I have a relationship:
y = a + b + c
I have the average and standard deviation of a, b and c
and I would like to obtain the probability distribution of y
from this by Monte Carlo simulation.
Is there a function or package or easy way that I can use to do this?
I assume that your are assuming your inputs a,b and c are normally distributed because you say you can define them with mean and standard deviation. If that is the case, you can do this pretty fast without any special package.
mu.a=33
mu.b=32
mu.c=13
sigma.a=22
sigma.b=22
sigma.c=222
n= a.large.number=10^5
a=rnorm(n,mu.a,sigma.a)
b=rnorm(n,mu.b,sigma.b)
c=rnorm(n,mu.c,sigma.c)
y=a+b+c
plot(density(y))
mean(y)
sd(y)
Make sure to be aware of all the assumptions we are making about y,a,b and c.
If you want to do something more complex like figure out the sampling variance of the mean of y. Then do this procedure many times collecting the mean and plot it.
mysimfun=function(n,mu,sigma,stat.you.want='mean')
# mu is length 3 and sigma is too.
{
n= a.large.number=10^5
a=rnorm(n,mu[1],sigma[1])
b=rnorm(n,mu[2],sigma[2])
c=rnorm(n,mu[3],sigma[3])
y=a+b+c
plot(density(y))
return(ifelse(stat.you.want=='mean',mean(y),sd(y))
}
mu=c(mu.a,my.b,mu.c)
sigma=c(sigma.a,sigma.b,sigma.c)
mi=rep(NA,100)
Then run it in a loop of some sort.
for(i in 1:100) {mi[i]=mysimfun(10,mu,sigma,stat.you.want='mean') }
par(mfrow=c(2,1)
hist(mi)
plot(density(mi))
mean(mi)
sd(mi)
There would be two approaches: bootstrapping which I think is what you might mean by MonteCarlo or if you are more interested in the theory than constructing estimates from empiric distributions, the 'distr' package and its friends 'distrSim" and "distrTEst".
require(boot)
ax <- rnorm(100); bx<-runif(100); cx<- rexp(100)
dat <- data.frame(ax=ax,bx=bx,cx=cx)
boot(dat, function(d){ with(d, mean(ax+bx+cx) )}, R=1000, sim="parametric")
boot(dat, function(d){ with(d, sd(ax+bx+cx) )}, R=1000, sim="parametric")
Related
I am trying to implement a Monte carlo simulation method to estimate an integral in R. However, I still get wrong answer. My code is as follows:
f <- function(x){
((cos(x))/x)*exp(log(x)-3)^3
}
t <- integrate(f,0,1)
n <- 10000 #Assume we conduct 10000 simulations
int_gral <- Monte_Car(n)
int_gral
You are not doing Monte-Carlo here. Monte-Carlo is a simulation method that helps you approximating integrals using sums/mean based on random variables.
You should do something in this flavor (you might have to verify that it's correct to say that the mean of the f output can approximates your integral:
f <- function(n){
x <- runif(n)
return(
((cos(x))/x)*exp(log(x)-3)^3
)
}
int_gral <- mean(f(10000))
What your code does is taking a number n and return ((cos(n))/n)*exp(log(n)-3)^3 ; there is no randomness in that
Update
Now, to get a more precise estimates, you need to replicate this step K times. Rather than using a loop, you can use replicate function:
K <- 100
dist <- data.frame(
int = replicate(K, mean(f(10000)))
)
You get a distribution of estimators for your integral :
library(ggplot2)
ggplot(dist) + geom_histogram(aes(x = int, y = ..density..))
and you can use mean to have a numerical value:
mean(dist$int)
# [1] 2.95036e-05
You can evaluate the precision of your estimates with
sd(dist$int)
# [1] 2.296033e-07
Here it is small because N is already large, giving you a good precision of first step.
I have managed to change the codes as follows. Kindly confirm to me that I am doing the right thing.
regards.
f <- function(x){
((cos(x))/x)*exp(log(x)-3)^3
}
set.seed(234)
n<-10000
for (i in 1:10000) {
x<-runif(n)
I<-sum(f(x))/n
}
I
I have a dataframe with observed and modelled data, and I would like to calculate the R2 value. I expected there to be a function I could call for this, but can't locate one. I know I can write my own and apply it, but am I missing something obvious? I want something like
obs <- 1:5
mod <- c(0.8,2.4,2,3,4.8)
df <- data.frame(obs, mod)
R2 <- rsq(df)
# 0.85
You need a little statistical knowledge to see this. R squared between two vectors is just the square of their correlation. So you can define you function as:
rsq <- function (x, y) cor(x, y) ^ 2
Sandipan's answer will return you exactly the same result (see the following proof), but as it stands it appears more readable (due to the evident $r.squared).
