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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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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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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.
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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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I'm having trouble solving a normal distribution problem in R. I'm unfamiliar with the syntax and would like some help.
If X~N(2,9), compute
a. P(X>=2)
b. P(1<=X<7)
c. P(-2.5<=X<-1)
d. P(-3<=X-2<3)
You are looking for the pnorm function. This is the normal CDF. So you want to do something like:
# A
1 - pnorm(2, mean = 2, sd = 9) # = 0.5
# B
pnorm(7, mean = 2, sd = 9) - pnorm(1, mean = 2, sd = 9) # = 0.255
I think you can figure out the last two yourself.
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T(n)=T(n-1) + lgn
My approach is:
Substituting n-1,n-2,n-3
Finally we get,
T(n)=T(1) + lg 2 +lg 3 and so lg n
=> T(n) = lg(2*3*4*5 n)
Hence T(n)=lg(n!).
But they give the answer as nlgn.
Is this a problem for computing complexity? If so then both you and "they" are correct.
O(lg(n!)) = O(lg(n^n)) = O(n lg(n))
More rigorously, from Stirling formula:
lg(n!) = n lg(n) - n + O(ln(n))
Therefore
O(lg(n!)) = O(n lg(n)) + O(n) + O(ln(n)) = O(n lg(n))