I am using R to run a simulation in which I use a likelihood ratio test to compare two nested item response models. One version of the LRT uses the joint likelihood function L(θ,ρ) and the other uses the marginal likelihood function L(ρ). I want to integrate L(θ,ρ) over f(θ) to obtain the marginal likelihood L(ρ). I have two conditions: in one, f(θ) is standard normal (μ=0,σ=1), and my understanding is that I can just pick a number of abscissa points, say 20 or 30, and use Gauss-Hermite quadrature to approximate this density. But in the other condition, f(θ) is a linearly transformed beta distribution (a=1.25,b=10), where the linear transformation B'=11.14*(B-0.11) is such that B' also has (approximately) μ=0,σ=1.
I am confused enough about how to implement quadrature for a beta distribution but then the linear transformation confuses me even more. My question is threefold: (1) can I use some variation of quadrature to approximate f(θ) when θ is distributed as this linearly transformed beta distribution, (2) how would I implement this in R, and (3) is this a ridiculous waste of time such that there is an obviously much faster and better method to accomplish this task? (I tried writing my own numerical approximation function but found that my implementation of it, being limited to the R language, was just too slow to suffice.)
Thanks!
First, I assume you can express your L(θ,ρ) and f(θ) in terms of actual code; otherwise you're kinda screwed. Given that assumption, you can use integrate to perform the necessary computations. Something like this should get you started; just plug in your expressions for L and f.
marglik <- function(rho) {
integrand <- function(theta, rho) L(theta, rho) * f(theta)
# set your lower/upper integration limits as appropriate
integrate(integrand, lower=-5, upper=5, rho=rho)
}
For this to work, your integrand has to be vectorized; ie, given a vector input for theta, it must return a vector of outputs. If your code doesn't fit the bill, you can use Vectorize on the integrand function before passing it to integrate:
integrand <- Vectorize(integrand, "theta")
Edit: not sure if you're also asking how to define f(θ) for the transformed beta distribution; that seems rather elementary for someone working with joint and marginal likelihoods. But if you are, then the density of B' = a*B + b, given f(B), is
f'(B') = f(B)/a = f((B' - b)/a) / a
So in your case, f(theta) is dbeta(theta/11.14 + 0.11, 1.25, 10) / 11.14
Related
I have been going through some random number simulation equations while i found out that as Pareto dosent have an inbuilt function.
RPareto is found as
rpareto <- function(n,a,l){
rp <- l*((1-runif(n))^(-1/a)-1)
rp
}
can someone explain the intuitive meaning behind this.
It's a well known result that if X is a continuous random variable with CDF F(.), then Y = F(X) has a Uniform distribution on [0, 1].
This result can be used to draw random samples of any continuous random variable whose CDF is known: generate u, a Uniform(0, 1) random variable and then determine the value of x for which F(x) = u.
In specific cases, there may well be more efficient ways of sampling from F(.), but this will always work as a fallback.
It's likely (I haven't checked the accuracy of the code myself, but it looks about right) that the body of your function solves f(x) = u for known u in order to generate a random variable with a Pareto distribution. You can check it with a little algebra after getting the CDF from this Wikipedia page.
I am working on optimization of an objective function which is a ratio of two linear functions given as mx + b/-mx+c. Can somebody comment about convexity of this function and/or give me some reference?
You might consider consulting Stephen Boyd's convex optimization book. Section 3.4 (example 3.32) is what you are interested in. Your example is called a linear fractional function and is indeed quasiconvex and quasiconcave if you restrict the domain of the denominator to be either greater or less than 0. Quasiconvex optimization problems can be solved using a method like bisection which involves solving a series of feasibility problems
The easiest litmus test for convexivity of a function is to take the derivative and consider the region where this derivative is zero - these are potential local minima, though they could be global minima or saddle points.
In this case, your derivative is: (d)/(dx)((m x + b)/(-m x + c)) = (m (b + c))/(c - m x)^2
There is no zero point at all that depends on x, except at infinity. There is no minimum.
