Is there any way for R to solve for the inverse of a given single variable function? The motivation is for me to later tell R to use a vector of values as inputs of the inverse function so that it can spit out the inverse function values.
For instance, I have the function y(x) = x^2, the inverse is y = sqrt(x). Is there a way R can solve for the inverse function?
I looked up uniroot(), but I am not solving for the zero of a function.
Any suggestions would be helpful.
Thanks!
What kind of inverse are you finding? If you're looking for a symbolic inverse (e.g., a function y that is identically equal to sqrt(x)) you're going to have to use a symbolic system. Look at ryacas for an R library to connect with a computer algebra system that can likely compute inverses, Yacas.
Now, if you need only to compute point-wise inverses, you can define your function in terms of uniroot as you've written:
> inverse = function (f, lower = -100, upper = 100) {
function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)[1]
}
> square_inverse = inverse(function (x) x^2, 0.1, 100)
> square_inverse(4)
[1] 1.999976
For a given y and f(x), this will compute x such that f(x) = y, also known as the inverse.
I cannot comment as my reputation is too low.
I am a newbie to R, and it took me a while to understand Mike's code as I was not used to the way functions are defined in his answer.
Below is Mike's code in a longer, but (to me) easier readable notation:
inverse <- function(f, lower, upper){
function(y){
uniroot(function(x){f(x) - y}, lower = lower, upper = upper, tol=1e-3)[1]
}
}
square_inverse <- inverse(function(x){x^2}, 0.1, 100)
square_inverse(4)
I hope it helps others newbies as well.
Related
I am having issues with an optimization problem involving numerical estimation of an integral which contains an unknown variable.
Numerical estimating an integral is simple enough, just use the integrate function in R. I am trying to estimate a rather unpleasant integral which requires optimization since it contains an unknown variable and a constraint. I am using the nlminb function but the result is highly incorrect. The idea is to evaluate the integral to constraint smaller or equal to 1-l, where l is between 0 and 1.
the code is the following:
integrand <- function(x, p) {dnorm(x,0,1)*(1-dnorm((qnorm(p)
-sqrt(0.12)*x)/(sqrt(1-0.12)), 0, 1))^800}
and it is the variable p which is unknown.
The objective function to be minimised is the following:
objective <- function(p){
PoD <- integrate(integrand, lower = -Inf, upper = Inf, p = p)$value
PoD - 0.5
}
test <- nlminb(0.015, objective = objective, lower = 0, upper = 1)$par*100
Edited to reflect mistakes in the objective function and the integral.
Same issue still remains.
I think my mistake is not specifying which variable to minimise. The optimisation just gives the starting value in nlminb multiplied by 100.
The authors of the paper used dummy variables and showed that a l = 0,5 should give p=0,15%.
Thank you for your time.
Of course, since your objective function does not depend on p. Do:
integrand <- function(x, p) {dnorm(x,0,1)*(1-dnorm((qnorm(p)
-sqrt(0.12)*x)/(sqrt(1-0.12)), 0, 1))^800}
objective <- function(p){
PoD <- integrate(integrand, lower = -Inf, upper = Inf, p = p)$value
PoD - 0.5
}
I'm working on an assignment for class and I'm a little lost with trying to write a function out which will calculate Newton's quotient.
This is what the questions is asking
The derivative of a function f(x) can be approximated by the Newton's quotient (f(x+h) - f(x))/h
where h is a small number. Write a function to calculate the Newton's quotient
for f(x) = exp(x). The function should take two scalar arguments, x and h.
Use a default value of h=1e-6.
Test your function at the point x=1 using the default value of h, and compare
to the true value of the derivative f'(1) = e^1.
So far I have written the code as so
x=1
newton = function(x, h = 1e-06){
quotiant = ((x+h) - x)/h
return(x = exp(x))
}
y = newton(1,h)
print(y)
I can see this is wrong, but I don't really understand how I can fix this, and what exactly I'm trying to calculate.
I have also tried this code
x=1
newton = function(x, h = 1e-06){
quotiant = ((x+h) - x)/h
}
y = newton(1,h)
print(y)
which I think gives me the right answer, but again I don't really understand what I'm calculating.