Let's do the statistics
Basically we fit a linear regression of y over x, and compute the ratio of regression sum of squares to total sum of squares.
lemma 1: a regression y ~ x is equivalent to y - mean(y) ~ x - mean(x)
lemma 2: beta = cov(x, y) / var(x)
lemma 3: R.square = cor(x, y) ^ 2
Warning
R squared between two arbitrary vectors x and y (of the same length) is just a goodness measure of their linear relationship. Think twice!! R squared between x + a and y + b are identical for any constant shift a and b. So it is a weak or even useless measure on "goodness of prediction". Use MSE or RMSE instead:
How to obtain RMSE out of lm result?
R - Calculate Test MSE given a trained model from a training set and a test set
I agree with 42-'s comment:
The R squared is reported by summary functions associated with regression functions. But only when such an estimate is statistically justified.
R squared can be a (but not the best) measure of "goodness of fit". But there is no justification that it can measure the goodness of out-of-sample prediction. If you split your data into training and testing parts and fit a regression model on the training one, you can get a valid R squared value on training part, but you can't legitimately compute an R squared on the test part. Some people did this, but I don't agree with it.
Here is very extreme example:
preds <- 1:4/4
actual <- 1:4
The R squared between those two vectors is 1. Yes of course, one is just a linear rescaling of the other so they have a perfect linear relationship. But, do you really think that the preds is a good prediction on actual??
In reply to wordsforthewise
Thanks for your comments 1, 2 and your answer of details.
You probably misunderstood the procedure. Given two vectors x and y, we first fit a regression line y ~ x then compute regression sum of squares and total sum of squares. It looks like you skip this regression step and go straight to the sum of square computation. That is false, since the partition of sum of squares does not hold and you can't compute R squared in a consistent way.
As you demonstrated, this is just one way for computing R squared:
preds <- c(1, 2, 3)
actual <- c(2, 2, 4)
rss <- sum((preds - actual) ^ 2) ## residual sum of squares
tss <- sum((actual - mean(actual)) ^ 2) ## total sum of squares
rsq <- 1 - rss/tss
#[1] 0.25
But there is another:
regss <- sum((preds - mean(preds)) ^ 2) ## regression sum of squares
regss / tss
#[1] 0.75
Also, your formula can give a negative value (the proper value should be 1 as mentioned above in the Warning section).
preds <- 1:4 / 4
actual <- 1:4
rss <- sum((preds - actual) ^ 2) ## residual sum of squares
tss <- sum((actual - mean(actual)) ^ 2) ## total sum of squares
rsq <- 1 - rss/tss
#[1] -2.375
Final remark
I had never expected that this answer could eventually be so long when I posted my initial answer 2 years ago. However, given the high views of this thread, I feel obliged to add more statistical details and discussions. I don't want to mislead people that just because they can compute an R squared so easily, they can use R squared everywhere.
Why not this:
rsq <- function(x, y) summary(lm(y~x))$r.squared
rsq(obs, mod)
#[1] 0.8560185
It is not something obvious, but the caret package has a function postResample() that will calculate "A vector of performance estimates" according to the documentation. The "performance estimates" are
RMSE
Rsquared
mean absolute error (MAE)
and have to be accessed from the vector like this
library(caret)
vect1 <- c(1, 2, 3)
vect2 <- c(3, 2, 2)
res <- caret::postResample(vect1, vect2)
rsq <- res[2]
However, this is using the correlation squared approximation for r-squared as mentioned in another answer. I'm not sure why Max Kuhn didn't just use the conventional 1-SSE/SST.
caret also has an R2() method, although it's hard to find in the documentation.
The way to implement the normal coefficient of determination equation is:
preds <- c(1, 2, 3)
actual <- c(2, 2, 4)
rss <- sum((preds - actual) ^ 2)
tss <- sum((actual - mean(actual)) ^ 2)
rsq <- 1 - rss/tss
Not too bad to code by hand of course, but why isn't there a function for it in a language primarily made for statistics? I'm thinking I must be missing the implementation of R^2 somewhere, or no one cares enough about it to implement it. Most of the implementations, like this one, seem to be for generalized linear models.
You can also use the summary for linear models:
summary(lm(obs ~ mod, data=df))$r.squared
Here is the simplest solution based on [https://en.wikipedia.org/wiki/Coefficient_of_determination]
# 1. 'Actual' and 'Predicted' data
df <- data.frame(
y_actual = c(1:5),
y_predicted = c(0.8, 2.4, 2, 3, 4.8))
# 2. R2 Score components
# 2.1. Average of actual data
avr_y_actual <- mean(df$y_actual)
# 2.2. Total sum of squares
ss_total <- sum((df$y_actual - avr_y_actual)^2)
# 2.3. Regression sum of squares
ss_regression <- sum((df$y_predicted - avr_y_actual)^2)
# 2.4. Residual sum of squares
ss_residuals <- sum((df$y_actual - df$y_predicted)^2)
# 3. R2 Score
r2 <- 1 - ss_residuals / ss_total
Not sure why this isn't implemented directly in R, but this answer is essentially the same as Andrii's and Wordsforthewise, I just turned into a function for the sake of convenience if somebody uses it a lot like me.
r2_general <-function(preds,actual){
return(1- sum((preds - actual) ^ 2)/sum((actual - mean(actual))^2))
}
I am use the function MLmetrics::R2_Score from the packages MLmetrics, to compute R2 it uses the vanilla 1-(RSS/TSS) formula.
