@Thomas Dean 

No offence, but I think your answer is useless.
What I need is to draw10^5 times a single sample from a Binomial distributions with parameters N=25000 and p itself drawn from a Uniform distribution of support a and b.

Just do this simple simulation  (here p is constant and equal to 1e-4)

CodeTools:-Usage(Sample(Binomial(25000, 1e-4), 10^5))

to convince yourself that the time it takes is already larger than the the one taken of my procedures in a more complex situation (p variable).
 

To complete tomleslie's reply :

If you want to continuously Explore these ranges (using tom's reply), replace 7 by 7. and 15 by 15. :-)

@acer 
Could it be that "RTABLE" is actually the name of a user procedure in a worksheet, for example:
RTABLE := proc(...) ... end proc: , 
that mapleatha didn't pay attention to this and thought that RTABLE was a Maple command.
(PS: I've been fooled with something like that before :-) )

@acer @vv
Thank you both (with some late)

@ogunmiloro 

Your first question concerned the determination of confidence intervals on the elements of a stochastic matrix. I have given you some elements but the problem remains: it is up to you to know what is the joint distribution of the elements in each row, not me. 
So, to end this question, it seems to me that I have only shown you a "way forward" but by no means a definitive answer.

Now you're asking a completely different question which boils down to "what can I do with this matrix P, the graph Carl drew, and what you [me] did".
I think that's for you to know, and no one can answer that question for you.
I think the reason you use stochastic matrices and Markov chains is because it's part of your work, so you have to know what these objects are, how you can manipulate them and what information you can get from them?

So, to sum up, I am no longer of any help unless you have a very specific question.

@ogunmiloro 

You mean the graph Carl Love built?

@ogunmiloro 

Great.
Here is a more elaborate version of my previous code which contains some plots. 
I hope it will work correctly with your Maple 18.

Some plots do not appear in Mapleprimes but will be displayed (should be displayed!) when you run the worksheet.

PS: do not forget to change "GG := add(...)" as I said earlier

 

Download Dirichlet_Markov_Chain.mw

@ogunmiloro 

I guess the answer could be the same that acer's.
PS: loops are authorized for directed graphs only in Maple 2020.

I wrote my code in Maple 2015, the version which comes just after Maple 18.
Unfortunately I no longer have this latter to discover where the errors come from.
Nevertheless I think you could replace the first occurence of GG:=add(G) by

GG := add(G[i], i=1..3)

and the second by
 

GG := add(G[i], i=1..2)

This could work, let me know.

@Carl Love 

Even if one can remove them by hand, I think an improvement of DrawGraph should be not to draw edges with a null weight .. maybe a drawing option?
In a flow problem an edge with weight 0 matters: going from one of its vertex to other simply costs nothing.
But here 0 means that it is impossible to reach some state from another one and thus the corresponding edge doesn't exist.

What is yout opinion about that?

@Scot Gould 

And with a smile on mine, I'll say that I do not know if the normal distribution is the most natural for physicists but it is surely the most widely used.

In your modelling I interpret u as the proportion of people that do not have already done a final choice, and these people can either vote Dem or GOP. From a metrologist point of view (maybe a physicist too?) one could say u is some "measurement error" and an alternative way to do the model could be to say that some amount of these "u" people will dep into the Dem basket or into the GOP basket in some random way.

For instance, for Ohio, u=0.06. Multiply this value by, lets say 100 to obtain 60. If you think that each of these "60" people is just as likely to vote for Dem as for GOP, you can simulate what could happen by sampling a binomial RV with parameters N=60, p=1/2.

If you do this you will obtain, for instance, 33 for Dem and 37 for GOP.
Divide these two numbers by 100 to obtain the projection Dem/Ohio = 0.47+0.033, GOP/Ohio = 0.47+0.027.
(WATCHOUT : I'm french, and this strategy will fit quite correctly our electoral system... which is not the US'. So "my " strategy is not necessarily adapted to this latter ???).
This is what I meant by "I would have chosen a Binomial one".