Your function doesn't evaluate the values of x and x+h using the exponential function. In your two examples you are either just returning the exponential of x, or not using the exponential function at all. What you want is something like this:
newton = function(x, h = 1e-06){
quotient = (exp(x+h) - exp(x))/h
quotient
}
newton(1)
I'm having trouble to compute and then plot multiple integral. It would be great if you could help me.
So I have this function
> f = function(x, mu = 30, s = 12){dnorm(x, mu, s)}
which i want to integrate multiple time between z(1:100) to +Inf to plot that with x=z and y = auc :
> auc = Integrate(f, z, Inf)
R return :
Warning message:
In if (is.finite(lower)) { :
the condition has length > 1 and only the first element will be used
I have tested to do a loop :
while(z < 100){
z = 1
auc = integrate(f,z,Inf)
z = z+1}
Doesn't work either ... don't know what to do
(I'm new to R , so I'm already sorry if it is really easy .. )
Thanks for your help :) !
There is no need to do the integrating by hand. pnorm gives the integral from negative infinity to the input for the normal density. You can get the upper tail instead by modifying the lower.tail parameter
z <- 1:100
y <- pnorm(z, mean = 30, sd = 12, lower.tail = FALSE)
plot(z, y)
If you're looking to integrate more complex functions then using integrate will be necessary - but if you're just looking to find probabilities for distributions then there will most likely be a function built in that does the integration for you directly.
Your problem is actually somewhat subtle, and in a certain sense gets to the core of how R works, so here is a slightly longer explanation.
R is a "vectorized" language, which means that just about everything works on vectors. If I have 2 vectors A and B, then A+B is the element-by-element sum of A and B. Nearly all R functions work this way also. If X is a vector, then Y <- exp(X) is also a vector, where each element of Y is the exponential of the corresponding element of X.
The function integrate(...) is one of the few functions in R that is not vectorized. So when you write:
f <- function(x, mu = 30, s = 12){dnorm(x, mu, s)}
auc <- integrate(f, z, Inf)
the integrate(...) function does not know what to do with z when it is a vector. So it takes the first element and complains. Hence the warning message.
There is a special function in R, Vectorize(...) that turns scalar functions into vectorized functions. You would use it this way:
f <- function(x, mu = 30, s = 12){dnorm(x, mu, s)}
auc <- Vectorize(function(z) integrate(f,z,Inf)$value)
z <- 1:100
plot(z,auc(z), type="l") # plot lines
I have a function which looks like:
g(x) = f(x) - a^b / f(x)^b
g(x) - known function, data vector provided.
f(x) - hidden process.
a,b - parameters of this function.
From the above we get the relation:
f(x) = inverse(g(x))
My goal is to optimize parameters a and b such that f(x) would be as close as possible
to a normal distribution. If we look on a f(x) Q-Q normal plot (attached), my purpose is to minimize the distance between f(x) to the straight line which represents the normal distribution, by optimizing parameters a and b.
I wrote the below code:
g_fun <- function(x) {x - a^b/x^b}
inverse = function (f, lower = 0, upper = 2000) {
function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)[1]
}
f_func = inverse(function(x) g_fun(x))
enter code here
# let's made up an example
# g(x) values are known
g <- c(-0.016339, 0.029646, -0.0255258, 0.003352, -0.053258, -0.018971, 0.005172,
0.067114, 0.026415, 0.051062)
# Calculate f(x) by using the inverse of g(x), when a=a0 and b=b0
for (i in 1:10) {
f[i] <- f_fun(g[i])
}
I have two question:
How to pass parameters a and b to the functions?
How to perform this optimization task, meaning find a and b such that f(x) would approximate normal distribution.
Not sure how you were able to produce the Q-Q plot since your provided examples do not work. You are not specifying the values of a and b and you are defining f_func but calling f_fun. Anyway here is my answer to your questions:
How to pass parameters a and b to the functions? - Just pass them as
arguments to the functions.
How to perform this optimization task, meaning find a and b such that f(x) would approximate normal distribution? - The same way any optimization task is done. Define a cost function, then minimize it.
