Approximation to maximum and minimum function : soft-min and soft-max - math

The approximation to the function max(x) can be written as a "noisy-OR" as given below:
maxk x = 1 - Πk(1 - x)
Are there any way to approximate min(x)?

If your values range between 0 and 1, then max is similar to the OR operator and min is similar to the AND operator. Similarly, the NOT operator can be thought of as 1 - x. Using De Morgan's laws, they're related by
x1 ∧ x2 ∧ x3 ... ∧ xk = (not x1) ∨ (not x2) ∨ ... ∨ (not xk)
Therefore, you should be able to approximate min{x1, x2, ..., xk} by computing 1 - max{1 - x1, 1 - x2, ..., 1 - xk}.
Hope this helps!

Related

Is it mathematically possible to solve this problem?

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!
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.
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.

How do I get all solutions from this system?

I am new to linear algebra and I am trying to solve a system of three equations with five unknowns. The system I have is the following:
x1 + x2 + x3 + x4 + x5 = 1
-x1 + x2 + x3 - 2x4 - 2x5 = 1
2x1 + 2x2 - x3 - x4 + x5 = 1
So what I did was set up the augmented matrix like this:
1 1 1 1 1 1
-1 1 1 -2 -2 1
2 2 -1 -1 1 1
Then I try to obtain an identity matrix on the left side and end up with the following:
1 0 0 3/2 3/2 0
0 1 0 -3/2 -5/6 2/3
0 0 1 1 1/3 1/3
So I think the answer is x1 = 0, x2 = 2/3 and x3 = 1/3
But when I look in my answer sheet it reads:
(x1, x2, x3, x4, x5) = (0, 2/3, 1/3, 0, 0) + s(−3/2, 3/2, −1, 1, 0) + t(−3/2, 5/6, −1/3, 0, 1)
I have no idea how to interpret this. My x1,x2,x3 seems to match the first three in the first five-tuple but what are the other two five-tuples? Can someone explain what I am missing here? I would highly appreciate it.
A system of equations can be represented in matrix form as
Ax = b
where A is the matrix of coefficients, x is the column vector (x1, ..., xn) and b is the column vector with as many entries as equations are.
When b is not 0 we say that the system is not homogeneous. The associated homogeneous system is
Ax = 0
where the 0 on the right is again a column vector.
When you have a non-homogeneous system, like in this case, the general solution has the form
P + G
where P is any particular solution and G is the generic solution of the homogeneous system.
In your case the vector
P = (0, 2/3, 1/3, 0, 0)
satisfies all the equations and is therefore a valid particular solution.
The other two vectors (−3/2, 3/2, −1, 1, 0) and (−3/2, 5/6, −1/3, 0, 1) satisfy the homogeneous equations (take a moment to check this). And since there are 3 (independent) equations with 5 unknowns (x1..x5), the space of homogenous solutions can be generated by these two vectors (again because they are independent).
So, to describe the space of all homogeneous solutions you need two scalar variables s and t. In other words
G = s(−3/2, 3/2, −1, 1, 0) + t(−3/2, 5/6, −1/3, 0, 1)
will generate all homogeneous solutions as s and t take all posible real values.

Constants in terms of ideal: "stdio:4:11:(3): error: can't promote number to ring" in Macaulay2

I am trying to demonstrate Handelman's theorem and the example 1 here with Macaulay2. I cannot understand the error in defining the ideal for the polytope restricted by the intervals.
R=QQ[x1,x2,x3,MonomialOrder=>Lex];
I=ideal(x1-0.2,-x1+0.5,x2,-x2+1,x3-1,-x3+1)
stdio:2:11:(3): error: can't promote number to ring
and what is the error for? How should I define the constants?
For some reason, Macaulay2 only accepts the computation for polynomial ring with RR not QQ:
i1 : R=RR[x1,x2,x3,MonomialOrder=>Lex]
o1 = R
o1 : PolynomialRing
i2 : I=ideal(x1-0.2,-x1+0.5,x2,-x2+1,x3-1,-x3+1)
o2 = ideal (x1 - .2, - x1 + .5, x2, - x2 + 1, x3 - 1, - x3 + 1)
o2 : Ideal of R
You get the error because M2 views decimals as real numbers as opposed to rationals:
i1 : .2
o1 = .2
o1 : RR (of precision 53)
So .2 isn't in your base ring.
Use fraction notation (as opposed to decimal notation) to input your ideal and you'll be in business.
i2 : R=QQ[x1,x2,x3, MonomialOrder => Lex];
i3 : I=ideal(x1-1/5,-x1+1/2,x2,-x2+1,x3-1,-x3+1)
o3 = ideal (x1 - 1/5, - x1 + 1/2, x2, - x2 + 1, x3 - 1, - x3 + 1)
o3 : Ideal of R

