I tried to calculate the mean and variance of two random variables X~F(m=2,n=5) and Y~F(m=10,n=5) from their density functions(df). It would be straightforward since R has df function already, however,
> X~F <- df(1,m=2,n=5)
[1] 0.3080008
> Y~F <- df(1,m=10,n=5)
[1] 0.4954798
Numerically, mean should equal to (n-2)/n, and var should be 2n^2(m+n-2)/(m(n-2)^2(n-4) which do not match the result.
It will be super painful to integrate the whole pdf since it involves with beta distribution. Any suggestions guys?
You have formulas for the mean and variances so why not compute the mean and variance that way?
What you are doing is finding the P(X = 1) given that X ~ F(m=2,n=5) when you run F <- df(1,m=2,n=5) in R.
You can randomly drawing values from the F distribution and then use the mean() and var() function, but these answers won't be exact.
rf(n, df1, df2, ncp)
so you would fill in
rand_values<-rf(100000,2,5)
mean(rand_values)
var(rand_values)
and you should get something close to the exact values.
Related
I have a dataset with several groups, where I want to calculate a median value for each group using dplyr. The data are weighted, and the weights need to be taken into account in calculating the median. I found the weighted.median function from spatstat which seems to work fine. Consider the following simplified example:
require(spatstat, dplyr)
tst <- data.frame(group = rep(c(1:5), each = 100))
tst$val = runif(500) * tst$group
tst$wt = runif(500) * tst$val
tst %>%
group_by(group) %>%
summarise(weighted.median(val, wt))
# A tibble: 5 × 2
group `weighted.median(val, wt)`
<int> <dbl>
1 1 0.752
2 2 1.36
3 3 1.99
4 4 2.86
5 5 3.45
However, I would also like to add 95% confidence intervals to these values, and this has me stumped. Things I've considered:
Spatstat also has a weighted.var function but there's no documentation, and it's not even clear to me whether this is variance around the median or mean.
This rcompanion post suggests various methods for calculating CIs around medians, but as far as I can tell none of them handle weights.
This blog post suggests a function for calculating CIs and a median for weighted data, and is the closest I can find to what I need. However, it doesn't work with my dplyr groupings. I suppose I could write a loop to do this one group at a time and build the output data frame, but that seems cumbersome. I'm also not totally sure I understand the function in the post and slightly suspicious of its results- for instance, testing this out I get wider estimates for alpha=0.1 than for alpha=0.05, which seems backwards to me. Edit to add: upon further investigation, I think this function works as intended if I use alpha=0.95 for 95% CIs, rather than alpha = 0.05 (at least, this returns values that feel intuitively about right). I can also make it work with dplyr by editing to return just a single moe value rather than a pair of high/low estimates. So this may be a good option- but I'm also considering others.
Is there an existing function in some library somewhere that can do what I want, or an otherwise straightforward way to implement this?
There are several approaches.
You could use the asymptotic formula for standard error of the sample median. The sample median is asymptotically normal with standard error 1/sqrt(4 n f(m)) where n is the number of observations, m is the true median, and f(x) is the probability density of the (weighted) random variable. You could estimate the probability density using the base R function density.default with the weights argument. If x is the vector of observed values and w the corresponding vector of weights, then
med <- weighted.median(x, w)
f <- density(x, weights=w)
fmed <- approx(f$x, f$y, xout=med)$y
samplesize <- length(x)
se <- 1/sqrt(4 * samplesize * fmed)
ci <- med + c(-1,1) * 1.96 * se
This relies on several asymptotic approximations so it may be inaccurate. Also the sample size depends on the interpretation of the weights. In some cases the sample size could be equal to sum(w).
If there is very little data in each group, you could use the even simpler normal reference approximation,
med <- weighted.median(x, w)
v <- weighted.var(x, w)
sdm <- sqrt(pi/2) * sqrt(v)
samplesize <- length(x)
se <- sdm/sqrt(samplesize)
ci <- med + c(-1,1) * 1.96 * se
Alternatively you could use bootstrapping - generate random resamples of the input data (by choosing random resamples of the indices 1, 2, ..., n), extract the corresponding weighted observations (x_i, w_i), compute the weighted median of each resampled dataset, and construct the 95% confidence interval.
(This approach implicitly assumes the sample size is equal to n)
Using R, it is trivial to calculate the quantiles for given probabilities in a sampled distribution:
x <- rnorm(1000, mean=4, sd=2)
quantile(x, .9) # results in 6.705755
However, I can't find an easy way to do the inverse—calculate the probability for a given quantile in the sample x. The closest I've come is to use pnorm() with the same mean and standard deviation I used when creating the sample:
pnorm(5, mean=4, sd=2) # results in 0.6914625
However, because this is calculating the probability from the full normal distribution, and not the sample x, it's not entirely accurate.