Here is the revised code: I have added a and b as parameters, removed the inverse function and incorporated it inside f_func, which can now take vector input so no need for a for loop.
g_fun <- function(x,a,b) {x - a^b/x^b}
f_func = function(y,a,b,lower = 0, upper = 2000){
sapply(y,function(z) { uniroot(function(x) g_fun(x,a,b) - z, lower = lower, upper = upper)$root})
}
# g(x) values are known
g <- c(-0.016339, 0.029646, -0.0255258, 0.003352, -0.053258, -0.018971, 0.005172,
0.067114, 0.026415, 0.051062)
f <- f_func(g,1,1) # using a = 1 and b = 1
#[1] 0.9918427 1.0149329 0.9873386 1.0016774 0.9737270 0.9905320 1.0025893
#[8] 1.0341199 1.0132947 1.0258569
f_func(g,2,10)
[1] 1.876408 1.880554 1.875578 1.878138 1.873094 1.876170 1.878304 1.884049
[9] 1.880256 1.882544
Now for the optimization part, it depends on what you mean by f(x) would approximate normal distribution. You can compare mean square error from the qq-line if you want. Also since you say approximate, how close is good enough? You can go with shapiro.test and keep searching till you find p-value below 0.05 (be ware that there may not be a solution)
shapiro.test(f_func(g,1,2))$p
[1] 0.9484821
cost <- function(x,y) shapiro.test(f_func(g,x,y))$p
Now that we have a cost function how do we go about minimizing it. There are many many different ways to do numerical optimization. Take a look at optim function http://stat.ethz.ch/R-manual/R-patched/library/stats/html/optim.html.
optim(c(1,1),cost)
This final line does not work, but without proper data and context this is as far as I can go. Hope this helps.
I have to solve the following convolution related numerical integration problem in R or perhaps computer algebra system like Maxima.
Integral[({k(y)-l(y)}^2)dy]
where
k(.) is the pdf of a standard normal distribution
l(y)=integral[k(z)*k(z+y)dz] (standard convolution)
z and y are scalars
The domain of y is -inf to +inf.
The integral in function l(.) is an indefinite integral. Do I need to add any additional assumption on z to obtain this?
Thank you.
Here is a symbolic solution from Mathematica:
R does not do symbolic integration, just numerical integration. There is the Ryacas package which intefaces with Yacas, a symbolic math program that may help.
See the distr package for possible help with the convolution parts (it will do the convolutions, I just don't know if the result will be integrable symbolicly).
You can numerically integrate the convolutions from distr using the integrate function, but all the parameters need to be specified as numbers not variables.
For the record, here is the same problem solved with Maxima 5.26.0.
(%i2) k(u):=exp(-(1/2)*u^2)/sqrt(2*%pi) $
(%i3) integrate (k(x) * k(y + x), x, minf, inf);
(%o3) %e^-(y^2/4)/(2*sqrt(%pi))
(%i4) l(y) := ''%;
(%o4) l(y):=%e^-(y^2/4)/(2*sqrt(%pi))
(%i5) integrate ((k(y) - l(y))^2, y, minf, inf);
(%o5) ((sqrt(2)+2)*sqrt(3)-2^(5/2))/(4*sqrt(3)*sqrt(%pi))
(%i6) float (%);
(%o6) .02090706601281356
Sorry for the late reply. Leaving this here in case someone finds it by searching.
I try to do something similar in matlab, where I convolute two random (Rayleigh distributed) variables. The result of fz_fun is equal to fy_fun, I don't know why. Maybe some here knows it?
sigma1 = 0.45;
sigma2 = 0.29;
fx_fun =#(x) [0*x(x<0) , (x(x>=0)./sigma1^2).*exp(-0.5*(x(x>=0)./sigma1).^2)];
fy_fun =#(y) [0*y(y<0) , (y(y>=0)./sigma2^2).*exp(-0.5*(y(y>=0)./sigma2).^2)];
% Rayleigh distribution of random var X,Y:
step = 0.1;
x= -2:step:3;
y= -2:step:3;
%% Convolution:
z= y;
fz = zeros(size(y));
for i = 1:length(y)
fz_fun(i) = integral(#(z) fy_fun(y(i)).*fx_fun(z-y(i)),0,Inf); % probability density of random variable z= x+y
end