Im new to statictics and received below question that need to be answered in R language:
Simulate an i.i.d process {Xt}t=1,···,n following standard normal Xt ∼ Normal(0,1) with
sample size n = 1000 and simulation time N = 500. Compute the sample mean ̄X(1),··· , ̄X(N),
where ̄X(i) is the sample mean from the i-th simulation. Plot the histogram for ̄X(1),··· , ̄X(N).
my thought is:
as sample size n=1000, then I should
set.seed(1) # Setting a seed
X1 <- rnorm(1000) # Simulating X1
to compute the sample mean of X1-XN
result.mean <- mean(x1)
plot the histogram for mean X1-XN
plot(result.mean, type = 'h')
However I'm not sure what to do with the simulation time N = 500? the plot i generated is just 1 bar histogram, so I'm pretty sure the simulation time should be used.
what is the purpose of simulation here? and if my thought correct in the case of iid? thank you
Using randomized numbers from a normal distribution, the base (stats) r code is rnorm, with default values having a mean of 0 and standard deviation of 1. We get 500 samples from this. Then, take the mean of a vector of those 1000 numbers. We repeat that with replicate 1000 times and throw the result into a histogram.
hist(replicate(500, mean(rnorm(1000)), simplify = "vector"))
Related
I want to identify the probability of certain events occurring for a range.
Min = 600 Max = 50,000 Most frequent outcome = 600
I generated a sequence of events: numbers <- seq(600,50000,by=1)
This is where I get stuck. Not sure if using the wrong distribution or attempt at execution is going down the wrong path.
qpois(numbers, lambda = 600) produces NaNs
So the outcome desired is to be able to get an output of weighted probabilities (weighted to the mean of 600). And then be able to assess the likelihood of an outlier event about 30000 is 5% or different cuts like that by summing the probabilities for those numbers.
A bit rusty, haven't used this for a few years so any online resources to refresh is also appreciated!
Firstly, I think you're looking for ppois rather than qpois. The function qpois(p, 600) takes a vector p of probabilities. If you do qpois(0.75, 600) you will get 616, meaning that 75% of observations will be at or below 616.
ppois is the opposite of qpois. If you do ppois(616, 600) you will get (approximately) 0.75.
As for your specific distribution, it can't be a Poisson distribution. Let's see what a Poisson distribution with a mean of 600 looks like:
x <- 500:700
plot(x, dpois(x, 600), type = "h")
Getting a value of greater than even 900 has (essentially) a zero probability:
1 - ppois(900, 600)
#> [1] 0
So if your data contains values of 30,000 or 50,000 as well as 600, it's certainly not a Poisson distribution.
Without knowing more about your actual data, it's not really possible to say what distribution you have. Perhaps if you include a sample of it in your question we could be of more help.
EDIT
With the sample of numbers provided in the comments, we can have a look at the actual empirical distribution:
hist(numbers, 200)
and if we want to know the probability at any point, we can create the empirical cumulative distribution function like this:
get_probability_of <- ecdf(numbers)
This allows us to do:
number <- 1:50000
plot(number, get_probability_of(number), ylab = "probability", type = "l")
and
get_probability_of(30000)
#> [1] 0.83588
Which means that the probability of getting a number higher than 30,000 is
1 - get_probability_of(30000)
#> [1] 0.16412
However, in this case, we know how the distribution is generated, so we can calculate the exact theoretical cdf just using some simple geometry (I won't show my working here because although it is simple, it is rather lengthy, dull, and not applicable to other distributions):
cdf <- function(x) ifelse(x < 600, 0, 1 - ((49400 - (x - 600)) / 49400)^2)
and
cdf(30000)
#> [1] 0.8360898
which is very close to, but more theoretically accurate than the empirical value.
Objective: The overall objective of the problem is to calculate the confidence interval (CI) of various sample sizes (n=2,4..1024) of rnorm, 10,000 times and then count the number of times each one fails (this likely requires a counter and an if/else statement). Finally the results are to be plotted
I am trying to calculate CI of the means for several simulations of a sample sizes, however, I am first trying to break down the code for one specific sample size a = 8.
The problem I have is that I do not know how to generate a linear model for each row. Would anyone know how I can do this? Here is what I have so far:
a <- 8
n.sim.3 <- 10000
for ( i in a) {
r.mat <- matrix(rnorm(i*n.sim.3), nrow=n.sim.3, ncol = a)
lm.tmp <- apply(three.mat,1,lm(n.sim.3~1) # The lm command is where I'm stuck I don't think this is correct)
confint.tmp <- confint(lm.tmp)
I want to simulate demand values that follows different distributions (ex above: starts of linear> exponential>invlog>etc) I'm a bit confused by the notion of probability distributions but thought I could use rnorm, rexp, rlogis, etc. Is there any way I could do so?