Is there a function that essentially does the inverse of quantile()? Something that essentially lets me do the same thing as pnorm() but with a sample? Something like this:
backwards_quantile(x, 5)
I've found the ecdf() function, but can't figure out a way to make it result in a single probability instead of a full equation object.
ecdf returns a function: you need to apply it.
f <- ecdf(x)
f( quantile(x,.91) )
# Equivalently:
ecdf(x)( quantile(x,.91) )
Just for convenience, this function helps:
quantInv <- function(distr, value) ecdf(distr)(value)
set.seed(1)
x <- rnorm(1000, mean=4, sd=2)
quantInv(x, c(4, 5, 6.705755))
[1] 0.518 0.685 0.904
You more or less have the answer yourself. When you want to write
backwards_quantile(x, 5)
just write
ecdf(x)(5)
This corresponds to the inverse of quantile() with type=1. However, if you want other types (I favour the NIST standard, corresponding to Excel's Percentile.exc, which is type=6), you have more work to do.
In these latter cases, consider which use you are going to put it to. If all you want is to plot it, for instance, then consider
yVals<-seq(0,1,0.01)
plot(quantile(x,yVals,type=6))
But if you want the inverse for a single value, like 5, then you need to write a solving function to find the P that makes
quantile(x,P,type=6) = 5
For instance this, which uses binary search between the extreme values of x:
inverse_quantile<-function(x,y,d=0.01,type=1) {
A<-min(x)
B<-max(x)
k<-(log((B-A)/d)/log(2))+1
P=0.5
for (i in 1:k) {
P=P+ifelse((quantile(x,P,type=type)<y),2^{-i-1},-2^{-i-1})
}
P
}
So if you wanted the type 4 quantile of your set x for the number 5, with precision 0.00001, then you would write
inverse_quantile<-function(x,5,d=0.00001,type=4)
i'm comparing different measures of distance and similarity for vector profiles (Subtest results) in R, most of them are easy to compute and/or exist in dist().
Unfortunately, one that might be interesting and is to difficult for me to calculate myself is Cattel's Rp. I can not find it in R.
Does anybody know if this exists already?
Or can you help me to write a function?
The formula (Cattell 1994) of Rp is this:
(2k-d^2)/(2k + d^2)
where:
k is the median for chi square on a sample of size n;
d is the sum of the (weighted=m) difference between the two profiles,
sth like: sum(m(x(i)-y(i)));
one thing i don't know is, how to get the chi square median in there
Thank you
What i get without defining the k is:
Rp.Cattell <- function(x,y){z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);return(z)}
Vector examples are:
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
They are measures by the same device, but related to different bodyparts. They don't need to be standartised or weighted, i would say.
This page gives a general formula for k, and then gives a more thorough method using SAS/IML which pretty much gives the same results. So I used the general formula, added calculation of degrees of freedom, which leads to this:
Rp.Cattell <- function(x,y) {
dof <- (2-1) * (length(y)-1)
k <- (1-2/(9*dof))^3
z <- (2*k-sum(sum(x-y))^2)/(2*k+sum(sum(x-y))^2)
return(z)
}
x <- c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758)
y <- c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925)
Rp.Cattell(x, y)
# [1] -0.9012083
Does this figure appear to make sense?
Trying to verify the function, I found out now that the median of chisquare is the chisquare value for 50% probability - relating to random. So the function should be:
Rp.Cattell <- function(x,y){
dof <- (2-1) * (length(y)-1)
k <- qchisq(.50, df=dof)
z <- (2k-(sum(x-y))^2)/(2k+(sum(x-y))^2);
return(z)}
It is necessary though to standardize the Values before, so the results are distributed correctly.
So:
library ("stringr")
# they are centered already
x <- as.vector(scale(c(-1.2357,-1.1999,-1.4727,-0.3915,-0.2547,-0.4758),center=F, scale=T))
y <- as.vector(scale(c(0.7785,0.9357,0.7165,-0.6067,-0.4668,-0.5925),center=F, scale=T))
Rp.Cattell(x, y) -0.584423
This sounds reasonable now - or not?
I consider calculation of z is incorrect.
You need to calculate the sum of the squared differences. Not the square of the sum of differences. Besides product operator is missing in 2k.
It should be
z <- (2*k-sum((x-y)^2))/(2*k+sum((x-y)^2))
Do you agree?
I'm trying a significance test using wilcox.test in R. I want to basically test if a value x is significantly within/outside a distribution d.