Concerning your last question "Would you be willing to recommend code that would substitute for win_prob as a function of the three variables?" I understand that you ask me if replacing win_prob by a code whose output would be a function of 3 RVs (f1, f2 and u) would be interesting?
Honestly I do not know.
One thing is that f1, f2 and u cannot be independant variables because of the relation f1+f2+u=1.
In such a situation a good statistical modelling uses the Dirichlet distribution Dirichlet_distribution. with an ad hoc vector parameter alpha (in wiki notation). By choosing carefully the 3 elements ot alpha you can set the mode of the  Dirichlet distribution at the point f1, f2, u. Simulating a random realization is very simple (reference above).

Watchout: If you do this, some people thar are willing to vote Dem (or Gop) can change your mind.

Here is an example of use of the Dirichlet distribution.
The adjustement parameter is alpha[3] (associated to the number of undecided people).
Dirichlet.mw

I hope the worksheet is clear enough. If not, don't hesitate to let me know.

Nevertheless I believe (a very personal opinion) a better way to handle three RVs proceed this way
Let f[1], f[2], f[3]=1-f[1]-f[2]the current estimations of the polls for some State.
Let's suppose a fraction g[n] of f[n] can change is mind and chose either of the two others options, lets's say f[i] and f[j], with probabilities p[n, i] and p[n, j].
Of course p[n,n] = 1-p[n, i]-p[n,j] represents the probability that someone aimed to vote f[n] will not change his mind.
You can build a 3 by 3 latriw whose all lines sum to 1, which representents Markovian transitions from one state (some f[n]) to any other (f[n] influded).
The stationnary state of this Stochastic_matrix will give an estimation of the final scores for the State you have considered.
The problem is to give values to the 9 p[n, n'] probabilities.
You could say, for instance, that 90% of Dems and Reps won't change your mind, that 7% of them will vote for the other side and 3% will join the mass of undecided people.
On the other way 60% of the undecided people will finally make a choice and 30% of them will vote for Dem while 70% will choice GOP.
 

@ogunmiloro 
1) The matrix P values are estimates, How do i obtain the standard deviation and 99% confidence interval values for each estimated values in the matrix P

As the sum of the lines must be equal to 1, the only possibility is that the random variables P[i,1], ....P[i, 5] follow some Dirichlet distribution (look here Dirichlet_distribution) with parameters a[i, 1], ...a[i, 5] for each value of i=1..5.
But if you do not know these 25 values of the a[1, 1]...a[5, 5] there is no way to obtain a confidence interval.

Here is an example: 

 

Download Dirichlet_Markov_Chain.mw

Very usefull, thanks for the job.
I'm not very familiar with DataFrames in Maple ("historically" I use R for this type of structure), so I find your work very interesting.

A very didactical introduction to Maple's DataFrames


PS: From a statistical point of view the choice of the Normal pdf seems unnatural (personnaly I would have chosen a Binomial one).

@Kitonum 

Thank you.
I blame myself for not having thought of it on my own.

@acer 

Hi, 
what I find destabilyzing is that

not [n,x,y]::[nonnegint$3]

seems to do implicitely a lot of operations.
Naively I would have written the things this way

`and`(op(zip((u,v)->not(is(u::v)), [n,x,y], [nonnegint$3])))

Could you please help me understand why the first syntax works.

TIA

@Preben Alsholm 

Hi Preben,
There is a slight modification of your code.
I names sol1 the solution you obtained and sol2 the one I obtained. sol2 has a simpler expression than sol1 but normal(sol2) clearly shows sol2=sol1.
My question is, having obtained sol1, how could we transformed it in order to obtain simpler sol2?

I tried a lot of things, freezing arctan(x) or 6*arctan(x)+3*Pi, , completing squares and so on but I wasn't able to transform sol1 into sol2.

Download simplification.mw