library(pbivnorm)
rho <- 0.5
f1 <- function(x, y) {
pbivnorm(log(x)-10, log(y)-10, rho)*(exp(-(log(x)-10)^2/2)/(sqrt(2*pi)*x))*(exp(-(log(y)-10)^2/2)/(sqrt(2*pi)*y))
}
integration1 <- round(integrate(function(y) {
sapply(y, function(y) {
integrate(function(x) f1(x,y), 0, Inf, rel.tol = 1e-12)$value
})
}, 0, Inf, rel.tol = 1e-12)$value, 10)
This integration should be around 0.3, but R gives 0. Could anyone point out the problem? What is the best function for integral in R? Many thanks.
Package cubature can solve the problem giving the expected result. The function must be rewritten as a one argument function, and the values for x and y set in the function body.
library(cubature)
f2 <- function(X) {
x <- X[1]
y <- X[2]
pbivnorm(log(x)-10, log(y)-10, rho)*(exp(-(log(x)-10)^2/2)/(sqrt(2*pi)*x))*(exp(-(log(y)-10)^2/2)/(sqrt(2*pi)*y))
}
hcubature(f2, c(0, 0), c(Inf, Inf))
#$integral
#[1] 0.2902153
#
#$error
#[1] 2.863613e-06
#
#$functionEvaluations
#[1] 7599
#
#$returnCode
#[1] 0
Edit.
Following the OP's second comment, here is the integral computed with hcubature
f3 <- function(x) {
pnorm(log(x)-10.2)*(exp(-(log(x)-10)^2/2)/(sqrt(2*pi)*x))
}
hcubature(f3, lowerLimit = 0, upperLimit = Inf, tol = 1e-12)$integral
#[1] 0.4437685
Related
I want to calculate the following integral in R.
I tried to use Vectorize and integrate functions but I got error
Error in (log(z)) * (InIntegl2) : non-numeric argument to binary operator
fxyz= function(x,y,z) { (x*y*z)+z+x+2*y}
InIntegl1 = Vectorize(function(x) { integrate(fxyz, 0,5)$value})
InIntegl2 = Vectorize(function(y) { integrate( InIntegl1, 0,3)$value})
InIntegl3 = Vectorize(function(z) { integrate((log(z))*(InIntegl2), 2,6)$value})
Integral = integrate(InIntegl3 , 2, 6)$value
The first integral must be parameterized by y and z and the second by z. Then we can perform the final integration.
int1 <- Vectorize(function(y, z) integrate(fxyz, 0, 5, y = y, z = z)$value)
int2 <- Vectorize(function(z) integrate(int1, 0, 3, z = z)$value)
integrate(function(z) log(z) * int2(z), 2, 6)$value
## [1] 2071.71
In the spirit of Numerical Triple Integration in R
integrate(Vectorize(function(z) {
log(z)*integrate(Vectorize(function(y) {
integrate(function(x) { x*y*z +x + 2*y + z}, 0, 5)$value }), 0,3)$value }), 2,6)
Package cubature can solve triple integrals with one call.
library(cubature)
f <- function(X){
x <- X[1]
y <- X[2]
z <- X[3]
log(z)*(x*y*z + x+ 2*y + z)
}
loLim <- c(0, 0, 2)
hiLim <- c(5, 3, 6)
tol <- .Machine$double.eps^0.5
hcubature(f, loLim, hiLim, tol = tol)
#$integral
#[1] 2071.71
#
#$error
#[1] 2.059926e-05
#
#$functionEvaluations
#[1] 165
#
#$returnCode
#[1] 0
If only the integral's value is needed,
hcubature(f, loLim, hiLim, tol = tol)$integral
#[1] 2071.71
I have a function f of two variables which I want to minimize under the constraint x[1]+x[2]=1.
Here,
f <- function(x){
y <- 4*sin(x[1])+3*cos(x[2])
return(y)
}
I have read here that optim() does the work, but how do I include my constraint?
After adding the constraint x[1] + x[2] = 1, the function becomes an univariate function and you can rewrite it as the following:
f <- function(x){
4*sin(x) + 3*cos(1-x)
}
optimize() can be used on one-dimensional optimization.
opt <- optimize(0, c(0, 10))
opt
# $minimum
# [1] 4.468871
#
# $objective
# [1] -6.722745
curve(f, 0, 10)
with(opt, points(minimum, objective, col = "red", pch = 16))
I'm trying to translate the following double integral in R.
n <- 1
InnerFunc <- function(x, y) {f(y - x)*g(x)}
InnerIntegral <- Vectorize(function(y) {integrate(InnerFunc, -Inf, Inf)$value})
integrate(InnerIntegral, n, Inf)
I tried to find inspiration from another question but I'm getting mixed up in the arguments. Thanks for the help.