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 want to generate 2 continuous random variables Q1, Q2 (quantitative traits, each are normal) and 2 binary random variables Z1, Z2 (binary traits) with given pairwise correlations between all possible pairs of them.
Say
(Q1,Q2):0.23
(Q1,Z1):0.55
(Q1,Z2):0.45
(Q2,Z1):0.4
(Q2,Z2):0.5
(Z1,Z2):0.47
Please help me generate such data in R.
This is crude but might get you started in the right direction.
library(copula)
options(digits=3)
probs <- c(0.5,0.5)
corrs <- c(0.23,0.55,0.45,0.4,0.5,0.47) ## lower triangle
Simulate correlated values (first two quantitative, last two transformed to binary)
sim <- function(n,probs,corrs) {
tmp <- normalCopula( corrs, dim=4 , "un")
getSigma(tmp) ## test
x <- rCopula(1000, tmp)
x2 <- x
x2[,3:4] <- qbinom(x[,3:4],size=1,prob=rep(probs,each=nrow(x)))
x2
}
Test SSQ distance between observed and target correlations:
objfun <- function(corrs,targetcorrs,probs,n=1000) {
cc <- try(cor(sim(n,probs,corrs)),silent=TRUE)
if (is(cc,"try-error")) return(NA)
sum((cc[lower.tri(cc)]-targetcorrs)^2)
}
See how bad things are when input corrs=target:
cc0 <- cor(sim(1000,probs=probs,corrs=corrs))
cc0[lower.tri(cc0)]
corrs
objfun(corrs,corrs,probs=probs) ## 0.112
Now try to optimize.
opt1 <- optim(fn=objfun,
par=corrs,
targetcorrs=corrs,probs=c(0.5,0.5))
opt1$value ## 0.0208
Stops after 501 iterations with "max iterations exceeded". This will never work really well because we're trying to use a deterministic hill-climbing algorithm on a stochastic objective function ...
cc1 <- cor(sim(1000,probs=c(0.5,0.5),corrs=opt1$par))
cc1[lower.tri(cc1)]
corrs
Maybe try simulated annealing?
opt2 <- optim(fn=objfun,
par=corrs,
targetcorrs=corrs,probs=c(0.5,0.5),
method="SANN")
It doesn't seem to do much better than the previous value. Two possible problems (left as an exercise for the reader are) (1) we have specified a set of correlations that are not feasible with the marginal distributions we have chosen, or (2) the error in the objective function surface is getting in the way -- to do better we would have to average over more replicates (i.e. increase n).
I am trying to determine whether there is a significant difference between two Gamm distributions. One distribution has (shape, scale)=(shapeRef,scaleRef) while the other has (shape, scale)=(shapeTarget,scaleTarget). I try to do analysis of variance with the following code
n=10000
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
glmm1 <- gam(y~x,family=Gamma(link=log))
anova(glmm1)
The resulting p values keep changing and can be anywhere from <0.1 to >0.9.
Am I going about this the wrong way?
Edit: I use the following code instead
f <- gl(2, n)
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
xy <- c(x, y)
anova(glm(xy ~ f, family = Gamma(link = log)),test="F")
But, every time I run it I get a different p-value.
You will indeed get a different p-value every time you run this, if you pick different realizations every time. Just like your data values are random variables, which you'd expect to vary each time you ran an experiment, so is the p-value. If the null hypothesis is true (which was the case in your initial attempts), then the p-values will be uniformly distributed between 0 and 1.
Function to generate simulated data:
simfun <- function(n=100,shapeRef=2,shapeTarget=2,
scaleRef=1,scaleTarget=2) {
f <- gl(2, n)
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
xy <- c(x, y)
data.frame(xy,f)
}
Function to run anova() and extract the p-value:
sumfun <- function(d) {
aa <- anova(glm(xy ~ f, family = Gamma(link = log),data=d),test="F")
aa["f","Pr(>F)"]
}
Try it out, 500 times:
set.seed(101)
r <- replicate(500,sumfun(simfun()))
The p-values are always very small (the difference in scale parameters is easily distinguishable), but they do vary:
par(las=1,bty="l") ## cosmetic
hist(log10(r),col="gray",breaks=50)