I have to compute the probability of Poisson distribution in Julia. I just know how to get Poisson distribution. But i have to compute the probability. also i have lambda from 20 to 100.
using Distributions
Poisson()
The objects in Distributions.jl are like random variables. If you declare a value to be of a distribution, you can sample from it using rand, but there are a whole lot of other methods you can apply to it. Among them is pdf:
julia> X = Poisson(30)
Distributions.Poisson{Float64}(λ=30.0)
julia> pdf(X, 2)
4.2109303359780846e-11
julia> pdf(X, 0:1:10)
11-element Array{Float64,1}:
9.35762e-14
2.80729e-12
4.21093e-11
4.21093e-10
3.1582e-9
1.89492e-8
9.47459e-8
4.06054e-7
1.5227e-6
5.07567e-6
1.5227e-5
Related
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.
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 am trying to calculate the density function of a continuos random variable in range in Julia using Distributions, but I am not able to define the range. I used Truncator constructor to construct the distribution, but I have no idea how to define the range. By density function I mean P(a
Would appreciate any help. The distribution I'm using is Gamma btw!
Thanks
To get the maximum and minimum of the support of distribution d just write maximum(d) and minimum(d) respectively. Note that for some distributions this might be infinity, e.g. maximum(Normal()) is Inf.
What version of Julia and Distributions du you use? In Distribution v0.16.4, it can be easily defined with the second and third arguments of Truncated.
julia> a = Gamma()
Gamma{Float64}(α=1.0, θ=1.0)
julia> b = Truncated(a, 2, 3)
Truncated(Gamma{Float64}(α=1.0, θ=1.0), range=(2.0, 3.0))
julia> p = rand(b, 1000);
julia> extrema(p)
(2.0007680527633305, 2.99864177354943)
You can see the document of Truncated by typing ?Truncated in REPL and enter.
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 want to compute the expected value of a multivariate function f(x) wrt to dirichlet distribution. My problem is "penta-nomial" (i.e 5 variables) so calculating the explicit form of the expected value seems unreasonable. Is there a way to numerically integrate it efficiently?
f(x) = \sum_{0,4}(x_i*log(n/x_i))
x = <x_0, x_1, x_2, x_3, x_4> and n is a constant