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I want to know R code for generating 1000 random values for x and y between 1 to 7 such that x<=y and all the numbers are identically distributed.
With float number
x <- c()
y <- c()
for (i in 1:1000){
x[i] <- runif(1, 1, 7)
y[i] <- runif(1, x[i], 7)
// print or do something you want here
}
With integer number
x <- c()
y <- c()
for (i in 1:1000){
x[i] <- sample(1:7, 1)
y[i] <- sample(x[i]:7, 1)
// print or do something you want here
}
You can try it.
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For example, I try to solve for x: 12*exp(-2x/4)=0.5, or x^2+3x^4+9x^5=10, is there any method in R to solve such equations?
Try with uniroot function
r1 <- uniroot(function(x) 12*exp(-x/2) - 0.5, c(-1000, 1000))
r2 <- uniroot(function(x) x^2 + 3*x^4 + 9*x^5 - 10, c(-1000, 1000))
r1$root
[1] 6.356108
r2$root
[1] 0.943561
You have to define:
a function of the type f(x) = 0 (and remove the second member of equality)
an interval within which to search for the solution
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I need a non time-series dataset for evaluating various forecasting techniques in R. Please help me find a suitable dataset. I can't find any. The requirements are: one dependent variable and at least 4 continuous independent variables no factor or binary columns. Please help me out if you have such data.
You could make one yourself:
var1 <- runif(100, 10, 100)
var2 <- rnorm(100, 20, 3)
var3 <- runif(100, 0.1, 0.9)
var4 <- rnorm(100, 1000, 2000)
odds <- (var1 + var2 + 50 * var3 + var4 / 100)/100
dv <- rbinom(100, 1, odds/(1 + odds))
df <- data.frame(dv, var1, var2, var3, var4)
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What is a more elegant way to write the below function. I am trying to practice my function development skills, and I am simply trying to manually recreate the CIs of a linear model. I am very well aware of the confint(model) function, but still...
jad<-function(x, y) {
model<-lm(y~x)
std.err<-coef(summary(model))[, 2]
coef.model1<-coef(summary(model))[, 1]
upper.ci<-coef.model1+1.96*std.err
lower.ci<-coef.model1-1.96*std.err
print(upper.ci)
print(lower.ci)
}
What about the code below?
jad <- function(x, y) {
`colnames<-`(
coef(summary(lm(y ~ x)))[,1:2] %*% matrix(c(1, 1.96, 1, -1.96), nrow = 2),
c("upper.ci", "lower.ci")
)
}
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Write a program(use R) to calculate P(X + Y + Z = k) for arbitrary discrete non-negative rv’s X, Y , and Z.(rv, random variable)
This is an exercise from my book. I have no idea how to start.
Thank you for your time.
Here's a start
First, define three different lists, each containing 30 draws from unique binomial distributions (you can define X,Y, and Z as any discrete distribution you want):
X = rbinom(30, 10, 0.8)
Y = rbinom(30, 5, 0.5)
Z = rbinom(30, 8, 0.3)
Then, create a function that calculates the probability of drawing a certain number k from the added lists:
probability <- function(k) {
combine <- X+Y+Z
return(sum(combine==k)/length(combine))
}
An example call with k=14:
> probability(k=14)
[1] 0.2333333
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How do we go about numerically solving equations of the sort below using R?
Please note, this can be shown to be convex and there is a separate thread on this.
https://stats.stackexchange.com/questions/158042/convexity-of-function-of-pdf-and-cdf-of-standard-normal-random-variable
This question has been posted on the Mathematics Forum to get Closed Form or other Theoretical Approaches, but it seems numerically solutions are the way to go?
https://math.stackexchange.com/questions/2689251/solving-equations-with-standard-normal-cdf-and-pdf-optimization
You can use the build in optimize function to directly optimize the original function:
g <- function(x, xi) {
(xi * x + dnorm(xi * x) / pnorm(xi * x))
}
fun <- function(x, xi, K) {
K * g(x, xi) + (K - x) * g((K - x), xi)
}
optimize(fun, interval = c(0, 10), xi = 1, K = 1)
#> $minimum
#> [1] 1.173975
#>
#> $objective
#> [1] 1.273246
Your original problem f(x) = g(x) can be formulated as a root finding problem f(x) - g(x) = 0. You can then use the uniroot function to solve that. See ?uniroot for details.