I would like to create a Student's t distribution density plot with a mean of 0.02 instead of 0. is that possible to do?
the distribtion should have 2 degrees of freedom.
is this possible to do?
I tried the following:
X<-rnorm(100000,mean=0.02, sd=(1/sqrt(878)))
pop.mean<-mean(X)
t<-sapply(1:10000, function(x) (mean(sample(X,100))-pop.mean)/(1/sqrt(878)))
plot(density(t))
Is this approach correct?
If it is correct, how can I get the real densities, not just the approximation?
Your statement and example contradict each other somewhat.
Do you want a non-central t distribution which is based on a normal with mean 0.02? This is what your example suggests, but note that the non-central t is not just a shifted t, it is now skewed.
If you want the non-central t then you can plot it with a command like:
curve(dt(x,2,0.02), from=-5, to=6)
Or, do you want a shifted t distribution? A distribution that is symmetric around 0.02 with the shape of a t distribution?
You can plot the curve shifted by using a command like:
curve(dt(x-0.02,2), from=-5, to=6 )
The curve function has an add argument that you could use to plot both on the same plot if you want to compare them (not much difference in this case), changing the color on one of them would be suggested.
Related
I calculated the cumulative probability (in English, cdf) of my data, based on the probability of exceedance (edf). No problem at all.
However, does anyone know if there is any command to transform this data into probability density (pdf)?
I have already tested using the histogram function, but it does not work correctly.
x <- c (0.00000000, 0.03505324, 0.07005407, 0.10512053, 0.14021308,
0.17533767, 0.21051443, 0.24570116, 0.28090087, 0.31592221,
0.35092739, 0.38591441,0.42085712, 0.45599341, 0.49119521, 0.52646341,
0.56159558, 0.59673546, 0.63172464, 0.66674853, 0.70177413, 0.73712542,
0.77225123, 0.80750715, 0.84250460, 0.87720473, 0.91172191, 0.94588810,
0.98056348)
Is the function you are looking for density() ?
You can plot with
plot(density(x))
You can see x and y values with:
density(x)$x
density(x)$y
uppose that i have a poisson distribution with mean of 6 i would like to plot a probability mass function which includes an overlay of the approximating normal density.
This is what i have tried
plot( dpois( x=0:10, lambda=6 ))
this produces
which is wrong since it doesnt contain an overlay of approxiamating noral density
How do i go about this?
Something like what you seem to be asking for (I'm outlining the commands and the basic ideas, but checking the help on the functions and trying should fill in the remaining details):
taking a wider range of x-values (out to at least 13 or so) and use xlim to extend the plot slightly into the negatives (maybe to -1.5) and
plotting the pmf of the Poisson with solid dots (similar to your command but with pch=16 as an argument to plot) with a suitable color, then
call points with the same x and y arguments as above and have type=h and lty=3 to get vertical dotted lines (to give a clear impression of the relative heights, somewhat akin to the appearance of a Cleveland dot-chart); I'd use the same colour as the dots or a slightly lighter/greyer version of the dot-colour
use curve to draw the normal curve with the same mean and standard deviation as the Poisson with mean 6 (see details at the Wikipedia page for the Poisson which gives the mean and variance), but across the wider range we plotted; I'd use a slightly contrasting colour for that.
I'd draw a light x-axis in (e.g. using abline with the h argument)
Putting all those suggestions together:
(However, while it's what you're asking for it's not strictly a suitable way to compare discrete and continuous variables since density and pmf are not on the same scale, since density is not probability -- the "right" comparison between a Poisson and an approximating normal would be on the scale of the cdfs so you compare like with like -- they'd both be on the scale of probabilities then)
I have an algorithm that uses an x,y plot of sorted y data to produce an ogive.
I then derive the area under the curve to derive %'s.
I'd like to do something similar using kernel density estimation. I like how the upper/lower bounds are smoothed out using kernel densities (i.e. the min and max will extend slightly beyond my hard coded input).
Either way... I was wondering if there is a way to treat an ogive as a type of cumulative distribution function and/or use kernel density estimation to derive a cumulative distribution function given y data?
