I am not familiar with function over a vector in R.
I would like a vector with the different values of cumulative probability of a bivariate when some parameters change value simultaneously according to different function. For example here:
library(mvtnorm)
m<-2
corr<-diag(2)
corr[2,1]<-0
vapply(2*1:3,function(x)
pmvnorm(mean=c(2,x),corr,lower=c(-Inf,-Inf), upper=c(1,2)),1)
[1] 7.932763e-02 3.609428e-03 5.024809e-06
I have the different value of cumulative probability when the mean of the second distribution takes value 2,4 and 6.
My problem is that I want simultaneously change also the value of the value of the mean of the first distribution. I can't write properly the vapply command with more than one function. What can I do?
Thank you very much
You will need to use mapply for this task
library(mvtnorm)
corr <- diag(2)
m1 <- c(3, 5, 7)
m2 <- c(2, 4, 6)
mapply(function(x, y)
pmvnorm(mean = c(x, y), corr, lower = c(-Inf, -Inf), upper = c(1, 2)),
m1, m2)
## [1] 1.1375e-02 7.2052e-07 3.1246e-14
Related
Using the mvrnorm() from the MASS package, now we can simulate realizations of multivariate normal distributions. This function works as follows:
library(MASS)
MASS::mvrnorm(
n = 10, # Number of realizations,
mu = c(1, 5), # Parameter vector mu,
Sigma = my_cov_matrix(1, 3, 0.2) # Parameter matrix Sigma
)
What does this output mean? Why are there two columns with ten random variables each?
The task is as follows:
Now, I created a function my_mvrnorm(n, mu_1, mu_2, sigma_1, sigma_2, rho), which simulates realizations of the corresponding multivariate normal distribution depending on mu and the matrix n and stores them in a tibble with the column names X and Y. In addition, this tibble is to contain a third column rho, in which all entries are filled with rho.
This should look like the following then:
But I couldn't write a function yet, because I don't quite understand what the values in table X and Y should be. Can someone help me?
Attempt:
my_mvrnorm <- function(n, mu_1, mu_2, sigma_1, sigma_2, rho){
mu = c(mu_1, mu_2)
sigma = my_cov_matrix(sigma_1, sigma_2, rho)
tb <- tibble(
X = ,
Y = ,
rho = rep(rho, n)
)
return(tb)
}
The n = 10 specification says do 10 samples. The mu = c(1, 5) specification says do two means. So, you get a 10 X 2 matrix as the result. If you check, the first column has a mean close to 2, and the second a mean close to 5. Is my_cov_matrix defined somewhere else?
Suppose I have
set.seed(2020) # make the results reproducible
a <- rnorm(100, 0, 1)
My probability density is estimated through kernel density estimator (gaussian) in R using the R built in function density. The question is how to integrate the square of the estimated density. It does not matter between which values, let us suppose between -Inf and +Inf. I have tried the following:
f <- approxfun(density(a)$x, density(a)$y)
integrate (f*f, min(density(a)$x), max(density(a)$x))
There are a couple of problems here. First you have the x and y round the wrong way in approxfun. Secondly, you can't multiply function names together. You need to specify a new function that gives you the square of your original function:
set.seed(2020)
a <- rnorm(100, 0, 1)
f <- approxfun(density(a)$x, density(a)$y)
f2 <- function(v) ifelse(is.na(f(v)), 0, f(v)^2)
integrate (f2, -Inf, Inf)
#> 0.2591153 with absolute error < 0.00011
We can also plot the original density function and the squared density function:
curve(f, -3, 3)
curve(f2, -3, 3, add = TRUE, col = "red")
I think you should write the objective function as function(x) f(x)**2, rather than f*f, e.g.,
> integrate (function(x) f(x)**2, min(density(a)$x), max(density(a)$x))
0.2331793 with absolute error < 6.6e-06
Here is a way using package caTools, function trapz. It computes the integral given a vector x and its corresponding image y using the trapezoidal rule.
I also include a function trapzf based on the original to have the integral computed with the function returned by approxfun
trapzf <- function(x, FUN) trapz(x, FUN(x))
set.seed(2020) # make the results reproducible
a <- rnorm(100, 0, 1)
d <- density(a)
f <- approxfun(d$x, d$y)
int1 <- trapz(d$x, d$y^2)
int2 <- trapzf(d$x, function(x) f(x)^2)
int1
#[1] 0.2591226
identical(int1, int2)
#[1] TRUE
I wrote a function to calculate the log-likelihood of a set of observations sampled from a mixture of two normal distributions. This function is not giving me the correct answer.
I will not know which of the two distributions any given sample is from, so the function needs to sum over possibilities.