I'm trying to generate chi-square random variables according to the following algorithm:
where a(i) are independent, standard normal random variables witn m even and odd
respectively.
Wikipedia gives the following definition:
https://en.wikipedia.org/wiki/Chi-squared_distribution#Definition
The code I wrote is:
dch = double(1000)
t = double(1)
for(i in 1:1000) {
for(j in 1:m) {
x = runif(1, 0, 1)
t = t + x*x
}
dch[i] = t
}
but I am getting the wrong density plot.
So, where is/are the mistake/s and how can I fix them?
As Gregor suggested in comments, you are misinterpreting the inputs to the algorithm. One way to get a Chi-squared with m degrees of freedom is to sum m independent squared standard normals, but that's not the only distributional relationship we know. It turns out that a Chi-squared(2) is the same as an exponential distribution with a mean of 2, and exponentials are straightforward to generate with inverse transform sampling, a.k.a. inversion. So in principle, if m is even you want to generate m/2 exponential(2)'s and sum them. If m is odd, do the same but add one additional standard normal squared.
What all that means is that a straightforward implementation would have you doing m/2 logarithmic evaluations to generate the exponentials. It turns out you can apply the superposition property of exponentials so you only have to do one log evaluation. Since transcendental functions are computationally expensive, this improves the efficiency of the algorithm.
When the dust settles - the z on the second line of your algorithm is a standard normal, but the a's are Uniform(0,1)'s, not normals.
I want to do some calculations on a binomial distribution Bin(100,0.55). In particular:
P[μₓ - σₓ ≤ X ≤ μₓ + σₓ]
I also want to calculate it using the normal approximation for the binomial:
f(x)=1/√(2πσ²) exp[-(x-μ)²/2σ²]
and attempt to do so using the Chebyshev inequality (say for k=2)
P[|X-μ| ≥ kσ] ≤ 1/k²
I would like to know if there is a relatively straightforward way to implement these calculations easily in R (that is, using in built functions rather than coding this manually).
please check the help documentation for
dbinom
pbinom
qbinom
In general, for other distributions, this pattern dDISTRIBUTION_NAME, pDISTRIBUTION_NAME, qDISTRIBUTION_NAME still holds.
I'm trying to analyze repairable systems reliability using growth models.
I have already fitted a Crow-Amsaa model but I wonder if there is any package or any code for fitting a Generalized Renewal Process (Kijima Model I) or type II
in R and find it's parameters Beta, Lambda(or alpha) and q.
(or some other model for the mean cumulative function MCF)
The equation number 15 of this article gives an expression for the
Log-likelihood
I tried to create the function like this:
likelihood.G1=function(theta,x){
# x is a vector with the failure times, theta vector of parameters
a=theta[1] #Alpha
b=theta[2] #Beta
q=theta[3] #q
logl2=log(b/a) # First part of the equation
for (i in 1:length(x)){
logl2=logl2 +(b-1)*log(x[i]/(a*(1+q)^(i-1))) -(x[i]/(a*(1+q)^(i-1)))^b
}
return(-logl2) #Negavite of the log-likelihood
}
And then use some rutine for minimize the -Log(L)
theta=c(0.5,1.2,0.8) #Start parameters (lambda,beta,q)
nlm(likelihood.G1,theta, x=Data)
Or also
optim(theta,likelihood.G1,method="BFGS",x=Data)
However it seems to be some mistake, since the parameters it returns has no sense
Any ideas of what I'm doing wrong?
Thanks
Looking at equation (16) of the paper you reference and comparing it with your code it looks like you are missing one term in the for loop. It seems that each data point contributes to three terms of the log-likelihood but in your code (inside the loop) you only have two terms (not considering the updating term)
Specifically, your code does not include the 4th term in equation (16):
and neither it does the 7th term, and so on. This is at least one error in the code. An extra consideration would be that α and β are constrained to be greater than zero. I am not sure if the solver you are using is considering this constraint.