I need to find the probability Pr(X = i), i = 2, . . . , 6, by simulation using R when two players A and B agree that the winner of a game will get 1 point and the loser 0 points; the match ends as one of the players is ahead by 2 points or the number of games reaches 6. Suppose that the probabilities of A and B winning a game are 2 3 y 1 3 , respectively, and each game is independent. Let X denote the number of games needed to end the game.
I am applying the following code:
juegos<-rbinom(6,1,2/3)
juegos
A<-cumsum(juegos)
B<-cumsum(1-juegos)
K<-abs(A-B)==2
R<-rep(0,1000)
for(i in 1:1000)
{R[i]<-which.max(K)}
R
However I don´t know what is the next step to find the probabilities when i=2, 4 and 6.
Here is one way that uses a function to simulate a single match:
# Function to simulate one match
one_match = function(p = 2/3){
g = 0
score = 0
while (g < 6){
g = g + 1
# Play one game & update score
if (runif(1) < p)
score = score + 1
else
score = score - 1
if (abs(score) == 2) break
}
return(g)
}
# Simulate matches
n_sims = 100000
outcomes = replicate(n_sims, one_match())
# Or, with a different winning probability, say p = 1/2
# outcomes = replicate(n_sims, one_match(p = 1/2))
# Estimate probabilities
probs = table(outcomes)/n_sims
print(probs)
Cheers!
Related
I try to implement a example using R in Simulation (2006, 4ed., Elsevier) by Sheldon M. Ross, which wants to generate a random permutation and reads as follows:
Suppose we are interested in generating a permutation of the numbers 1,2,... ,n
which is such that all n! possible orderings are equally likely.
The following algorithm will accomplish this by
first choosing one of the numbers 1,2,... ,n at random;
and then putting that number in position n;
it then chooses at random one of the remaining n-1 numbers and puts that number in position n-1 ;
it then chooses at random one of the remaining n-2 numbers and puts it in position n-2 ;
and so on
Surely, we can achieve a random permutation of the numbers 1,2,... ,n easily by
sample(1:n, replace=FALSE)
For example
> set.seed(0); sample(1:5, replace=FALSE)
[1] 1 4 3 5 2
However, I want to get similar results manually according to the above algorithmic steps. Then I try
## write the function
my_perm = function(n){
x = 1:n # initialize
k = n # position n
out = NULL
while(k>0){
y = sample(x, size=1) # choose one of the numbers at random
out = c(y,out) # put the number in position
x = setdiff(x,out) # the remaining numbers
k = k-1 # and so on
}
out
}
## test the function
n = 5; set.seed(0); my_perm(n) # set.seed for reproducible
and have
[1] 2 2 4 5 1
which is obviously incorrect for there are two 2 . How can I fix the problem?
You have implemented the logic correctly but there is only one thing that you need to be aware which is related to R.
From ?sample
If x has length 1, is numeric (in the sense of is.numeric) and x >= 1, sampling via sample takes place from 1:x
So when the last number is remaining in x, let's say that number is 4, sampling would take place from 1:4 and return any 1 number from it.
For example,
set.seed(0)
sample(4, 1)
#[1] 2
So you need to adjust your function for that after which the code should work correctly.
my_perm = function(n){
x = 1:n # initialize
k = n # position n
out = NULL
while(k>1){ #Stop the while loop when k = 1
y = sample(x, size=1) # choose one of the numbers at random
out = c(y,out) # put the number in position
x = setdiff(x,out) # the remaining numbers
k = k-1 # and so on
}
out <- c(x, out) #Add the last number in the output vector.
out
}
## test the function
n = 5
set.seed(0)
my_perm(n)
#[1] 3 2 4 5 1
Sample size should longer than 1. You can break it by writing a condition ;
my_perm = function(n){
x = 1:n
k = n
out = NULL
while(k>0){
if(length(x)>1){
y = sample(x, size=1)
}else{
y = x
}
out = c(y,out)
x = setdiff(x,out)
k = k-1
}
out
}
n = 5; set.seed(0); my_perm(n)
[1] 3 2 4 5 1
Let's say I have a population like {1,2,3, ..., 23} and I want to generate a sample so that the sample's mean equals 6.
I tried to use the sample function, using a custom probability vector, but it didn't work:
population <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23)
mean(population)
minimum <- min(population)
maximum <- max(population)
amplitude <- maximum - minimum
expected <- 6
n <- length(population)
prob.vector = rep(expected, each=n)
for(i in seq(1, n)) {
if(expected > population[i]) {
prob.vector[i] <- (i - minimum) / (expected - minimum)
} else {
prob.vector[i] <- (maximum - i) / (maximum - expected)
}
}
sample.size <- 5
sample <- sample(population, sample.size, prob = prob.vector)
mean(sample)
The mean of the sample is about the mean of the population (oscillates around 12), and I wanted it to be around 6.
