Is it mathematically possible to solve this problem? - r

Question

x <- abs(rnorm(8))
C <- (x[1]*x[2]*x[3])^(1/3)
y <- log(x/C)
Is it mathematically possible to determine x[1:3] given you only have y? Here, x and y are always vectors of length 8. I should note that x is known for some of my dataset, which could be useful to find a solution for the other portion of the data where x is unknown. All of my code is implemented in R, so R code would be appreciated if this is solvable!

Answer 1

Defining f as
f <- function(x) {
C <- (x[1]*x[2]*x[3])^(1/3)
log(x/C)
}
we first note that if k is any scalar constant then f(x) and f(k*x) give the same result so if we have y = f(x) we can't tell whether y came from x or from k*x. That is, y could have come from any scalar multiple of x; therefore, we cannot recover x from y.
Linear formulation
Although we cannot recover x we can determine x up to a scalar multiple. Define the matrix A:
ones <- rep(1, 8)
a <- c(1, 1, 1, 0, 0, 0, 0, 0)
A <- diag(8) - outer(ones, a) / 3
in which case f(x) equals:
A %*% log(x)
Inverting formula
From this formula, given y and solving for x, the value of x would equal
exp(solve(A) %*% y) ## would equal x if A were invertible
if A were invertible but unfortunately it is not. For example, rowSums(A) equals zero which shows that the columns of A are linearly dependent which implies non-invertibility.
all.equal(rowSums(A), rep(0, 8))
## [1] TRUE
Rank and nullspace
Note that A is a projection matrix. This follows from the fact that it is idempotent, i.e. A %*% A equals A.
all.equal(A %*% A, A)
## [1] TRUE
It also follows from the fact that its eigenvalues are all 0 and 1:
zapsmall(eigen(A)$values)
## [1] 1 1 1 1 1 1 1 0
From the eigenvalues we see that A has rank 7 (the number of nonzero eigenvalues) and the dimension of the nullspace is 1 (the number of zero eigenvalues).
Another way to see this is that knowing that A is a projection matrix its rank equals its trace, which is 7, so its nullspace must have dimension 8-7=1.
sum(diag(A)) # rank of A
## [1] 7
Taking scalar multiples spans a one dimensional space so from the fact that the nullspace has dimension 1 it must be the entirely of the values that map into the same y.
Key formula
Now replacing solve in ## above with the generalized inverse, ginv, we have this key formula for our approximation to x given that y = f(x) for some x:
library(MASS)
exp(ginv(A) %*% y) # approximation to x accurate up to scalar multiple
or equivalently if y = f(x)
exp(y - mean(y))
While these do not give x they do determine x up to a scalar multiple. That is if x' is the value produced by the above expressions then x equals k * x' for some scalar constant k.
For example, using x and y from the question:
exp(ginv(A) %*% y)
## [,1]
## [1,] 1.2321318
## [2,] 0.5060149
## [3,] 3.4266146
## [4,] 0.1550034
## [5,] 0.2842220
## [6,] 3.7703442
## [7,] 1.0132635
## [8,] 2.7810703
exp(y - mean(y)) # same
## [1] 1.2321318 0.5060149 3.4266146 0.1550034 0.2842220 3.7703442 1.0132635
## [8] 2.7810703
exp(y - mean(y))/x
## [1] 2.198368 2.198368 2.198368 2.198368 2.198368 2.198368 2.198368 2.198368
Note
Note that y - mean(y) can be written as
B <- diag(8) - outer(ones, ones) / 8
B %*% y
and if y = f(x) then y must be in the range of A so we can verify that:
all.equal(ginv(A) %*% A, B %*% A)
## [1] TRUE
It is not true that the matrix ginv(A) equals B. It is only true that they act the same on the range of A which is all that we need.

Answer 2

No, it's not possible. You have three unknowns. That means you need three independent pieces of information (equations) to solve for all three. y gives you only one piece of information. Knowing that the x's are positive imposes a constraint, but doesn't necessarily allow you to solve. For example:
x1 + x2 + x3 = 6
Doesn't allow you to solve. x1 = 1, x2 = 2, x3 = 3 is one solution, but so is x1 = 1, x2 = 1, x3 = 4. There are many other solutions. [Imposing your "all positive" constraint would rule out solutions such as x1 = 100, x2 = 200, x3 = -294, but in general would leave more than one remaining solution.]
x1 + x2 + x3 = 6,
x1 + x2 - x3 = 0
Constrains x3 to be 3, but allows arbitrary solutions for x1 and x2, subject to x1 + x2 = 3.
x1 + x2 + x3 = 6,
x1 + x2 - x3 = 0,
x1 - x2 + x3 = 2
Gives the unique solution x1 = 1, x2 = 2, x3 = 3.