whats the meaning of setting up a new source element from the PML library with the following parameters?
Arrivals defined by: interarrival time;
Interarrival time: exponential(1) - minutes;
how often it generates new agents? I'have read the documentation but I don't understood.
I'm not sure if you're interested in all the statistical part of the distribution so I'll keep it simple.
When you have an exponential distribution exponential(lambda) you have the following:
Expected value E(X) (mean): 1/lambda
Variance V(X): 1/lambda^2
Standard Deviation = sqrt(Variance)
Therefore the standard deviation is the same as the mean.
Those are the basics about the exponential distribution.
In your case, you are using an exponential distribution with lambda = 1 which means the mean will also be 1 so, in average, your interarrival time will be 1.
If you had lambda = 2 your average interarrival time would be 1/2 = 0.5.
Hope that helps
Related
I study the effect of a drug on the variability of a continuous dependent variable. The study includes two groups, one of them receives the drug. The dependent variable is repeatedly measured 6 times during the study. The variability is assessed by residual standard deviation and mean absolute difference.
Any idea how to perform the analysis in SPSS?
I study the effect of a drug on the variability of a continuous dependent variable. The study includes two groups, one of them receives the drug. The dependent variable is repeatedly measured 6 times during the study. The variability is assessed by residual standard deviation and mean absolute difference.
Any idea how to perform the analysis in SPSS?
I'm studying Anylogic. I'm curious about something.
Some people explain that arrival rate follows Exponential Distribution.
I wanna know 'How can prove that?'
Any kind guidance from you would be very helpful and much appreciated.
Thank you so much.
The arrival rate doesn't follow an exponential distribution, it follows a poisson distribution, so there's nothing to prove on that regard.
What follows an exponential distribution is the inter-arrival time between agents.
To prove that this thing actually follows a particular distribution, you can use one of the many distribution fitting techniques out there, my favorite and the one is the Cullen and Frey Graph. You can see an answer about it here:
https://stats.stackexchange.com/questions/333495/fitting-a-probability-distribution-and-understanding-the-cullen-and-frey-graph
You can also check the wikipedia page on distribution fitting:
https://en.wikipedia.org/wiki/Probability_distribution_fitting
Have in mind that distribution fitting is kinda an art, and no technique gives you the correct distribution, but maybe a good enough approximation of a distribution. But in this case it should be quite easy.
You can't really prove that a distribution fits the data though, you can just have maybe an error estimation when you compare the distribution function with the actual data, and you can have a confidence interval for that... I'm not sure if that's what you want.
not exactly sure what you mean by "prove" that it is exponential... But anyway, it is not "some people" that explain that, it is actually mentioned in AnyLogic help under the "Source" topic as follows:
Rate - agents are generated at the specified arrival rate (which is
equivalent to exponentially distributed interarrival time with mean =
1/rate).
What you can do is collect the interval time between arrivals and plot that distribution to see that it actually looks like an exponential distribution.
To do that:
Create a typical DES process (e.g. source, queue, delay, sink)
Set the arrival type to rate and specify for example 1 per hour
Create a variable in main called "prevTime"
Create a histogram data element called "data"
In the "On exit" of the source write the following code:
data.add(time() - prevTime);
prevTime = time();
Look at the plot of the histogram and its mean.
I am trying to understand the best practices regarding AnyLogic's source arrival rates. I know that Exponential and Poisson are two different probability distributions. When using "Arrival Rate" in AnyLogic and choosing a rate of 10/hour for example, does this generate 10 agents per hour exponentially or according to a Poisson distribution or is it the same thing?
I really need guidance on understanding the best practices in this matter. To simplify the question, if I have an arrival rate of 10/hour following a Poisson distribution, what is the right way to model that in AnyLogic?
Many thanks!