A good sample would be:
{3,5,6,8,9}, mean=6.2
{2,3,4,8,9}, mean=5.6
The problem is different from sample integer values in R with specific mean because I have a specific population and I can't just generate arbitrary real numbers, they must be inside the population.
The plot of the probability vector:
You can try this:
m = local({b=combn(1:23,5);
d = colMeans(b);
e = b[,d>5.5 &d<6.5];
function()sample(e[,sample(ncol(e),1)])})
m()
[1] 8 5 6 9 3
m()
[1] 6 4 5 3 13
breakdown:
b=combn(1:23,5) # combine the numbers into 5
d = colMeans(b) # find all the means
e = b[,d>5.5 &d<6.5] # select only the means that are within a 0.5 range of 6
sample(e[,sample(ncol(e),1)]) # sample the values the you need
I would like to generate N random positive integers that sum to M. I would like the random positive integers to be selected around a fairly normal distribution whose mean is M/N, with a small standard deviation (is it possible to set this as a constraint?).
Finally, how would you generalize the answer to generate N random positive numbers (not just integers)?
I found other relevant questions, but couldn't determine how to apply their answers to this context:
https://stats.stackexchange.com/questions/59096/generate-three-random-numbers-that-sum-to-1-in-r
Generate 3 random number that sum to 1 in R
R - random approximate normal distribution of integers with predefined total
Normalize.
rand_vect <- function(N, M, sd = 1, pos.only = TRUE) {
vec <- rnorm(N, M/N, sd)
if (abs(sum(vec)) < 0.01) vec <- vec + 1
vec <- round(vec / sum(vec) * M)
deviation <- M - sum(vec)
for (. in seq_len(abs(deviation))) {
vec[i] <- vec[i <- sample(N, 1)] + sign(deviation)
}
if (pos.only) while (any(vec < 0)) {
negs <- vec < 0
pos <- vec > 0
vec[negs][i] <- vec[negs][i <- sample(sum(negs), 1)] + 1
vec[pos][i] <- vec[pos ][i <- sample(sum(pos ), 1)] - 1
}
vec
}
For a continuous version, simply use:
rand_vect_cont <- function(N, M, sd = 1) {
vec <- rnorm(N, M/N, sd)
vec / sum(vec) * M
}
Examples
rand_vect(3, 50)
# [1] 17 16 17
rand_vect(10, 10, pos.only = FALSE)
# [1] 0 2 3 2 0 0 -1 2 1 1
rand_vect(10, 5, pos.only = TRUE)
# [1] 0 0 0 0 2 0 0 1 2 0
rand_vect_cont(3, 10)
# [1] 2.832636 3.722558 3.444806
rand_vect(10, -1, pos.only = FALSE)
# [1] -1 -1 1 -2 2 1 1 0 -1 -1
Just came up with an algorithm to generate N random numbers greater or equal to k whose sum is S, in an uniformly distributed manner. I hope it will be of use here!
First, generate N-1 random numbers between k and S - k(N-1), inclusive. Sort them in descending order. Then, for all xi, with i <= N-2, apply x'i = xi - xi+1 + k, and x'N-1 = xN-1 (use two buffers). The Nth number is just S minus the sum of all the obtained quantities. This has the advantage of giving the same probability for all the possible combinations. If you want positive integers, k = 0 (or maybe 1?). If you want reals, use the same method with a continuous RNG. If your numbers are to be integer, you may care about whether they can or can't be equal to k. Best wishes!
Explanation: by taking out one of the numbers, all the combinations of values which allow a valid Nth number form a simplex when represented in (N-1)-space, which lies at one vertex of a (N-1)-cube (the (N-1)-cube described by the random values range). After generating them, we have to map all points in the N-cube to points in the simplex. For that purpose, I have used one method of triangulation which involves all possible permutations of coordinates in descending order. By sorting the values, we are mapping all (N-1)! simplices to only one of them. We also have to translate and scale the numbers vector so that all coordinates lie in [0, 1], by subtracting k and dividing the result by S - kN. Let us name the new coordinates yi.