This function takes a vector of five model parameters as its first argument (μ1, σ1, μ2, σ2 and p) where μi and σi are the mean and standard deviation of the ith distribution and p is the probability a sample is from the first distribution. For the second argument, the function takes a vector of observations.
I have written the following function:
mixloglik <- function(p, v) {
sum(log(dnorm(v, p[1], p[2])*p[5] + dnorm(v,p[3],p[4]))*p[5])
}
I can create test data, for which I know the solution should be ~ -854.6359:
set.seed(42)
v<- c(rnorm(100), rnorm(200, 8, 2))
p <- c(0, 1, 6, 2, 0.5)
When I test this function on the test data I do not get the correct solution
> mixloglik(p, v)
[1] -356.7194
I know the solution should be ~ -854.6359. Where am I going wrong in my function?
The correct expression for the log-likelihood is the following.
mixloglik <- function(p, v) {
sum(log(p[5]*dnorm(v, p[1], p[2]) + (1 - p[5])*dnorm(v, p[3], p[4])))
}
Now try it:
set.seed(42)
v<- c(rnorm(100), rnorm(200, 8, 2))
p <- c(0, 1, 6, 2, 0.5)
mixloglik(p, v)
#[1] -854.6359
In cases like this, the best way to solve the error is to restart by rewriting the expression on paper and recode it.
I am rather new to R and could use some basic help. I'd like to generate sums of two normal random variables (variance = 1 for each) as their means move apart and plot the results. The basic idea: if the means are sufficiently far apart, the distribution will be bimodal. Here's the code I'm trying:
x <- seq(-3, 3, length=500)
for(i in seq(0, 3, 0.25)) {
y <- dnorm(x, mean=0-i, sd=1)
z <- dnorm(x, mean=0+i, sd=1)
plot(x,y+z, type="l", xlim=c(-3,3))
}
Several questions:
Are there better ways to do this?
I'm only getting one PDF on my plot. How can I put multiple PDFs on the same plot?
Thank you in advance!
It is not difficult to do this using basic R features. We first define a function f to compute the density of this mixture of normal:
## `x` is an evaluation grid
## `dev` is deviation of mean from 0
f <- function (x, dev) {
(dnorm(x, -dev) + dnorm(x, dev)) / 2
}
Then we use sapply to loop through various dev to get corresponding density:
## `dev` sequence to test
dev <- seq(0, 3, 0.25)
## evaluation grid; extending `c(-1, 1) * max(dev)` by 4 standard deviation
x <- seq(-max(dev) -4, max(dev) + 4, by = 0.1)
## density matrix
X <- sapply(dev, f, x = x)
## a comment on 2022-07-31: X <- outer(x, dev, f)
Finally we use matplot for plotting:
matplot(x, X, type = "l", lty = 1)
Explanation of sapply:
During sapply, x is not changed, while we pick up and try one element of dev each iteration. It is like
X <- matrix(0, nrow = length(x), ncol = length(dev))
for (i in 1:length(dev)) X[, i] <- f(x, dev[i])
matplot(x, X) will plot columns of X one by one, against x.
A comment on 2022-07-31: Just use outer. Here are more examples:
Run a function of 2 arguments over a span of parameter values in R
Plot of a Binomial Distribution for various probabilities of success in R
I wrote a function that calculates the deciles of each row in a vector. I am doing this with the intention of creating graphics to evaluate the efficacy of a predictive model. There has to be a easier way to do this, but I haven't been able to figure it out for a while. Does anyone have any idea how I could score a vector in this way without having so many nested ifelse() statements? I included the function as well as some code to copy my results.
# function
decile <- function(x){
deciles <- vector(length=10)
for (i in seq(0.1,1,.1)){
deciles[i*10] <- quantile(x, i)
}
return (ifelse(x<deciles[1], 1,
ifelse(x<deciles[2], 2,
ifelse(x<deciles[3], 3,
ifelse(x<deciles[4], 4,
ifelse(x<deciles[5], 5,
ifelse(x<deciles[6], 6,
ifelse(x<deciles[7], 7,
ifelse(x<deciles[8], 8,
ifelse(x<deciles[9], 9, 10))))))))))
}
# check functionality
test.df <- data.frame(a = 1:10, b = rnorm(10, 0, 1))
test.df$deciles <- decile(test.df$b)
test.df
# order data frame
test.df[with(test.df, order(b)),]
You can use quantile and findInterval
# find the decile locations
decLocations <- quantile(test.df$b, probs = seq(0.1,0.9,by=0.1))
# use findInterval with -Inf and Inf as upper and lower bounds
findInterval(test.df$b,c(-Inf,decLocations, Inf))
Another solution is to use ecdf(), described in the help files as the inverse of quantile().
round(ecdf(test.df$b)(test.df$b) * 10)
Note that #mnel's solution is around 100 times faster.