I'm doing the following:
d = c(90,99,60,80,80,90,90,54,65,100,90,90,90,90,90)
wilcox.test(60,d)
Wilcoxon rank sum test with continuity correction
data: 60 and d
W = 4.5, p-value = 0.5347
alternative hypothesis: true location shift is not equal to 0
Warning message:
In wilcox.test.default(60, d) : cannot compute exact p-value with ties
and basically the p-value is the same for a big range of numbers i test.
I've tried wilcox_test() from the coin package, but i can't get it to work testing a value against a distribution.
Is there an alternative to this test that does the same and knows how to deal with ties?
How worried are you about the non-exact results? I would guess that the approximation is reasonable for a data set this size. (I did manage to get coin::wilcox_test working, and the results are not hugely different ...)
d <- c(90,99,60,80,80,90,90,54,65,100,90,90,90,90,90)
pfun <- function(x) {
suppressWarnings(w <- wilcox.test(x,d)$p.value)
return(w)
}
testvec <- 30:120
p1 <- sapply(testvec,pfun)
library("coin")
pfun2 <- function(x) {
dd <- data.frame(y=c(x,d),f=factor(c(1,rep(2,length(d)))))
return(pvalue(wilcox_test(y~f,data=dd)))
}
p2 <- sapply(testvec,pfun2)
library("exactRankTests")
pfun3 <- function(x) {wilcox.exact(x,d)$p.value}
p3 <- sapply(testvec,pfun3)
Picture:
par(las=1,bty="l")
matplot(testvec,cbind(p1,p2,p3),type="s",
xlab="value",ylab="p value of wilcoxon test",lty=1,
ylim=c(0,1),col=c(1,2,4))
legend("topright",c("stats::wilcox.test","coin::wilcox_test",
"exactRankTests::wilcox.exact"),
lty=1,col=c(1,2,4))
(exactRankTests added by request, but given that it's not maintained any more and recommends the coin package, I'm not sure how reliable it is. You're on your own for figuring out what the differences among these procedures are and which would be best to use ...)
The results make sense here -- the problem is just that your power is low. If your value is completely outside the range of the data, for n=15, that will be a probability of something like 2*(1/16)=0.125 [i.e. probability of your sample ending up as the first or the last element in a permutation], which is not quite the same as the minimum value here (wilcox.test: p=0.105, wilcox_test: p=0.08), but that might be an approximation issue, or I might have some detail wrong. Nevertheless, it's in the right ballpark.
You can do this.
wilcox.test(60,d, exact=FALSE)
Using R, it is trivial to calculate the quantiles for given probabilities in a sampled distribution:
x <- rnorm(1000, mean=4, sd=2)
quantile(x, .9) # results in 6.705755
However, I can't find an easy way to do the inverse—calculate the probability for a given quantile in the sample x. The closest I've come is to use pnorm() with the same mean and standard deviation I used when creating the sample:
pnorm(5, mean=4, sd=2) # results in 0.6914625
However, because this is calculating the probability from the full normal distribution, and not the sample x, it's not entirely accurate.
Is there a function that essentially does the inverse of quantile()? Something that essentially lets me do the same thing as pnorm() but with a sample? Something like this:
backwards_quantile(x, 5)
I've found the ecdf() function, but can't figure out a way to make it result in a single probability instead of a full equation object.
ecdf returns a function: you need to apply it.
f <- ecdf(x)
f( quantile(x,.91) )
# Equivalently:
ecdf(x)( quantile(x,.91) )
Just for convenience, this function helps:
quantInv <- function(distr, value) ecdf(distr)(value)
set.seed(1)
x <- rnorm(1000, mean=4, sd=2)
quantInv(x, c(4, 5, 6.705755))
[1] 0.518 0.685 0.904
You more or less have the answer yourself. When you want to write
backwards_quantile(x, 5)
just write
ecdf(x)(5)
This corresponds to the inverse of quantile() with type=1. However, if you want other types (I favour the NIST standard, corresponding to Excel's Percentile.exc, which is type=6), you have more work to do.
In these latter cases, consider which use you are going to put it to. If all you want is to plot it, for instance, then consider
yVals<-seq(0,1,0.01)
plot(quantile(x,yVals,type=6))
But if you want the inverse for a single value, like 5, then you need to write a solving function to find the P that makes
quantile(x,P,type=6) = 5
For instance this, which uses binary search between the extreme values of x:
inverse_quantile<-function(x,y,d=0.01,type=1) {
A<-min(x)
B<-max(x)
k<-(log((B-A)/d)/log(2))+1
P=0.5
for (i in 1:k) {
P=P+ifelse((quantile(x,P,type=type)<y),2^{-i-1},-2^{-i-1})
}
P
}
So if you wanted the type 4 quantile of your set x for the number 5, with precision 0.00001, then you would write
inverse_quantile<-function(x,5,d=0.00001,type=4)