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How can we find E(X^n) for a discrete random variable X in R?
Random sample from given bivariate discrete distributionGrouping functions (tapply, by, aggregate) and the *apply familyDrop data frame columns by nameGenerate a random point within a circle (uniformly)How to make a great R reproducible exampleComputing SNR with discrete random variablescalculate mean and variance for weighted discrete random variables in RHow can I view the source code for a function?Discrete probability distribution with a given maximumCreate discrete random variables to find the probability that both events A and B happen
Suppose values of a discrete random variable X, randomNumbers, and its distribution prob is given.
I can find E(X) using the following code:
weighted.mean(randomNumbers, prob)
How can we find E(X^n) in R?
Would this code work?
weighted.mean(randomNumbers^n, prob)
r statistics probability weighted-average
edited Mar 8 at 21:28
divibisan
5,03781834
asked Mar 8 at 4:44
user366312user366312
3,79946159319
add a comment |
Suppose values of a discrete random variable X, randomNumbers, and its distribution prob is given.
I can find E(X) using the following code:
weighted.mean(randomNumbers, prob)
How can we find E(X^n) in R?
Would this code work?
weighted.mean(randomNumbers^n, prob)
r statistics probability weighted-average
edited Mar 8 at 21:28
divibisan
5,03781834
asked Mar 8 at 4:44
user366312user366312
3,79946159319
add a comment |
Suppose values of a discrete random variable X, randomNumbers, and its distribution prob is given.
I can find E(X) using the following code:
weighted.mean(randomNumbers, prob)
How can we find E(X^n) in R?
Would this code work?
weighted.mean(randomNumbers^n, prob)
r statistics probability weighted-average
edited Mar 8 at 21:28
divibisan
5,03781834
asked Mar 8 at 4:44
user366312user366312
3,79946159319
Suppose values of a discrete random variable X, randomNumbers, and its distribution prob is given.
I can find E(X) using the following code:
weighted.mean(randomNumbers, prob)
How can we find E(X^n) in R?
Would this code work?
weighted.mean(randomNumbers^n, prob)
r statistics probability weighted-average
r statistics probability weighted-average
edited Mar 8 at 21:28
divibisan
5,03781834
asked Mar 8 at 4:44
user366312user366312
3,79946159319
edited Mar 8 at 21:28
divibisan
5,03781834
asked Mar 8 at 4:44
user366312user366312
3,79946159319
edited Mar 8 at 21:28
divibisan
5,03781834
edited Mar 8 at 21:28
divibisan
5,03781834
edited Mar 8 at 21:28
divibisan
5,03781834
5,03781834
asked Mar 8 at 4:44
user366312user366312
3,79946159319
asked Mar 8 at 4:44
user366312user366312
3,79946159319
asked Mar 8 at 4:44
user366312user366312
3,79946159319
3,79946159319
add a comment |
add a comment |
1 Answer
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votes
Take Poisson random variable X ~ Poisson(2) for example.
probabilistic method
f1 <- function (N)
x <- 0:N
p <- dpois(x, 2)
## approximate E[X]
m1 <- weighted.mean(x, p)
## approximate E[X ^ 2]
m2 <- weighted.mean(x ^ 2, p)
## approximate E[X ^ 3]
m3 <- weighted.mean(x ^ 3, p)
## return
c(m1, m2, m3)
As N gets bigger, approximation is more and more accurate, in the sense that the sequence converges analytically.
N <- seq(10, 200, 10)
m123_prob <- t(sapply(N, f1))
matplot(m123_prob, type = "l", lty = 1)
statistical method (sampling based method)
f2 <- function (sample_size)
x <- rpois(sample_size, 2)
## unbiased estimate of E[x]
m1 <- mean(x)
## unbiased estimate of E[x ^ 2]
m2 <- mean(x ^ 2)
## unbiased estimate of E[x ^ 3]
m3 <- mean(x ^ 3)
## return
c(m1, m2, m3)
As sample_size grows, estimation is more and more accurate, in the sense that the sequence converges in probability.
sample_size <- seq(10, 200, 10)
m123_stat <- t(sapply(sample_size, f2))
matplot(m123_stat, type = "l", lty = 1)
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Take Poisson random variable X ~ Poisson(2) for example.
probabilistic method
f1 <- function (N)
x <- 0:N
p <- dpois(x, 2)
## approximate E[X]
m1 <- weighted.mean(x, p)
## approximate E[X ^ 2]
m2 <- weighted.mean(x ^ 2, p)
## approximate E[X ^ 3]
m3 <- weighted.mean(x ^ 3, p)
## return
c(m1, m2, m3)
As N gets bigger, approximation is more and more accurate, in the sense that the sequence converges analytically.
