For the following example how do I compute the percentile / probability values / tail area in julia
Example : N(1100, 200) #Normally distributed with mean 1100 & standard deviation 200 for lets say SAT score
x = 1030 #Lets say students SAT score
#manual calculation
z-score = (x-mean)/std.dev = 1030-1100 / 200 = -0.35
#using the probability table the tail area corresponding to this is 0.3632
The zscore can be computed using the stats base package.
using StatsBase
zscore([1030], 1100, 200)
# Out > 0.35
How do i compute the corresponding probability (0.3632) obtained from the statistical tables?
Turning #DNF's comment into an answer:
You can use the cdf function from Distributions.jl:
julia> using Distributions
julia> cdf(Normal(1100, 200), 1030)
0.3631693488243809
Related
I have a data set and one of columns contains random numbers raging form 300 to 400. I'm trying to find what proportion of this column in between 320 and 350 using R. To my understanding, I need to standardize this data and creates a bell curve first. I have the mean and standard deviation but when I do (X - mean)/SD and get histogram from this column it's still not a bell curve.
This the code I tried.
myData$C1 <- (myData$C1 - C1_mean) / C1_SD
If you are simply counting the number of observations in that range, there's no need to do any standardization and you may directly use
mean(myData$C1 >= 320 & myData$C1 <= 350)
As for the standardization, it definitely doesn't create any "bell curves": it only shifts the distribution (centering) and rescales the data (dividing by the standard deviation). Other than that, the shape itself of the density function remains the same.
For instance,
x <- c(rnorm(100, mean = 300, sd = 20), rnorm(100, mean = 400, sd = 20))
mean(x >= 320 & x <= 350)
# [1] 0.065
hist(x)
hist((x - mean(x)) / sd(x))
I suspect that what you are looking for is an estimate of the true, unobserved proportion. The standardization procedure then would be applicable if you had to use tabulated values of the standard normal distribution function. However, in R we may do that without anything like that. In particular,
pnorm(350, mean = mean(x), sd = sd(x)) - pnorm(320, mean = mean(x), sd = sd(x))
# [1] 0.2091931
That's the probability P(320 <= X <= 350), where X is normally distributed with mean mean(x) and standard deviation sd(x). The figure is quite different from that above since we misspecified the underlying distribution by assuming it to be normal; it actually is a mixture of two normal distributions.
I am comparing Bayes estimators to MLE in multinomial distributions. I am drawing random samples using rmultinom from a particular multinomial distribution using
rmultinom(400, size = 30, prob = c(5,7,10,8,14,10,15,12,10,9))
For each of the 400 samples, I compute the MLE and Bayes estimators for the ten probability parameters. I now want to find in each case the total variation distance between the true distribution and the one defined by the estimators.
Since for size 30 and 10 bins there are over 200 million possible arrangements, I don't think that using the theoretical definition is a good idea.
The package distrEx has a function "TotalVarDist()", but it can only be used with distributions defined in the distr package, and multinomial is not one of them. There are directions for defining them (see here and here) but the options are either to define a discrete distribution by explicitly listing the support (again, I don't think this is a good option since the support has a size of over 200 million) or starting from scratch using the same methods as how the distr package was created, which is beyond my current ability.
Any thoughts on how to do this, either using the packages mentioned or in a completely different way?
My answer is about how to calculate this using base R.
We have two multinomial parameter vectors, θ and η. The total variation distance is equivalent to P_θ(E) - P_η(E), where E={ω | P_θ({ω})>P_η({ω})}, and ω is a vector of sample counts.
I know of two ways to evaluate P(E) in base R. One is a very simple simulation-based method. The other reframes the problem in terms of a linear combination of the counts, which is approximately normally distributed, and uses the pnorm function.
Simulation-based method
You simulate samples from each distribution, check whether they're in E using the probability mass functions, and count how often they are. I'll go through an example here. We'll assume the true distribution from your question:
unnormalized.true <- c(5,7,10,8,14,10,15,12,10,9)
true <- unnormalized.true / sum(unnormalized.true)
We'll draw a sample and estimate a new distribution using a Bayes estimator:
set.seed(921)
result <- as.vector(rmultinom(1, size = 30, prob = true))
result
## [1] 3 6 2 0 5 3 3 4 1 3
dirichlet <- (result+1)/(30+length(true))
Calculating the probability of E under the true distribution:
set.seed(939)
true.dist <- rmultinom(10^6, 30, true)
p.true.e <- mean(apply(true.dist, 2, function(x)
dmultinom(x, 30, true) - dmultinom(x, 30, dirichlet) > 0))
Calculating the probability of E under the estimated distribution from the Bayes estimator:
dirichlet.dist <- rmultinom(10^6, 30, dirichlet)
p.dirichlet.e <- mean(apply(dirichlet.dist, 2, function(x)
dmultinom(x, 30, true) - dmultinom(x, 30, dirichlet) > 0))
And we can subtract to get the total variation distance.
