dexp | Dataset EXploration & Processing | Data Manipulation library

Add an await client.process_commands(message) at the end of your on_message event

https://discordpy.readthedocs.io/en/latest/faq.html#why-does-on-message-make-my-commands-stop-working

There are a bunch of issues here. The primary one is that your likelihood expression is wrong (you can't log the components separately and then add them, you have to add the components and then take the log). Your bounds are also funny: the mixture probability should be [0,1] and the means should be [0, Inf].

The other problem you have is that with the current simulation design (n=20, prob=0.01), you have a high probability of getting no points in the first mixture component (the probability of a point being in the second component is 1-0.01=0.99, so the probability that all of the points are in the second component is 0.99^20 = 82%). In this case the MLE will be degenerate (i.e., you're trying to fit a two-component mixture to a data set that essentially only has one component); in this case any of these solutions will give equivalent likelihoods:

• prob=0, mu2=mean of the data, mu1=anything
• prob=1, mu1=mean of the data, mu2=anything
• mu1=mu2=mean of the data, prob=anything

With all these solutions, where you end up will depend very sensitively on starting conditions and optimization algorithm.

For this problem I would encourage you to use the built-in dmixexp2 function from the Renext package (which correctly implements the log-likelihood as log(p*Prob(X|exp1) + (1-p)*Prob(X|exp2))) and the formula interface to mle2:

You should mention the operating system you are using.

You need to include the appropriate headers. And tell the linker where and which libraries you want to use.

So your source should be

The original code is prone to warnings and errors since the parameters may go to invalid values easily. For example, we need w in [0, 1] and lambda > 0. Also, if a is larger than a data point, then the density becomes zero, hence infinite log likelihood.

The code below uses some tricks to handle these cases.

• w is converted to the range [0, 1] by the logistic function
• lambda are converted to positive values by the exponential function.
• Added tiny value to the likelihood to deal with cases of zero likelihood.

Also, the data generation process has been changed so that samples are generated from one of the exponential distributions with the given probability w.

Finally, increased the sample size since the result was not stable with n=500.

The function you pass to optim needs to return a scalar value, and your kld_optim function returns the result of a call to integrate() which according to the ?integrate help page, returns a list, not a numeric value. The "value" in contained in that list under the name "value". So change your integrate() call to

You seem to have several issues in your code:

• shannon_portf does not exist
• input$rate does not exist; but input$lambda does
• the pivot_longer statement returns an error - perhaps you wanted to sort by p?
• the functions in the sourced file should return the value (return(mk)), instead of printing it to console (print(mk))

Below is a version that gives a plot; you may still want to tweak it a little bit... some sliders don't seem to do anything, and you could isolate calculations that don't need to be redone.

eqn2 and eqn3 encode, in the limit of infinitesimal time and space steps, the same equation. However, in their discretization they single out different terms for a more active role in the construction of the time step propagator.

In the short term this should only lead to insignificant differences in the result, in the long term there may be more pronounced differences in stability and accuracy.

That the "more natural" equation eq3 is present at all seems to be an indirect invitation to experiment with it, and perhaps to explore if the transformation in eq2 with its possible advantages is a coincidence or result of some more deeply founded idea.