Then we apply the transformation by multiplying the inverse matrix of the original basis, something like this:
/ 1 1 1 \ / 1 -1 0 \
B = | 0 1 1 |, B^-1 = | 0 1 -1 |, Y' = B^-1 Y
\ 0 0 1 / \ 0 0 1 /
Which gives y'i = yi - yi+1. When we rescale the coordinates, we get:
x'i = y'i(S - kN) + k = yi(S - kN) - yi+1(S - kN) + k = (xi - k) - (xi+1 - k) + k = xi - xi+1 + k, hence the above formula. This is applied to all elements except the last one.
Finally, we should take into account the distortion that this transformation introduces into the probability distribution. Actually, and please correct me if I'm wrong, the transformation applied to the first simplex to obtain the second should not alter the probability distribution. Here is the proof.
The probability increase at any point is the increase in the volume of a local region around that point as the size of the region tends to zero, divided by the total volume increase of the simplex. In this case, the two volumes are the same (just take the determinants of the basis vectors). The probability distribution will be the same if the linear increase of the region volume is always equal to 1. We can calculate it as the determinant of the transpose matrix of the derivative of a transformed vector V' = B-1 V with respect to V, which, of course, is B-1.
Calculation of this determinant is quite straightforward, and it gives 1, which means that the points are not distorted in any way that would make some of them more likely to appear than others.
I figured out what I believe to be a much simpler solution. You first generate random integers from your minimum to maximum range, count them up and then make a vector of the counts (including zeros).
Note that this solution may include zeros even if the minimum value is greater than zero.
Hope this helps future r people with this problem :)
rand.vect.with.total <- function(min, max, total) {
# generate random numbers
x <- sample(min:max, total, replace=TRUE)
# count numbers
sum.x <- table(x)
# convert count to index position
out = vector()
for (i in 1:length(min:max)) {
out[i] <- sum.x[as.character(i)]
}
out[is.na(out)] <- 0
return(out)
}
rand.vect.with.total(0, 3, 5)
# [1] 3 1 1 0
rand.vect.with.total(1, 5, 10)
#[1] 4 1 3 0 2
I would like to generate N random positive integers that sum to M. I would like the random positive integers to be selected around a fairly normal distribution whose mean is M/N, with a small standard deviation (is it possible to set this as a constraint?).
Finally, how would you generalize the answer to generate N random positive numbers (not just integers)?
I found other relevant questions, but couldn't determine how to apply their answers to this context:
https://stats.stackexchange.com/questions/59096/generate-three-random-numbers-that-sum-to-1-in-r
Generate 3 random number that sum to 1 in R
R - random approximate normal distribution of integers with predefined total
Normalize.
rand_vect <- function(N, M, sd = 1, pos.only = TRUE) {
vec <- rnorm(N, M/N, sd)
if (abs(sum(vec)) < 0.01) vec <- vec + 1
vec <- round(vec / sum(vec) * M)
deviation <- M - sum(vec)
for (. in seq_len(abs(deviation))) {
vec[i] <- vec[i <- sample(N, 1)] + sign(deviation)
}
if (pos.only) while (any(vec < 0)) {
negs <- vec < 0
pos <- vec > 0
vec[negs][i] <- vec[negs][i <- sample(sum(negs), 1)] + 1
vec[pos][i] <- vec[pos ][i <- sample(sum(pos ), 1)] - 1
}
vec
}
For a continuous version, simply use:
rand_vect_cont <- function(N, M, sd = 1) {
vec <- rnorm(N, M/N, sd)
vec / sum(vec) * M
}
Examples
rand_vect(3, 50)
# [1] 17 16 17
rand_vect(10, 10, pos.only = FALSE)
# [1] 0 2 3 2 0 0 -1 2 1 1
rand_vect(10, 5, pos.only = TRUE)
# [1] 0 0 0 0 2 0 0 1 2 0
rand_vect_cont(3, 10)
# [1] 2.832636 3.722558 3.444806
rand_vect(10, -1, pos.only = FALSE)
# [1] -1 -1 1 -2 2 1 1 0 -1 -1
Just came up with an algorithm to generate N random numbers greater or equal to k whose sum is S, in an uniformly distributed manner. I hope it will be of use here!
First, generate N-1 random numbers between k and S - k(N-1), inclusive. Sort them in descending order. Then, for all xi, with i <= N-2, apply x'i = xi - xi+1 + k, and x'N-1 = xN-1 (use two buffers). The Nth number is just S minus the sum of all the obtained quantities. This has the advantage of giving the same probability for all the possible combinations. If you want positive integers, k = 0 (or maybe 1?). If you want reals, use the same method with a continuous RNG. If your numbers are to be integer, you may care about whether they can or can't be equal to k. Best wishes!