I think it may be this but in R: Generating smoothed randoms that follow a distribution
Simulating random values from commonly-used probability distributions in R is fairly trivial using rnorm(), rexp(), etc, if you know what distribution you want to use, as well as its parameters. For example, rnorm(10, mean=5, sd=2) returns 10 draws from a normal distribution with mean 5 and sd 2.
rnorm(10, mean = 5, sd = 2)
## [1] 5.373151 7.970897 6.933788 5.455081 6.346129 5.767204 3.847219 7.477896 5.860069 6.154341
## or here's a histogram of 10000 draws...
hist(rnorm(10000, 5, 2))
You might be interested in an exponential distribution - check out hist(rexp(10000, rate=1)) to get the idea.
The easiest solution will be to investigate what probability distribution(s) you're interested in and their implementation in R.
It is still possible to return random draws from some custom function, and there are a few techniques out there for doing it - but it might get messy. Here's a VERY rough implementation of drawing randomly from probabilities defined by the region of x^3 - 3x^2 + 4 between zero and 3.
## first a vector of random uniform draws from the region
unifdraws <- runif(10000, 0, 3)
## assign a probability of "keeping" draws based on scaled probability
pkeep <- (unifdraws^3 - 3*unifdraws^2 + 4)/4
## randomly keep observations based on this probability
keep <- rbinom(10000, size=1, p=pkeep)
draws <- unifdraws[keep==1]
## and there it is!
hist(draws)
## of course, it's less than 10000 now, because we rejected some
length(draws)
## [1] 4364
My question is quite simple, I'm trying to simulate 500 draws from any distribution using sample().
Let's take binomial distribution B(10,0.5) for example. Using the sample() function in R.
This is what I did:
draw = 1:500
data = sample(x=draw, size=10, replace=TRUE, prob=rep(0.5, each=500))
However whenever I draw the historgram, it looks like it's random and doesn't have a binomial distribution. What am I doing wrong?
Note: I know there is rbinom() function in r that does this. I am trying to understand how the sample() function works.
data = rbinom(n=500,size=10,prob=.5)
hist(data)
sample(x = c(1,0),size = 10,replace=TRUE,prob = c(0.5,0.5))
You may want to histogram the sum of this vector generated multiple times to see your binomial distribution.
draws=c()
for(i in 1:500){
draws=c(draws,sum(sample(x = c(1,0),size = 10,replace=TRUE,prob = c(0.5,0.5))))
}
hist(draws)
In this example sample is returning 10 (size = 10) samples of the values 1 or 0 (x = c(1,0)), with equal probability for each (prob = c(0.5,0.5)). replace=TRUE just means that either item can be draw more than once. These 1's and 0's are the results of 10 bernoulli trials with probability 0.5. The binomial distribution is the probability distribution of the number of successes (1's) in a series of n Bernoulli trials each with probability p. So (n=10 and p=0.5). Calling sample once gives the 10 draws, summing that vector gives a draw from the binomial. We take 500 draws from that binomial distribution and draw a histogram.
I'm very new to coding in R(coding in general). I've created a distribution using a random walk within the following code:
set.seed(124)
norm <- rnorm(1000)
mean(norm)
mean(norm)^2
sd(norm)
d <- density(norm)
plot(d)
Now I want to create a function of n-steps using the above distribution. The function calculates the expected values based on the probability of moving n-steps to the left or right from the center. I have no idea where to begin.
Any direction would be greatly appreciated.
Thanks
If each normally distributed variate is your step size (positive moves right and negative moves left), then the cumulative sum of your random draws represents your current position. You can compute that with the cumsum function in R:
set.seed(144)
pos <- cumsum(rnorm(1000))
plot(seq_along(pos), pos, xlab="Step Number", ylab="Current Position")
Using replicate and logical operations, you can simulate any number of different questions about random walks. For instance "with what probability does the value of the random walk exceed 100 within the first 1000 steps" could be simulated with:
set.seed(144)
exceed.100 <- replicate(100000, any(cumsum(rnorm(1000)) >= 100))
mean(exceed.100)
# [1] 0.00173
From these 100k replicates, it looks like the probability is around 0.17% that the random walk will exceed 100 during the first 1000 steps.