Databases Reference
In-Depth Information
2
4
3
5
2
4
3
5
2.6
1
5.204
3.6
3.6
3.6
3.6
»
-
1
1
2.617
2
2
2
2
1.617
1
1
1
1
1.178
1
=
2.905
2.178
2.178
2.178
2.178
×
1
1
1
1
1
1
1
2.617
2
2
2
2
1
1
2.617
2
2
2
2
Figure 9.16: Replace z by 1.178
that M is an n-by-m utility matrix with some entries blank, while U and V
are matrices of dimensions n-by-d and d-by-m, for some d. We shall use m
ij
,
u
ij
, and v
ij
for the entries in row i and column j of M , U , and V , respectively.
Also, let P = U V , and use p
ij
for the element in row i and column j of the
product matrix P .
Suppose we want to vary u
rs
and find the value of this element that mini-
mizes the RMSE between M and U V . Note that u
rs
affects only the elements
in row r of the product P = U V . Thus, we need only concern ourselves with
the elements
d
p
rj
=
u
rk
v
kj
=
u
rk
v
kj
+ xv
sj
k=1
k=s
for all values of j such that m
rj
is nonblank. In the expression above, we have
replaced u
rs
, the element we wish to vary, by a variable x, and we use the
convention
•
is shorthand for the sum for k = 1, 2, . . . , d, except for k = s.
k=s
If m
rj
is a nonblank entry of the matrix M , then the contribution of this
element to the sum of the squares of the errors is
m
rj
2
−p
rj
)
2
(m
rj
=
−
u
rk
v
kj
−xv
sj
k=s
We shall use another convention:
•
is shorthand for the sum over all j such that m
rj
is nonblank.
j
Then we can write the sum of the squares of the errors that are affected by
the value of x = u
rs
as
2
m
rj
−
u
rk
v
kj
−xv
sj
j
k=s
Take the derivative of the above with respect to x, and set it equal to 0, in
order to find the value of x that minimizes the RMSE. That is,
m
rj
−2v
sj
−
u
rk
v
kj
−xv
sj
= 0
j
k=s
