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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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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 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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Is there a way that I can run 100 different regressions together and get the output of all equations together in a table format?
Any software will work.
I need to find growth rates of 100 commodities using log-linear model. So I have 100 equations with dependent variable being ln(value of exports) and independent variables being time (0 to 30).
So running regression individually for 100 equations is lot of manual work.
I just require the coefficients of t for all the 100 equations. Any way to shorten the time spent doing so?
For example, assuming you have a data frame commodity_data in R with each commodity as a different column:
n <- ncol(commodity_data)
logslopes <- numeric(n)
tvec <- 0:(nrow(n)-1)
for (i in 1:n) {
m <- lm(log(commodity_data[,i]) ~ tvec)
slope <- coef(m)["tvec"]
logslopes[i] <- slope
}
There are slicker ways of doing this, but this one should work fine.
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I have distribution of parameter (natural gas mixture composition) expressed in percents. How to test such data for distribution parameters (it should be gamma, normal or lognormal distribution) and generate random composition based on that parameters in R?
This might be a better question for CrossValidated, but:
it is not generally a good idea to choose from among a range of possible distributions according to goodness of fit. Instead, you should choose according to the qualitative characteristics of your data, something like this:
Frustratingly, this chart doesn't actually have the best choice for your data (composition, continuous, bounded between 0 and 1 [or 0 and 100]), which is a Beta distribution (although there are technical issues if you have values of exactly 0 or 100 in your sample).
In R:
## some arbitrary data
z <- c(2,8,40,45,56,58,70,89)
## fit (beta values must be in (0,1), not (0,100), so divide by 100)
(m <- MASS::fitdistr(z/100,"beta",start=list(shape1=1,shape2=1)))
## sample 1000 new values
z_new <- 100*rbeta(n=1000,shape1=m$estimate["shape1"],
shape2=m$estimate["shape2"])
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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))