Thanks G5W for referring me to this post:
Numerical Triple Integration in R
I think I am able to answer my own question now.
n <- 1
integrate(Vectorize(function(y) {
integrate(Vectorize(function(x) {f(y - x)*g(x)}), -Inf, Inf)$value }), n, Inf)
I tested the syntax with
f <- function(x) {x^2}
g <- function(x) {sin(x)}
integrate(Vectorize(function(y) {
integrate(Vectorize(function(x) {f(y-x)*g(y)}), 0, 1)$value }), 0, pi)
Which gives 3.394678 as in https://www.wolframalpha.com/input/?i=integrate+(y-x)%5E2+sin+y+dx+dy,+x%3D0+to+1,+y%3D0+to+pi
At first, I have two functions like the following:
ef <- function(x, a){
if(a == 0){
return(x)
} else {
return(1-exp(-a*(5+x)))
}
}
f1 <- function(x) ef(x,a)-0.75*ef(2.5,a)-0.25*ef(-1,a)
If a is 2 (i.e. a <- 2), then the root should be:
uniroot(f1, c(-5, 0), tol = 0.0001)$root
Now my question is how to calculate the root of x of the function when a change from 0.05 to 3 by 0.05?
I think it's more flexible to put a into f1() as an argument.
f1 <- function(x, a) ef(x, a)-0.75*ef(2.5, a)-0.25*ef(-1, a)
Then use sapply() to operate each value in the sequence seq(0.05, 3, 0.05):
sapply(seq(0.05, 3, 0.05), function(A){
uniroot(f1, c(-10, 10), tol = 0.0001, extendInt = "yes", a = A)$root
})
# [1] 1.565924900 1.503659791 1.438426382 1.370549617 1.300423929
# [6] 1.228478774 1.155273229 1.081323809 1.007194271 0.933431003 ...
The argument extendInt = "yes" can conquer the error when f1() does not have different signs at the endpoints. In addition, I prefer apply family rather than a for loop in this case. You can read this for the reason.
Edit: for loop solution
a <- seq(0.05, 3, 0.05)
root <- numeric()
for(i in 1:length(a)){
root[i] <- uniroot(f1, c(-10, 10), tol = 0.0001, extendInt = "yes", a = a[i])$root
}
At the end of the loop, the variable root will store all the roots. You can check whether the outputs of the two solutions are equal.
With the goal of turning the following into a function, I was wondering how I can write the following double integral in terms of R codes?: ($\bar{x} = \mu$):
Assuming pi0 and pi1 implement your functions $\pi_0$ and $\pi_1$ in a vectorized way, a possible solution is:
integral <- function(n, mu, s, pi0, pi1) {
C <- (2 * pi)^(-n/2)
C * integrate(f = function(sigmavec) sapply(sigmavec, function(sigma) {
integrate(f = function(delta) {
exp(-n/2 * ((mu / sigma - delta)^2 + (s / sigma)^2)) * pi1(delta)
}, lower = -Inf, upper = Inf)$value
}) * pi0(sigmavec) / (sigmavec^n), lower = 0, upper = Inf)$value
}
# Tests
integral(n = 1, mu = 0, s = 1, pi0 = dnorm, pi1 = dnorm)
# [1] 0.0473819
integral(n = 1, mu = 0, s = 1, pi0 = function(sigma) 1/sigma, pi1 = dcauchy)
# [1] 0.2615783
Note sure if this question is on topic, but I am open to answer.
May be you should ask a more general question, how to write/computing integral
using computer program (code)? There at least are two ways
Using numerical integration, such as Monte Carlo method
Using symbolic toolbox to solve the problem analytically and plugin the numerical value.
Examples on $\int_0^1 x^2$
f<-function(x){
x^2
}
curve(f,0,1)
# method 1
integrate(f,lower=0,upper = 1)
# method 2
library(Ryacas)
x <- Sym("x")
f <- function(x) {
x^2
}
f2=yacas(yacas(Integrate(f(x), x)))
f2
x <- 1
Eval(f2)