I apologize if this is a confusing question. I know there is a way to derive a cumulative frequency graph (i.e. ogive). However, I can't determine how to derive a % given this cumulative frequency graph.
What I don't want is an ecdf. I know how to do that, and I am not quite trying to capture an ecdf. But, rather integration of an ogive given two intervals.
I'm not exactly sure what you have in mind, but here's a way to calculate the area under the curve for a kernel density estimate (or more generally for any case where you have the y values at equally spaced x-values (though you can, of course, generalize to variable x intervals as well)):
library(zoo)
# Kernel density estimate
# Set n to higher value to get a finer grid
set.seed(67839)
dens = density(c(rnorm(500,5,2),rnorm(200,20,3)), n=2^5)
# How to extract the x and y values of the density estimate
#dens$y
#dens$x
# x interval
dx = median(diff(dens$x))
# mean height for each pair of y values
h = rollmean(dens$y, 2)
# Area under curve
sum(h*dx) # 1.000943
# Cumulative area
# cumsum(h*dx)
# Plot density, showing points at which density is calculated
plot(dens)
abline(v=dens$x, col="#FF000060", lty="11")
# Plot cumulative area under curve, showing mid-point of each x-interval
plot(dens$x[-length(dens$x)] + 0.5*dx, cumsum(h*dx), type="l")
abline(v=dens$x[-length(dens$x)] + 0.5*dx, col="#FF000060", lty="11")
UPDATE to include ecdf function
To address your comments, look at the two plots below. The first is the empirical cumulative distribution function (ECDF) of the mixture of normal distributions that I used above. Note that the plot of this data looks the same below as it does above. The second is a plot of the ECDF of a plain vanilla normal distribution, mean=0, sd=1.
set.seed(67839)
x = c(rnorm(500,5,2),rnorm(200,20,3))
plot(ecdf(x), do.points=FALSE)
plot(ecdf(rnorm(1000)))
I'm trying to plot an histogram of the Cauchy distribution in R using the following code:
X = rcauchy(10^5)
hist(X)
and no matter what options I try in the hist() function, I can never see more than two bars on my histogram (basically one for negative values and one for positive values).
It works fine, however, when I use the normal distribution (or others).
This results from the properties of the distribution.
Most values are relatively close to zero, but very large absolute values are much more probable than for the normal distribution. There are about 1 % values with an absolute value greater than 50, and 0.1 % greater than 500.
Try plotting only part of the values:
hist(X[abs(X)<1])
hist(X[abs(X)<5])
hist(X[abs(X)<50])
hist(X)
You can also look at the cumulative distribution function:
plot(ecdf(X))
And check the boxplot:
boxplot(X)
I have
probability values: 0.06,0.06,0.1,0.08,0.12,0.16,0.14,0.14,0.08,0.02,0.04 ,summing up to 1
the corresponding intervals where a stochastic variable may take its value with the corresponding probability from the above list:
126,162,233,304,375,446,517,588,659,730,801,839
How can I plot the probability distribution?
On the x axis, the interval values, between the intervals histogram with the probability value?
Thanks.
How about
x <- c(126,162,233,304,375,446,517,588,659,730,801,839)
p <- c(0.06,0.06,0.1,0.08,0.12,0.16,0.14,0.14,0.08,0.02,0.04)
plot(x,c(p,0),type="s")
lines(x,c(0,p),type="S")
rect(x[-1],0,x[-length(x)],p,col="lightblue")
for a quick answer? (With the rect included you might not need the lines call and might be able to change it to plot(x,p,type="n"). As usual I would recommend par(bty="l",lty=1) for my preferred graphical defaults ...)
(Explanation: "s" and "S" are two different stair-step types (see Details in ?plot): I used them both to get both the left and right boundaries of the distribution.)
edit: In your comments you say "(it) doesn't look like a histogram". It's not quite clear what you want. I added rectangles in the example above -- maybe that does it? Or you could do
b <- barplot(p,width=diff(x),space=0)
but getting the x-axis labels right is a pain.