Explanation: by taking out one of the numbers, all the combinations of values which allow a valid Nth number form a simplex when represented in (N-1)-space, which lies at one vertex of a (N-1)-cube (the (N-1)-cube described by the random values range). After generating them, we have to map all points in the N-cube to points in the simplex. For that purpose, I have used one method of triangulation which involves all possible permutations of coordinates in descending order. By sorting the values, we are mapping all (N-1)! simplices to only one of them. We also have to translate and scale the numbers vector so that all coordinates lie in [0, 1], by subtracting k and dividing the result by S - kN. Let us name the new coordinates yi.
Then we apply the transformation by multiplying the inverse matrix of the original basis, something like this:
/ 1 1 1 \ / 1 -1 0 \
B = | 0 1 1 |, B^-1 = | 0 1 -1 |, Y' = B^-1 Y
\ 0 0 1 / \ 0 0 1 /
Which gives y'i = yi - yi+1. When we rescale the coordinates, we get:
x'i = y'i(S - kN) + k = yi(S - kN) - yi+1(S - kN) + k = (xi - k) - (xi+1 - k) + k = xi - xi+1 + k, hence the above formula. This is applied to all elements except the last one.
Finally, we should take into account the distortion that this transformation introduces into the probability distribution. Actually, and please correct me if I'm wrong, the transformation applied to the first simplex to obtain the second should not alter the probability distribution. Here is the proof.
The probability increase at any point is the increase in the volume of a local region around that point as the size of the region tends to zero, divided by the total volume increase of the simplex. In this case, the two volumes are the same (just take the determinants of the basis vectors). The probability distribution will be the same if the linear increase of the region volume is always equal to 1. We can calculate it as the determinant of the transpose matrix of the derivative of a transformed vector V' = B-1 V with respect to V, which, of course, is B-1.
Calculation of this determinant is quite straightforward, and it gives 1, which means that the points are not distorted in any way that would make some of them more likely to appear than others.
I figured out what I believe to be a much simpler solution. You first generate random integers from your minimum to maximum range, count them up and then make a vector of the counts (including zeros).
Note that this solution may include zeros even if the minimum value is greater than zero.
Hope this helps future r people with this problem :)
rand.vect.with.total <- function(min, max, total) {
# generate random numbers
x <- sample(min:max, total, replace=TRUE)
# count numbers
sum.x <- table(x)
# convert count to index position
out = vector()
for (i in 1:length(min:max)) {
out[i] <- sum.x[as.character(i)]
}
out[is.na(out)] <- 0
return(out)
}
rand.vect.with.total(0, 3, 5)
# [1] 3 1 1 0
rand.vect.with.total(1, 5, 10)
#[1] 4 1 3 0 2
So let me start off by saying that I do not have the statistics toolbox for Matlab so I am trying to find a way to work around this. In any case, what I am trying to do is to replicate the R sample function. For example, in R
> x = sample(1:5,20,replace=T,prob=c(.1,.1,.1,.1,.6))
> x
[1] 5 5 5 4 5 2 5 5 1 5 5 5 5 5 5 3 5 1 5 5
so I am sampling the integers 1,2,3,4,5 with replacement. But furthermore, I am sampling each integer with a certain proportion, i.e., the integer 5 should be sampled about 60% of the time.
So my question that I would like to find a solution to is how to achieve this in Matlab?
Here's how you can perform weighted sampling with replacement (something Matlab's randsample doesn't support, btw);
function r = sample(pop,n,weights)
%# each weight creates a "bin" of defined size. If the value of a random number
%# falls into the bin, we pick the value
%# turn weights into a normed cumulative sum
csWeights = cumsum(weights(:))/sum(weights);
csWeights = [0;csWeights(1:end-1)];
%# for each value: pick a random number, check against weights
idx = sum(bsxfun(#ge,rand(1,n),csWeights),1);
r = pop(idx);
The unweighted case is easy using randi.
function r = sample(pop, n)
imax = length(pop);
index = randi(imax, n, 1);
r = pop(index);
end
In the weighted case, something like this should do the trick:
function r = sample(pop, n, prob)
cumprob = cumsum(prob);
r = zeros(1, n);
for i = 1:n
index = find(rand < cumprob, 1, 'last');
r(i) = pop(index);
end
end
Here's one way to make your own sample function:
function x = sample(v, n, p)
pc = cumsum(p) / sum(p);
r = rand(1,n);
x = zeros(1,n);
for i = length(pc):-1:1
x(r<pc(i)) = v(i);
end
It's not exactly efficient, but it does what you want. Call it like so:
v = [1 2 3 4 5];
p = [.1 .1 .1 .1 .6];
n = 20;
x = sample(v,n,p);