I have been trying to integrate the following function over r in [0,1], but to no avail:
brownian_bridge <- function(r){X[r*(length(X)-1)+1]-r*X[length(X)]}
X is a vector of length 1000, and r is defined as
r=seq(from=0,to=1,length=1000)
Furthermore,
X=cumsum(rnorm(1000,mean=0,sd=sqrt(1/1000)))
Now my question is: How can I integrate browian_bridge over r in [0,1]? Is there a built-in R function to do this? Standard tools such as integrate2() don't seem to work, because r is a sequence and not a function that holds for all r in [0,1].
Maybe you should use Vectorize first and then apply integrate
f <- Vectorize(brownian_bridge,"r")
res <- integrate(f,0,1,subdivisions = 1e5)
such that
set.seed(1)
X=cumsum(rnorm(1000,mean=0,sd=sqrt(1/1000)))
brownian_bridge <- function(r){
X[r*(length(X)-1)+1]-r*X[length(X)]
}
f <- Vectorize(brownian_bridge,"r")
res <- integrate(f,0,1,subdivisions = 1e5)
gives
> res
0.2478581 with absolute error < 1e-04
Related
I am trying to implement the head/tail breaks classification algorithm in R (see here). This relatively new algorithm is a less computationally expensive alternative to other classification methods used in Cartography for highly skewed data.
So far, I have been looking as template a code in Python (see here) with relatively success. Here is my implementation in R:
# fake data to classify
pareto_data <- c()
for (i in 1:100){
pareto_data[i] <- (1.0/i)^1.16
}
# head/tail breaks algorithm
ht <- function(data){
ln <- length(data)
mn <- mean(data)
res <- append(c(),mn) # this is where I was hopping to store my output
head <- subset(data,data>=mn)
while (length(head)>=1 & length(head)/ln <= 0.40){
print(res)
return(ht(head))
}
#return(res)
}
ht(pareto_data)
As a result of running above code, I have been able to print the following:
[1] 0.03849691
[1] 0.1779904
[1] 0.4818454
This output is very likely the same of running the original Python code I have been using as template. However, I have not been successful in storing it in either a vector or a list.
I would be really thankful if you can give hints to overcome this problem and also to improve my code (which is not exactly the same as the original one in Python, particularly in the conditions of the while statement).
A possible recursive version of the algorithm could be the following.
ht_breaks <- function(x){
ht_inner <- function(x, mu){
n <- length(x)
mu <- c(mu, mean(x))
h <- x[x > mean(x)]
if(length(h) > 1 && length(h)/n <= 0.4){
ht_inner(h, mu)
} else mu
}
ht_inner(x, NULL)
}
pareto_data <- (1.0/(1:100))^1.16
ht_breaks(pareto_data)
#[1] 0.03849691 0.17799039 0.48184535
I am trying to write the R code for the expression given in the image for calculation purposes. I tried to use two loops and sapply function but I failed. Can anyone suggest a suitable code for the calculation of this expression?
I tried below lines given in the image.
R Code tried:
Please see as below:
gamma <- 1.5
s <- 1
k <- 3
i <- s:k
j <- lapply(i, function(x) 0:x)
prod_i <- sapply(j, function(x) prod(k + gamma - x))
f <- sum(factorial(k) / factorial(k - i) * prod_i)
f
# [1] 637.875
in numerical analysis we students are obligated to implement code in R that given a function f(x) finds its Fourier interpolation tN(x) and computes the interpolation error
$||f(x)-t_{N}(x)||=\int_{0}^{2\pi}$ $|f(x)-t_{N}(x)|^2$
or a variety of different $N$
I first tried to compute the d-coefficients according to this formular:
$d = \frac 1N M y$
with M denoting the DFT matrix and y denoting a series of equidistant function values with
$y_j = f(x_j)$ and
$x_j = e^{\frac{2*pi*i}N*j}$
for $j = 1,..,N-1$.
My goal was to come up with a sum that can be described by:
$t_{N}(x) = \Sigma_{k=0}^{N-1} d_k * e^{i*k*x}$
Which would be easier to later integrate in sort of a subsequently additive notation.
f <- function(x) 3/(6+4*cos(x)) #first function to compare with
g <- function(x) sin(32*x) #second one
xj <- function(x,n) 2*pi*x/n
M <- function(n){
w = exp(-2*pi*1i/n)
m = outer(0:(n-1),0:(n-1))
return(w^m)
}
y <- function(n){
f(xj(0:(n-1),n))
}
transformFunction <- function(n, f){
d = 1/n * t(M(n)) %*% f(xj(0:(n-1),n))
script <- paste(d[1])
for(i in 2:n)
script <- paste0(script,paste0("+",d[i],"*exp(1i*x*",i,")"))
#trans <- sum(d[1:n] * exp(1i*x*(0:(n-1))))
return(script)
}
The main purpose of the transform function was, initially, to return a function - or rather: a mathematical expression - which could then be used in order to declarate my Fourier Interpolation Function. Problem is, based on my fairly limited knowledge, that I cannot integrate functions that still have sums nested in them (which is why I commented the corresponding line in the code).
