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I can't figure out how to plot the F distribution in R, given two degrees of freedom using standard normal variates. Any suggestions?
You can use curve()
curve(df(x, df1=1, df2=2), from=0, to=5)
Here is the documentation of curve()
df is the density of the F distribution. This can be found in ?distributions and follows the standard naming conventions dnorm for normal distribution, dt for t distribution, etc. The F distribution has two degrees of freedom parameters. Use pf if you want the CDF.
x = seq(0, 5, length = 100)
plot(x, df(x = x, df1 = 1, df2 = 1))
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Generate a vector of 1000 Poisson random numbers with λ = 3. Make a histogram and a boxplot of the 1000 numbers. Find the expected value of the vector in Rstudio
Try with this:
set.seed(123)
#Code
v <- rpois(1000,lambda = 3)
#Hist
hist(v)
#Boxplot
boxplot(v)
#Mean
mean(v)
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Good evening,
Even though I know it will "destroy" an actual normal distribution, I need to set a maximum and a minimum to a rnorm function in R.
I'm using survival rates in vultures to calculate population trends and although I need it to fluctuate, for logic reasons, survival rates can't be over 1 or under 0.
I tried doing it with if's and else's but I think there should be a better way to do it.
Thanks!
You could sample from a large normalized rnorm draw:
rbell <- function(n) {
r <- rnorm(n * 1000)
sample((r - min(r)) / diff(range(r)), n)
}
For example:
rbell(10)
#> [1] 0.5177806 0.5713479 0.5330545 0.5987649 0.3312775 0.5508946 0.3654235 0.3897417
#> [9] 0.1925600 0.6043243
hist(rbell(1000))
This will always be curtailed to the interval (0, 1).
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I would like to find a maximum economic stress scenario restricted by a limit of the mahalanobis distance of this scenario. For this, I have to consider two functions in the optimization.
To make it easier, we can work with a simplifying problem: We have a simple linear model: y=a+bx. For this I want to minimize: sum(a+bx-y)^2. But also, I have for example the restriction that: (ab*5)/2<30.
To calculate this problem with the excel solver is not a problem. But, how I get this in r?
You could try to incorporate the constraint into the objective function, like this
# example data whose exact solution lies outside the constraint
x <- runif(100, 1, 10)
y <- 3 + 5*x + rnorm(100, mean=0, sd=.5)
# big but not too big
bigConst <- sum(y^2) * 100
# if the variables lie outside the feasible region, add bigConst
f <- function(par, x, y)
sum((par["a"]+par["b"]*x-y)^2) +
if(par["a"]*par["b"]>12) bigConst else 0
# simulated annealing can deal with non-continous objective functions
sol <- optim(par=c(a=1, b=1), fn=f, method="SANN", x=x, y=y)
# this is how it looks like
plot(x,y)
abline(a=sol$par["a"], b=sol$par["b"])
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I have a book with this equation, but I am not sure how to translate this to code in R. I was wondering if someone could provide an example.
I want to generate random values from this distribution.
You can use rnorm to generate random values:
set.seed(100)
mu <- 5 # or whatever your mean is
n <- 10 # the number of random values you wish to generate.
x <- rnorm(n, mean = mu) #the function's default standard deviation is already 1
x
[1] 4.497808 5.131531 4.921083 5.886785 5.116971 5.318630 4.418209 5.714533
[9] 4.174741 4.640138
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How do I set two equations equal to each other in R to solve?
For example:
xlog(x)=8273
Find X?
Use equation in the form: x*log(x)-8273 = 0
You should have some idea of the range in which the answer lies. Then use uniroot function:
f <- function(x) (x*log(x)-8273)
uniroot(f, lower=0.1, upper=100000000)$root
[1] 1170.897
Or a more general form:
f <- function(x,y) (x*log(x)-y)
uniroot(f, y=8273, lower=0.1, upper=100000000)$root
[1] 1170.897
It turns out (with a little help from Wolfram Alpha) that this particular solution is related to the Lambert W function (which Wolfram Alpha calls the "product log" function):
library(emdbook)
exp(lambertW(8273)) ## 1170.897
The Lambert W is available in several other R packages (LambertW, spatstat, pracma, condmixt, VGAM) as well.