N <- seq(10, 200, 10)
m123_prob <- t(sapply(N, f1))
matplot(m123_prob, type = "l", lty = 1)
statistical method (sampling based method)
f2 <- function (sample_size)
x <- rpois(sample_size, 2)
## unbiased estimate of E[x]
m1 <- mean(x)
## unbiased estimate of E[x ^ 2]
m2 <- mean(x ^ 2)
## unbiased estimate of E[x ^ 3]
m3 <- mean(x ^ 3)
## return
c(m1, m2, m3)
As sample_size grows, estimation is more and more accurate, in the sense that the sequence converges in probability.
sample_size <- seq(10, 200, 10)
m123_stat <- t(sapply(sample_size, f2))
matplot(m123_stat, type = "l", lty = 1)
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
add a comment |
Take Poisson random variable X ~ Poisson(2) for example.
probabilistic method
f1 <- function (N)
x <- 0:N
p <- dpois(x, 2)
## approximate E[X]
m1 <- weighted.mean(x, p)
## approximate E[X ^ 2]
m2 <- weighted.mean(x ^ 2, p)
## approximate E[X ^ 3]
m3 <- weighted.mean(x ^ 3, p)
## return
c(m1, m2, m3)
As N gets bigger, approximation is more and more accurate, in the sense that the sequence converges analytically.
N <- seq(10, 200, 10)
m123_prob <- t(sapply(N, f1))
matplot(m123_prob, type = "l", lty = 1)
statistical method (sampling based method)
f2 <- function (sample_size)
x <- rpois(sample_size, 2)
## unbiased estimate of E[x]
m1 <- mean(x)
## unbiased estimate of E[x ^ 2]
m2 <- mean(x ^ 2)
## unbiased estimate of E[x ^ 3]
m3 <- mean(x ^ 3)
## return
c(m1, m2, m3)
As sample_size grows, estimation is more and more accurate, in the sense that the sequence converges in probability.
sample_size <- seq(10, 200, 10)
m123_stat <- t(sapply(sample_size, f2))
matplot(m123_stat, type = "l", lty = 1)
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
add a comment |
Take Poisson random variable X ~ Poisson(2) for example.
probabilistic method
f1 <- function (N)
x <- 0:N
p <- dpois(x, 2)
## approximate E[X]
m1 <- weighted.mean(x, p)
## approximate E[X ^ 2]
m2 <- weighted.mean(x ^ 2, p)
## approximate E[X ^ 3]
m3 <- weighted.mean(x ^ 3, p)
## return
c(m1, m2, m3)
As N gets bigger, approximation is more and more accurate, in the sense that the sequence converges analytically.
N <- seq(10, 200, 10)
m123_prob <- t(sapply(N, f1))
matplot(m123_prob, type = "l", lty = 1)
statistical method (sampling based method)
f2 <- function (sample_size)
x <- rpois(sample_size, 2)
## unbiased estimate of E[x]
m1 <- mean(x)
## unbiased estimate of E[x ^ 2]
m2 <- mean(x ^ 2)
## unbiased estimate of E[x ^ 3]
m3 <- mean(x ^ 3)
## return
c(m1, m2, m3)
As sample_size grows, estimation is more and more accurate, in the sense that the sequence converges in probability.
sample_size <- seq(10, 200, 10)
m123_stat <- t(sapply(sample_size, f2))
matplot(m123_stat, type = "l", lty = 1)
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
Take Poisson random variable X ~ Poisson(2) for example.
probabilistic method
f1 <- function (N)
x <- 0:N
p <- dpois(x, 2)
## approximate E[X]
m1 <- weighted.mean(x, p)
## approximate E[X ^ 2]
m2 <- weighted.mean(x ^ 2, p)
## approximate E[X ^ 3]
m3 <- weighted.mean(x ^ 3, p)
## return
c(m1, m2, m3)
As N gets bigger, approximation is more and more accurate, in the sense that the sequence converges analytically.
N <- seq(10, 200, 10)
m123_prob <- t(sapply(N, f1))
matplot(m123_prob, type = "l", lty = 1)
statistical method (sampling based method)
f2 <- function (sample_size)
x <- rpois(sample_size, 2)
## unbiased estimate of E[x]
m1 <- mean(x)
## unbiased estimate of E[x ^ 2]
m2 <- mean(x ^ 2)
## unbiased estimate of E[x ^ 3]
m3 <- mean(x ^ 3)
## return
c(m1, m2, m3)
As sample_size grows, estimation is more and more accurate, in the sense that the sequence converges in probability.
sample_size <- seq(10, 200, 10)
m123_stat <- t(sapply(sample_size, f2))
matplot(m123_stat, type = "l", lty = 1)
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
answered Mar 8 at 5:45
李哲源李哲源
48.8k1498152
48.8k1498152
add a comment |
add a comment |
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