p.true.e - p.dirichlet.e
## [1] 0.83737
Repeating this with the maximum likelihood estimate, we get a comparison of the estimators.
mle <- result/30
mle.dist <- rmultinom(10^6, 30, mle)
p.true.e2 <- mean(apply(true.dist, 2, function(x)
dmultinom(x, 30, true) - dmultinom(x, 30, mle) > 0))
p.mle.e2 <- mean(apply(mle.dist, 2, function(x)
dmultinom(x, 30, true) - dmultinom(x, 30, mle) > 0))
p.true.e2 - p.mle.e2
## [1] 0.968301
(edited to fix a serious mistake. Previously I had re-used p.true.e in the comparison with the MLE. I forgot that the event E is defined in terms of the estimated distribution.)
Normal approximation
I think this method is actually more accurate than the simulation based method, despite the normal approximation. As you'll see, we're not taking a normal approximation to the multinomial counts, which would be unlikely to be accurate for n=30. We're taking a normal approximation to a linear combination of these counts, which is close to normal. The weakness of this method will turn out to be that it can't handle zero probabilities in the estimated distribution. That's a real problem, since handling zeros gracefully is, to me, part of the point of using total variation distance rather than Kullback-Leibler divergence. But here it is.
The following derivation yields a restatement of E:
Define
where N_i is one cell of the multinomial sample, and
Then, E is the event that L>0.
The reason we have a problem with a zero probability is that it causes one of the λ_i's to be infinite.
I want to verify that L is close to normally distributed, in the example from before. I'll do that by getting samples from the distribution of L, using the previous multinomial simulation:
lambda <- log(true/dirichlet)
L.true.dist <- apply(true.dist, 2, function(x) sum(lambda*x))
L.dirichlet.dist <- apply(dirichlet.dist, 2, function(x) sum(lambda*x))
Note that I'm doing the comparison between the true distribution and the Bayes estimated distribution. I can't do the one with the MLE, because my sample had a zero count.
Plotting the distribution of L and comparing to a normal fit:
par(mfrow=c(1,2))
L.true.dist.hist <- hist(L.true.dist)
L.true.dist.fit <- function(x)
length(L.true.dist) * diff(L.true.dist.hist$breaks)[1] *
dnorm(x, mean(L.true.dist), sd=sd(L.true.dist))
curve(L.true.dist.fit, add=TRUE, n=1000, col='red')
L.dirichlet.dist.hist <- hist(L.dirichlet.dist)
L.dirichlet.dist.fit <- function(x)
length(L.dirichlet.dist) * diff(L.dirichlet.dist.hist$breaks)[1] *
dnorm(x, mean(L.dirichlet.dist), sd=sd(L.dirichlet.dist))
curve(L.dirichlet.dist.fit, add=TRUE, n=1000, col='red')
par(mfrow=c(1,1))
The distribution of L appears normal. So, instead of using simulations, we can just use pnorm. However, we need to calculate the mean and standard deviation of L. This can be done as follows.
The mean of L is
where p_i is the cell probability of cell i in the distribution p. The variance is
where
is the covariance matrix of the multinomial distribution. I'll calculate these moments for this example, and check them against the empirical moments in the simulation. First, for the distribution of L under the true distribution:
n <- 30
k <- length(true)
mean.L.true <- sum(lambda * n * true)
# Did we get the mean right?
c(mean.L.true, mean(L.true.dist))
## [1] 3.873509 3.875547
# Covariance matrix assuming the true distribution
sigma.true <- outer(1:k, 1:k, function(i,j)
ifelse(i==j, n*true[i]*(1-true[i]), -n*true[i]*true[j]))
var.L.true <- t(lambda) %*% sigma.true %*% lambda
# Did we get the standard deviation right?
c(sqrt(var.L.true), sd(L.true.dist))
## [1] 2.777787 2.776945
Then, the mean and variance of L under the Bayes estimate of the distribution:
mean.L.dirichlet <- sum(lambda * n * dirichlet)