Out of absolute desperation I then tried to paste each of the summands in form of text subsequently, only to parse them again as an expression.
So the main question that remains is: how do I return mathmatical expressions in a manner that allow me to use them as a function and later on integrate them?
I am sincerely sorry for any misunderstanding or confusion, as well as my seemingly amateurish coding.
Thanks in advance!
A function in R can return any class, so specifically also objects of class function. Hence, you can make trans a function of x and return that.
Since the integrate function requires a vectorized function, we use Vectorize before outputting.
transformFunction <- function(n, f){
d = 1/n * t(M(n)) %*% f(xj(0:(n-1),n))
## Output function
trans <- function(x) sum(d[1:n] * exp(1i*x*(0:(n-1))))
## Vectorize output for the integrate function
Vectorize(trans)
}
To integrate, now simply make a new variable with the output of transformFunction:
myint <- transformFunction(n = 10,f = f)
Test: (integrate can only handle real-valued functions)
integrate(function(x) Re(myint(x)),0,2)$value +
1i*integrate(function(x) Im(myint(x)),0,2)$value
# [1] 1.091337-0.271636i
I'm looking for a way to properly integrate my function:
lik = function(par, x){
cl = c()
for(i in 1:ncluster){
sub = c()
for(j in 1:nsub){
times = t[[i]][[j]]
m = c(1,t[[i]][j],t(cov[[i]][j,]))
repmat = cbind(1,1:t[[i]][j],matrix(rep(cov[[i]][j,], times),times, 3,byrow=T))
sub[j] = d[[i]][j]*m%*%c(par[-5],x)-sum(log((1+exp(repmat%*%c(par[-5],x)))))
}
cl[i] = sum(sub)
}
return(exp(cl))
}
function lik (which is likelihood) takes x, vector par of length 5, and yields a vector of likelihood at x at each cluster. For example,
> lik(1:5,1)
[1] 4.640101e-30 3.632315e-44 5.348611e-09 1.121790e-27 1.696704e-98
> #number of clusters=5
I want to integrate out x so that I can obtain the vector of marginalized pdf at each cluster, but function integrate or any other numerical integration packages are only capable of integrating scalar function. I've searched questions relating to this, and maybe Vectorization is the key to solving this problem, but I just do not know how.
I will really appreciate if you can give me any help. Thanks
Typically I recommend converting a function of 2 variables to a function of one variable prior to integrating as follows.
myfunc <-function(x,y){ stuff}
intfunc <-function(x){myfunc(x,y)}
integrate(intfunc,x, etc)
I am using R to do some multivariate analysis. For this work I need to integrate the trivariate PDF.Since I want to use this in a MLE, a want a vector of integration. Is there a way to make Integratebring a vector instead of one value.
Here is simple example:
f1=function(x, y, z) {dmvnorm(x=as.matrix(cbind(x,y,z)), mean=c(0,0,0), sigma=sigma)}
f1(x=c(1,1,1), y=c(1,1,1), z=c(1,1,1))
integrate(Vectorize(function(x) {f1(x=c(1,1,1), y=c(1,1,1), z=c(1,1,1))}), lower = - Inf, upper = -1)$value
Error in integrate(Vectorize(function(x) { : evaluation of function gave a result of wrong length
To integrate a function of one variable, with vector values,
you can transform the function into n functions with real values,
and integrate each of them.
This is very inefficient (when integrating the i-th function,
I evaluate all the functions, and discard all but one value).
# Function to integrate
d <- rnorm(10)
f <- function(x) dnorm(d, mean=x)
# Integrate those n functions separately.
n <- length(f(1))
r <- sapply( 1:n,
function(i) integrate(
Vectorize(function(x) f(x)[i]),
lower=-Inf, upper=0
)$value
)
r
For 2-dimensional integrals, you can check pracma::integral2,
but the same manipulation (transforming a bivariate function with vector values
into n bivariate functions with real values) will probably be needed.