In any source in AnyLogic, if you choose a rate, it will automatically be poisson where your rate will be the lambda parameter of your poisson distribution... this means that in average you will get lambda agents per time unit generated
The exponential distribution is equivalent to the poisson distribution, except that it takes into consideration the time between each arrival instead. (this means that you need to use arrivals defined by inter-arrival time in your source, otherwise it wouldn't make much sense)
poisson(lambda) arrivals per time unit is equivalent to exponential(lambda) time units per arrival, it doesn't really matter which one you use
In one hierarchical model, we have two hyer parameters: dnorm(A_mu, 0.25^-2) and dnorm (B_mu, 0.25^-2). In this case, 0.25 is the sd, I use the fixed number. A_mu and B_mu represent the mean of group level. After fitting the data by rjags, we get the distributions for each parameter. So I just directly compare the highest posterior density interval (HDI) of A_mu and B_mu? Do I need to calculate something using the sd(0.25)?
In another case, if the sd of two hyper parameters is not fixed, like that: dnorm(A_mu, A_sd) and dnorm (B_mu, B_sd). How can I compare the two hyper parameters and make a decision, e.g. this group is significantly different another group?
Remember that you are getting posterior distributions for A_mu and B_mu. This makes your comparison easy as you can have a look at 95% confidence intervals (CI) for the parameters (or pick a confidence value that satisfies your needs). I believe JAGS uses Gibbs sampling and so you should be able to get the raw samples from the posteriors for A_mu and B_mu. You can then ask "what is the probability that B_mu is greater than some value?" by calculating the percentage of posterior samples that are greater than that value. Alternatively, and in a similar way to frequentist Hypothesis testing, you can ask what is the probability that the mean of B_mu is a draw from the posterior of A_mu. So the key is just to directly use the samples from your posterior. I would recommend taking a look at Andrew Gelman's BDA3 textbook (Chapter 4) for a really good reference on these concepts.
A few things to keep in mind before drawing conclusions from the data: (1) you should always check the validity of your Markov Chains by evaluating things like autocorrelation (2) try to do a posterior predictive check to make sure your model is well fit to the data. If your model is poorly fit to the data then you can get very misleading results from the procedure above.
I am using Perl to model a random variable (Y) which is the sum of some ~15-40k independent Bernoulli random variables (X_i), each with a different success probability (p_i). Formally, Y=Sum{X_i} where Pr(X_i=1)=p_i and Pr(X_i=0)=1-p_i.
I am interested in quickly answering queries such as Pr(Y<=k) (where k is given).
Currently, I use random simulations to answer such queries. I randomly draw each X_i according to its p_i, then sum all X_i values to get Y'. I repeat this process a few thousand times and return the fraction of times Pr(Y'<=k).
Obviously, this is not totally accurate, although accuracy greatly increases as the number of simulations I use increases.
Can you think of a reasonable way to get the exact probability?
First, I would avoid using the rand built-in for this purpose which is too dependent on the underlying C library implementation to be reliable (see, for example, my blog post pointing out that the range of rand on Windows has cardinality 32,768).
To use the Monte-Carlo approach, I would start with a known good random generator, such as Rand::MersenneTwister or just use one of Random.org's services and pre-compute a CDF for Y assuming Y is pretty stable. If each Y is only used once, pre-computing the CDF is obviously pointless.
To quote Wikipedia:
In probability theory and statistics, the Poisson binomial distribution is the discrete probability distribution of a sum of independent Bernoulli trials.
In other words, it is the probability distribution of the number of successes in a sequence of n independent yes/no experiments with success probabilities p1, …, pn. (emphasis mine)
Closed-Form Expression for the Poisson-Binomial Probability Density Function might be of interest. The article is behind a paywall:
and we discuss several of its advantages regarding computing speed and implementation and in simplifying analysis, with examples of the latter including the computation of moments and the development of new trigonometric identities for the binomial coefficient and the binomial cumulative distribution function (cdf).
As far as I recall, shouldn't this end up asymptotically as a normal distribution? See also this newsgroup thread: http://newsgroups.derkeiler.com/Archive/Sci/sci.stat.consult/2008-05/msg00146.html
If so, you can use Statistics::Distrib::Normal.
To obtain the exact solution you can exploit the fact that the probability distribution of the sum of two or more independent random variables is the convolution of their individual distributions. Convolution is a bit expensive but must be calculated only if the p_i change.
Once you have the probability distribution, you can easily obtain the CDF by calculating the cumulative sum of the probabilities.