# Did we get the mean right?
c(mean.L.dirichlet, mean(L.dirichlet.dist))
## [1] -3.893836 -3.895983
# Covariance matrix assuming the estimated distribution
sigma.dirichlet <- outer(1:k, 1:k, function(i,j)
ifelse(i==j, n*dirichlet[i]*(1-dirichlet[i]), -n*dirichlet[i]*dirichlet[j]))
var.L.dirichlet <- t(lambda) %*% sigma.dirichlet %*% lambda
# Did we get the standard deviation right?
c(sqrt(var.L.dirichlet), sd(L.dirichlet.dist))
## [1] 2.796348 2.793421
With these in hand, we can calculate the total variation distance with pnorm:
pnorm(0, mean.L.true, sd=sqrt(var.L.true), lower.tail=FALSE) -
pnorm(0, mean.L.dirichlet, sd=sqrt(var.L.true), lower.tail=FALSE)
## [1] 0.8379193
# Previous result was 0.83737
We get three digits of agreement with the simulation.
I don't know of any easy way to extend the normal approximation method to handle zero probabilities, though. I had an idea, but I got stuck trying to calculate the covariance matrix of the counts conditional on a specific cell having 0 count. I could share my progress if you think you could make something of it.
I have the following question regarding the mc2d package for Monte Carlo simulations.
Given a mc node, i.e. a mc object. How can we get the uncertainty for the values of the distribution?
For instance, as an input distribution I am using an uniform distribution, where the min is e.g. equal to 2, and the max equal to 8. Given this, we produce a mc object, apply it to mc.
The summary function produces values such as the median, mean, 97.5% etc. etc.
But as I said, how can be get an estimate of uncertainty for a given value?
Thanks in advance!
Well, you'd have to collect second momentum
Then
v = <x^2> - <x>^2
u = sqrt(v)/sqrt(N-1)
a = <x> +-u
To make things more clear, you sample events
x = 2 + (8-2)*U(0,1)
somewhere in the summary function you compute sum of events
m = m + x
so after running N events you report mean=m/N
You have to add code to collect second momentum, something like
m2 = m2 + x*x
So after run you could compute
v = m2/N - mean*mean
u = sqrt(v)/sqrt(N-1)
and report mean with uncertainty as mean +-u
I can't seem to find the correct way to simulate an AR(1) time series with a mean that is not zero.
I need 53 data points, rho = .8, mean = 300.
However, arima.sim(list(order=c(1,0,0), ar=.8), n=53, mean=300, sd=21)
gives me values in the 1500s. For example:
1480.099 1480.518 1501.794 1509.464 1499.965 1489.545 1482.367 1505.103 (and so on)
I have also tried arima.sim(n=52, model=list(ar=c(.8)), start.innov=300, n.start=1)
but then it just counts down like this:
238.81775870 190.19203239 151.91292491 122.09682547 96.27074057 [6] 77.17105923 63.15148491 50.04211711 39.68465916 32.46837830 24.78357345 21.27437183 15.93486092 13.40199333 10.99762449 8.70208879 5.62264196 3.15086491 2.13809323 1.30009732
and I have tried arima.sim(list(order=c(1,0,0), ar=.8), n=53,sd=21) + 300 which seems to give a correct answer. For example:
280.6420 247.3219 292.4309 289.8923 261.5347 279.6198 290.6622 295.0501
264.4233 273.8532 261.9590 278.0217 300.6825 291.4469 291.5964 293.5710
285.0330 274.5732 285.2396 298.0211 319.9195 324.0424 342.2192 353.8149
and so on..
However, I am in doubt that this is doing the correct thing? Is it still auto-correlating on the correct number then?
Your last option is okay to get the desired mean, "mu". It generates data from the model:
(y[t] - mu) = phi * (y[t-1] - mu) + \epsilon[t], epsilon[t] ~ N(0, sigma=21),
t=1,2,...,n.
Your first approach sets an intercept, "alpha", rather than a mean:
y[t] = alpha + phi * y[t-1] + epsilon[t].
Your second option sets the starting value y[0] equal to 300. As long as |phi|<1 the influence of this initial value will vanish after a few periods and will have no effect
on the level of the series.
Edit
The value of the standard deviation that you observe in the simulated data is correct. Be aware that the variance of the AR(1) process, y[t], is not equal the variance of the innovations, epsilon[t]. The variance of the AR(1) process, sigma^2_y, can be obtained obtained as follows:
Var(y[t]) = Var(alpha) + phi^2 * Var(y[t-1]) + Var(epsilon[t])
As the process is stationary Var(y[t]) = Var(t[t-1]) which we call sigma^2_y. Thus, we get:
sigma^2_y = 0 + phi^2 * sigma^2_y + sigma^2_epsilon
sigma^2_y = sigma^2_epsilon / (1 - phi^2)
For the values of the parameters that you are using you have:
sigma_y = sqrt(21^2 / (1 - 0.8^2)) = 35.
