How do I use the Binomial function to solve this experiment:
number of trials -> n=18,
p=10%
success x=2
The answer is 28% . I am using Binomial(18, 0.1) but how I pass the n=2?
julia> d=Binomial(18,0.1)
Binomial{Float64}(n=18, p=0.1)
pdf(d,2)
How can I solve this in Julia?
What you want is the Probability Mass Function, aka the probability, that in a binomial experiment of n Bernoulli independent trials with a probability p of success on each individual trial, we obtain exactly x successes.
The way to answer this question in Julia is, using the Distribution package, to first create the "distribution" object with parameters n and p, and then call the function pdf to this object and the variable x:
using Distributions
n = 18 # number of trials in our experiments
p = 0.1 # probability of success of a single trial
x = 2 # number of successes for which we want to compute the probability/PMF
binomialDistribution = Binomial(n,p)
probOfTwoSuccesses = pdf(binomialDistribution,x)
Note that all the other probability related functions (like cdf, quantile, .. but also rand) work in the same way.. you first build the distribution object, that embed the specific distribution parameters, and then you call the function over the distribution object and the variable you are looking for, e.g. quantile(binomialDistribution,0.9) for 90% quantile.
Related
I am following the book (Statistical Rethinking) which has code in R and want to reproduce the same in code in Julia. In the book, they compute the likelihood of six successes out of 9 trials where a success, has a probability of 0.5. They achieve this using the following R code.
#R Code
dbinom(6, size = 9, prob=0.5)
#Out > 0.1640625
I am wondering how to do the same in Julia,
#Julia
using Distributions
b = Binomial(9,0.5)
# Its possible to look at random value,
rand(b)
#Out > 5
But how do I look at a specific value such as six successes?
I'm sure you know this but just to be sure the r dbinom function is the probability density (mass) function for the Binomial distribution.
Julia's Distributions package makes use of multiple dispatch to just have one generic pdf function that can be called with any type of Distribution as the first argument, rather than defining a bunch of methods like dbinom, dnorm (for the Normal distribution). So you can do:
julia> using Distributions
julia> b = Binomial(9, 0.5)
Binomial{Float64}(n=9, p=0.5)
julia> pdf(b, 6)
0.1640625000000001
There is also cdf which works in the same way to calculate (maybe unsurprisingly) for the cumulative density function.
I'm new to R, and it's been a while for me and statistics. I don't understand what the parameters of rbinom are. What's is the difference between n (="number of observations") and size (="number of trials")?
This question is related, but I don't understand the answer there.
Thank you very much.
A binomial distribution usually has two parameters, an integer which indicates the number of attempts and so the maximum possible value (here called the size) and a success probability for each attempt between 0 and 1. The expectation is then the product of these two parameters.
For a random sample from this distribution, you are also interested in having a particular number of observations.
So in rbinom(n, size, prob) you have
n being the number of sample observations
size being the integer parameter of the binomial distribution, using 1 if you want a Bernoulli distribution
prob for the probability parameter of the binomial distribution
As an example, you might get
set.seed(2021)
rbinom(5, 100, 0.2)
# 19 23 22 19 21
I have a bunch of random variables (X1,....,Xn) which are i.i.d. Exp(1/2) and represent the duration of time of a certain event. So this distribution has obviously an expected value of 2, but I am having problems defining it in R. I did some research and found something about a so-called Monte-Carlo Stimulation, but I don't seem to find what I am looking for in it.
An example of what i want to estimate is: let's say we have 10 random variables (X1,..,X10) distributed as above, and we want to determine for example the probability P([X1+...+X10<=25]).
Thanks.
You don't actually need monte carlo simulation in this case because:
If Xi ~ Exp(λ) then the sum (X1 + ... + Xk) ~ Erlang(k, λ) which is just a Gamma(k, 1/λ) (in (k, θ) parametrization) or Gamma(k, λ) (in (α,β) parametrization) with an integer shape parameter k.
From wikipedia (https://en.wikipedia.org/wiki/Exponential_distribution#Related_distributions)
So, P([X1+...+X10<=25]) can be computed by
pgamma(25, shape=10, rate=0.5)
Are you aware of rexp() function in R? Have a look at documentation page by typing ?rexp in R console.
A quick answer to your Monte Carlo estimation of desired probability:
mean(rowSums(matrix(rexp(1000 * 10, rate = 0.5), 1000, 10)) <= 25)
I have generated 1000 set of 10 exponential samples, putting them into a 1000 * 10 matrix. We take row sum and get a vector of 1000 entries. The proportion of values between 0 and 25 is an empirical estimate of the desired probability.
Thanks, this was helpful! Can I use replicate with this code, to make it look like this: F <- function(n, B=1000) mean(replicate(B,(rexp(10, rate = 0.5)))) but I am unable to output the right result.
replicate here generates a matrix, too, but it is an 10 * 1000 matrix (as opposed to a 1000* 10 one in my answer), so you now need to take colSums. Also, where did you put n?
The correct function would be
F <- function(n, B=1000) mean(colSums(replicate(B, rexp(10, rate = 0.5))) <= n)
For non-Monte Carlo method to your given example, see the other answer. Exponential distribution is a special case of gamma distribution and the latter has additivity property.
I am giving you Monte Carlo method because you name it in your question, and it is applicable beyond your example.
I have the pdf of a distribution. This distribution is not a standard distribution and no functions exist in R to sample from it. How to I sample from this pdf using R?