Functions for multivariate numerical integration in R [duplicate]

I am using the following R code, taken from a published paper (citation below). This is the code:
int2=function(x,r,n,p) {
(1+x)^((n-1-p)/2)*(1+(1-r^2)*x)^(-(n-1)/2)*x^(-3/2)*exp(-n/(2*x))}
integrate(f=int2,lower=0,upper=Inf,n=530,r=sqrt(.245),p=3, stop.on.error=FALSE)
When I run it, I get the error "non-finite function value". Yet Maple is able to compute this as 4.046018765*10^27.
I tried using "integral" in package pracma, which gives me a different error:
Error in if (delta < tol) break : missing value where TRUE/FALSE needed
The overall goal is to compute a ratio of two integrals, as described in Wetzels & Wagenmakers (2012) "A default Bayesian hypothesis test for correlations" (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3505519/). The entire function is as follows:
jzs.pcorbf = function(r0, r1, p0, p1, n) {
int = function(r,n,p,g) {
(1+g)^((n-1-p)/2)*(1+(1-r^2)*g)^(-(n-1)/2)*g^(-3/2)*exp(-n/(2*g))};
bf10=integrate(int, lower=0,upper=Inf,r=r1,p=p1,n=n)$value/
integrate(int,lower=0,upper=Inf,r=r0,p=p0,n=n)$value;
return(bf10)
}
Thanks!
The issue is that your integral function is generating NaN values when called with x values in its domain. You're integrating from 0 to Infinity, so let's check a valid x value of 1000:
int2(1000, sqrt(0.245), 530, 3)
# [1] NaN
Your objective multiplies four pieces:
x <- 1000
r <- sqrt(0.245)
n <- 530
p <- 3
(1+x)^((n-1-p)/2)
# [1] Inf
(1+(1-r^2)*x)^(-(n-1)/2)
# [1] 0
x^(-3/2)
# [1] 3.162278e-05
exp(-n/(2*x))
# [1] 0.7672059
We can now see that the issue is that you're multiplying infinity by 0 (or rather something numerically equal to infinity times something numerically equal to 0), which is causing the numerical issues. Instead of calculating a*b*c*d, it will be more stable to calculate exp(log(a) + log(b) + log(c) + log(d)) (using the identity that log(a*b*c*d) = log(a)+log(b)+log(c)+log(d)). One other quick note -- the value x=0 needs a special case.
int3 = function(x, r, n, p) {
loga <- ((n-1-p)/2) * log(1+x)
logb <- (-(n-1)/2) * log(1+(1-r^2)*x)
logc <- -3/2 * log(x)
logd <- -n/(2*x)
return(ifelse(x == 0, 0, exp(loga + logb + logc + logd)))
}
integrate(f=int3,lower=0,upper=Inf,n=530,r=sqrt(.245),p=3, stop.on.error=FALSE)
# 1.553185e+27 with absolute error < 2.6e+18

calibration of the posterior probabilities

currently i work on calibration of probability. i use the calibration approach, called rescaling algorithm - the source http://lem.cnrs.fr/Portals/2/actus/DP_201106.pdf (page 7).
the algorithm i wrote is:
rescaling_fun = function(x, y, z) {
P_korg = z # yhat_test_prob$BAD
P_k_C1 = sum(as.numeric(y) - 1)/length(y) # testset$BAD
P_kt_C1 = sum(as.numeric(x) - 1)/length(x) # trainset$BAD
P_k_C0 = sum(abs(as.numeric(y) - 2))/length(y)
P_kt_C0 = sum(abs(as.numeric(x) - 2))/length(x)
P_new <- ((P_k_C1/P_kt_C1) * P_korg)/((P_k_C0/P_k_C0) * (1 - P_korg) + (P_k_C0/P_k_C1) * (P_korg))
return(P_new)
}
the input values are:
1. x - train_set$BAD (actuals of `train set`)
2. y - test_set$BAD (actuals of `test set`)
3. z - yhat_test_prob$BAD (prediction on `test set`)
the problem - the result values are not within range of 0 and 1. Could you please help to solve the problem?
Your formulas to obtain probs (P_k_C1 ...) need to be modified. For example, according to the paper, y is a binary variable (0, 1) and the formula is sum(y - 1)/length(y) which is most likely to be negative - it converts y values to be -1 or 0, followed by adding them. I consider it should be (sum(y)-1)/length(y). Below is an example.
set.seed(1237)
y <- sample(0:1, 10, replace = T)
y
[1] 0 1 0 0 0 1 1 0 1 1
# it must be negative as it is sum(y - 1) - y is 0 or 1
sum(as.numeric(y) - 1)/length(y)
[1] -0.5
# modification
(sum(as.numeric(y)) - 1)/length(y)
[1] 0.4

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