer” matrix.
But that doesn't necessarily matter, because you're just look‐
ing for
a solution
.
Theorem: The resulting latent features will be uncorrelated
We already discussed that correlation was an issue with k-NN, and
who wants to have redundant information going into their model? So
a nice aspect of these latent features is that they're uncorrelated. Here's
a sketch of the proof:
Say we've found matrices
U
and
V
with a fixed product
U
·
V
=
X
such
that the squared error term is minimized. The next step is to find the
best
U
and
V
such that their entries are small—actually we're mini‐
mizing the sum of the squares of the entries of
U
and
V
.
But we can
modify
U
with any invertible
d
×
d
matrix
G
as long as we modify
V
by its inverse:
U
·
V
=
U
·
G
·
G
−1
·
V
=
X
.