Use the rGARMA function in the ts.extend package
You can generate random vectors from any stationary Gaussian ARMA model using the ts.extend package. This package generates random vectors directly form the multivariate normal distribution using the computed autocorrelation matrix for the random vector, so it gives random vectors from the exact distribution and does not require "burn-in" iterations. Here is an example of generating multiple independent time-series vectors all from an AR(1) model.
#Load the package
library(ts.extend)
#Set parameters
MEAN <- 300
ERRORVAR <- 21^2
AR <- 0.8
m <- 53
#Generate n = 16 random vectors from this model
set.seed(1)
SERIES <- rGARMA(n = 16, m = m, mean = MEAN, ar = AR, errorvar = ERRORVAR)
#Plot the series using ggplot2 graphics
library(ggplot2)
plot(SERIES)
As you can see, the generated time-series vectors in this plot use the appropriate mean and error variance that were specified in the inputs.
Background
A PostgreSQL database uses PL/R to call R functions. An R call to calculate Spearman's correlation looks as follows:
cor( rank(x), rank(y) )
Also in R, a naïve calculation of a fitted generalized additive model (GAM):
data.frame( x, fitted( gam( y ~ s(x) ) ) )
Here x represents the years from 1900 to 2009 and y is the average measurement (e.g., minimum temperature) for that year.
Problem
The fitted trend line (using GAM) is reasonably accurate, as you can see in the following picture:
The problem is that the correlations (shown in the bottom left) do not accurately reflect how closely the model fits the data.
Possible Solution
One way to improve the accuracy of the correlation is to use a root mean square error (RMSE) calculation on binned data.
Questions
Q.1. How would you implement the RMSE calculation on the binned data to get a correlation (between 0 and 1) of GAM's fit to the measurements, in the R language?
Q.2. Is there a better way to find the accuracy of GAM's fit to the data, and if so, what is it (e.g., root mean square deviation)?
Attempted Solution 1
Call the PL/R function using the observed amounts and the model (GAM) amounts: correlation_rmse := climate.plr_corr_rmse( v_amount, v_model );
Define plr_corr_rmse as follows (where o and m represent the observed and modelled data): CREATE OR REPLACE FUNCTION climate.plr_corr_rmse(
o double precision[], m double precision[])
RETURNS double precision AS
$BODY$
sqrt( mean( o - m ) ^ 2 )
$BODY$
LANGUAGE 'plr' VOLATILE STRICT
COST 100;
The o - m is wrong. I'd like to bin both data sets by calculating the mean of every 5 data points (there will be at most 110 data points). For example:
omean <- c( mean(o[1:5]), mean(o[6:10]), ... )
mmean <- c( mean(m[1:5]), mean(m[6:10]), ... )
Then correct the RMSE calculation as:
sqrt( mean( omean - mmean ) ^ 2 )
How do you calculate c( mean(o[1:5]), mean(o[6:10]), ... ) for an arbitrary length vector in an appropriate number of bins (5, for example, might not be ideal for only 67 measurements)?
I don't think hist is suitable here, is it?
Attempted Solution 2
The following code will solve the problem, however it drops data points from the end of the list (to make the list divisible by 5). The solution isn't ideal as the number "5" is rather magical.
while( length(o) %% 5 != 0 ) {
o <- o[-length(o)]
}
omean <- apply( matrix(o, 5), 2, mean )
What other options are available?
Thanks in advance.
You say that:
The problem is that the correlations (shown in the bottom left) do not accurately reflect how closely the model fits the data.
You could calculate the correlation between the fitted values and the measured values:
cor(y,fitted(gam(y ~ s(x))))
I don't see why you want to bin your data, but you could do it as follows:
mean.binned <- function(y,n = 5){
apply(matrix(c(y,rep(NA,(n - (length(y) %% n)) %% n)),n),
2,
function(x)mean(x,na.rm = TRUE))
}
It looks a bit ugly, but it should handle vectors whose length is not a multiple of the binning length (i.e. 5 in your example).
You also say that:
One way to improve the accuracy of the
correlation is to use a root mean
square error (RMSE) calculation on
binned data.
I don't understand what you mean by this. The correlation is a factor in determining the mean squared error - for example, see equation 10 of Murphy (1988, Monthly Weather Review, v. 116, pp. 2417-2424). But please explain what you mean.