This is more of a statistics question, as it requires sampling, but in general, you can take this approach to the problem:
Find a distribution f, whose pdf, when multiplied by any given constant k, is always greater than the pdf of the distribution in question, g.
For each sample, do the following steps:
Sample a random number x from the distribution f.
Calculate C = f(x)*k/g(x). This should be equal to or less than 1.
Draw a random number u from a uniform distribution U(0,1). If C < u, then go back to step 3. Otherwise keep x as the number and continue sampling if desired.
This process is known as rejection sampling, and is often used in random number generators that are not uniform.
The normal distribution and the uniform distribution are some of the more common distributions to sample from, but you can do other ones. Generally you want the shapes of k*f(x) and g(x) to be very close, so you don't have to reject a lot of samples.
Here's an example implementation:
#n is sample size
#g is pdf you want to sample from
#rf is sampling function for f
#df is density function for f
#k is multiplicative constant
#... is any necessary parameters for f
function.sample <- function(n,g,rf,df,k,...){
results = numeric(n)
counter = 0
while(counter < n){
x = rf(1,...)
x.pdf = df(x,...)
if (runif(0,1) >= x.pdf * k/g(x)){
results[counter+1] = x
counter = counter + 1
}
}
}
There are other methods to do random sampling, but this is usually the easiest, and it works well for most functions (unless their PDF is hard to calculate but their CDF isn't).
suppose I have the following 2 random variables :
X where mean = 6 and stdev = 3.5
Y where mean = -42 and stdev = 5
I would like to create a new random variable Z based on the first two and knowing that : X happens 90% of the time and Y happens 10% of the time.
It is easy to calculate the mean for Z : 0.9 * 6 + 0.1 * -42 = 1.2
But is it possible to generate random values for Z in a single function?
Of course, I could do something along those lines :
if (randIntBetween(1,10) > 1)
GenerateRandomNormalValue(6, 3.5);
else
GenerateRandomNormalValue(-42, 5);
But I would really like to have a single function that would act as a probability density function for such a random variable (Z) that is not necessary normal.
sorry for the crappy pseudo-code
Thanks for your help!
Edit : here would be one concrete interrogation :
Let's say we add the result of 5 consecutives values from Z. What would be the probability of ending with a number higher than 10?
But I would really like to have a
single function that would act as a
probability density function for such
a random variable (Z) that is not
necessary normal.
Okay, if you want the density, here it is:
rho = 0.9 * density_of_x + 0.1 * density_of_y
But you cannot sample from this density if you don't 1) compute its CDF (cumbersome, but not infeasible) 2) invert it (you will need a numerical solver for this). Or you can do rejection sampling (or variants, eg. importance sampling). This is costly, and cumbersome to get right.
So you should go for the "if" statement (ie. call the generator 3 times), except if you have a very strong reason not to (using quasi-random sequences for instance).
If a random variable is denoted x=(mean,stdev) then the following algebra applies
number * x = ( number*mean, number*stdev )
x1 + x2 = ( mean1+mean2, sqrt(stdev1^2+stdev2^2) )
so for the case of X = (mx,sx), Y= (my,sy) the linear combination is
Z = w1*X + w2*Y = (w1*mx,w1*sx) + (w2*my,w2*sy) =
( w1*mx+w2*my, sqrt( (w1*sx)^2+(w2*sy)^2 ) ) =
( 1.2, 3.19 )
link: Normal Distribution look for Miscellaneous section, item 1.
PS. Sorry for the wierd notation. The new standard deviation is calculated by something similar to the pythagorian theorem. It is the square root of the sum of squares.
This is the form of the distribution:
ListPlot[BinCounts[Table[If[RandomReal[] < .9,
RandomReal[NormalDistribution[6, 3.5]],
RandomReal[NormalDistribution[-42, 5]]], {1000000}], {-60, 20, .1}],
PlotRange -> Full, DataRange -> {-60, 20}]
It is NOT Normal, as you are not adding Normal variables, but just choosing one or the other with certain probability.
Edit
This is the curve for adding five vars with this distribution:
The upper and lower peaks represent taking one of the distributions alone, and the middle peak accounts for the mixing.
The most straightforward and generically applicable solution is to simulate the problem:
Run the piecewise function you have 1,000,000 (just a high number) of times, generate a histogram of the results (by splitting them into bins, and divide the count for each bin by your N (1,000,000 in my example). This will leave you with an approximation for the PDF of Z at every given bin.
Lots of unknowns here, but essentially you just wish to add the two (or more) probability functions to one another.
For any given probability function you could calculate a random number with that density by calculating the area under the probability curve (the integral) and then generating a random number between 0 and that area. Then move along the curve until the area is equal to your random number and use that as your value.
This process can then be generalized to any function (or sum of two or more functions).
Elaboration:
If you have a distribution function f(x) which ranges from 0 to 1. You could calculate a random number based on the distribution by calculating the integral of f(x) from 0 to 1, giving you the area under the curve, lets call it A.
Now, you generate a random number between 0 and A, let's call that number, r. Now you need to find a value t, such that the integral of f(x) from 0 to t is equal to r. t is your random number.
This process can be used for any probability density function f(x). Including the sum of two (or more) probability density functions.
I'm not sure what your functions look like, so not sure if you are able to calculate analytic solutions for all this, but worse case scenario, you could use numeric techniques to approximate the effect.