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Category: Statistics

The general theory, methods, and philosophy of the Science of Guessing What Is.

August 20, 2018 | 8 Comments

Does It Mean I’m Psychic?

Years ago I wrote and self-published for fun the book So, You Think You’re Psychic? You can download a free PDF of it on the Books page.

There’s nothing wrong with the book—except that now I would do all the probability tables differently. None of them in the book are wrong. But I wrote the explanations when I was still a p-value believer, as all fledgling statisticians were trained to be. And still are. That has to stop. But that’s a subject for another day.

I also wrote it back in my atheist days, when I was a mistaken believer in materialism. That means the discussions about physical mechanisms of purported psychic phenomenon are incomplete, and in part flawed.

Except for these weaknesses, the rest of the book stands up pretty well. It badly needs updating, of course, and if I find myself with unexpected free time, or lucrative financial incentive, I’ll do so.

For now, let me answer an email I recently received from a person with the wonderful name of Swapnil Kamble.

Hello sir,

I have read your book ‘so you think you are psychic?’. Its a great book. I am not very good at statistics. I had a question.

1) If try guessing numbers from a pool of 1 to 100, using (pseudo)RNG app on my mobile or pc, what is the probability of getting it right 3-4 times in a session of 100 guesses?

2) Does it mean am i psychic?

3) Also if i guess 99 out of 100 times, then am i psychic?

4) in context of guessing the numbers, at what point will i be called psychic, i mean what is the minimum probability that will prove that there is some extrasensory phenomenon involved?

Thank you

1) The probability of guessing a number from 1-100, when all you know is that the number will be 1-100, is 1%, or 0.01. The probability of getting k = 3-4 right in a session of n = 100 is had by a binomial calculation. For k = 3, it’s 0.061 and for k = 4 it’s 0.015. That means getting 3 or 4 right is 0.076.

That’s not so small. Especially if you consider you might repeat the session. The probability of getting 3 or 4 right if you repeat the session just once, for a total of 2 sessions, is 0.15. If you do 3 sessions, the probability is 0.21 that at least one of the 3 sessions you’ll get 3 or 4 right. By the time you repeat it just 10 times, the probability is 0.55, or 55% that at least in one session you’ll get 3 or 4 right. Better than a coin flip!

2) Now 10 sessions isn’t a lot if you consider more than one person around the globe has done them. If just 2 people did 10 sessions, the probability at least one of them sees at least one session with 3 or 4 right is 0.79. For 3 people it’s 0.91, for 4 it’s 0.96%, and for 5 it’s 0.98, or 98%! I certainly sold more than 5 copies of the book, so maybe at least this meany sessions were completed.

You can see it’s really easy for at least one “successful” psychic session to be reported using these criteria as a “success.” Even if people are just guessing—by which I mean not using any psychic powers.

In order to prove psychic ability using these criteria, you’re going to have to do a lot better. Guessing only 3 to 4 in 100 is indicative of very weak powers. What’s stopping you from guessing all 100? Or something in the high 90s?

One answer is that you’re a very weak psychic. Hey, not every ball player hits 400, so this is possible. Now if you can consistently hit 3-4 in every session, then you might be on to something. You must keep careful, careful track, not forgetting any sessions, or partial sessions, and you must not allow yourself any excuses about why a failed session (or partial session) “doesn’t really count.” Ball players don’t get those excuses, and neither do you.

3) The probability if guessing 99 out of 100 is about 1 times 10 to the negative 197. A very, very, exceptionally small number. So, yes, if you can in test conditions, under the watchful eye of people like myself, who can spot mistakes (people often fool themselves with sensory leakage), then I’d say you’d have psychic powers.

4) This is an excellent question. There is no excellent answer. The problem is that no session of the types you are attempting are ever considered in isolation. We have had long experience of people cheating, and amazing reports come under immediate suspicion. That’s why testing under controlled conditions are mandatory.

Then some paranormal powers don’t need probability at all. Like coming back from the dead. Or turning water into wine. Do these things and we’ll know you’ve got something.

August 16, 2018 | 3 Comments

How To Do Predictive Statistics: Part VI New (Free) Software Poisson Regression

Previous post in the series (or click above on Class). REVIEW!

Download the code: mcmc.pred.R, mcmc.pred.examples.R. If you downloaded before, download again. This is version 0.22! Only the example code changed since last time.

For an explanation of the theory behind all this, which is markedly different than classical views, get this book: Uncertainty.

We’re downloading data from another gent’s site, so you have to be on-line for today’s lesson.


# https://stats.idre.ucla.edu/r/dae/poisson-regression/
x <- read.csv("https://stats.idre.ucla.edu/stat/data/poisson_sim.csv")
x <- within(x, {
  prog <- factor(prog, levels=1:3, labels=c("General", "Academic", 
                                                     "Vocational"))
  id <- factor(id)
})
summary(x)

It's fictional data: "the outcome variable [num_awards] and indicates the number of awards earned by students at a high school in a year, math is a continuous predictor variable and represents students' scores on their math final exam, and prog is a categorical predictor variable with three levels indicating the type of program in which the students were enrolled."

Number of awards can be 0, 1, 2, ..., where that "ellipsis" means awards can go up to infinity, which is exactly the hope of Millennials. In reality, of course, awards cap out. How many are possible depends on the situation. About which here we now knowing, since it's all made up anyway.

Poisson distributions allow infinity. Thus they will always be approximations to reality, which does not have infinities. How much an approximation this is in any case, and how loose the approximation, again depends on the situation. Really, an ordered multinomial-type situation is what always pertains. Number of anything falls into progressive buckets from zero to some fixed number.

But here we shall pretend, as everybody does, that the approximation is better than reality. What's the Deadly Sin of Reification among friends?

We already know how to do this. Lucky for us, MCMCpack creators did the work for us as they did with ordinary regression, but not for multinomial regression.

fit = MCMCpoisson(num_awards ~ prog*math,data=x)

We now (and you reviewed so you already know this) move to finding probabilities of scenarios; i.e. of finding

     Pr (Y | new X, old X&Y, Model & Assumptions)

where here Y = "0 awards" or "1 award", etc., X is the program and math score, and the model, including its assumptions about priors and so forth, we already discussed.

As ever, we have built-in scenarios in the old data. Hence

# single new data
p = MCMCpoisson.pred(fit,x[1,])
plot(p,xlab='num_awards', ylab='Pr(num_awards|D,M)')

I won't here show the plot, but it gives a spike at 0, with estimated probability of about 0.87 , a probability of 0.12 for 1 award, about 0.01 for 2 awards, and close-enough-for-most-folks 0 for 3 and 4 awards. This is for an (as typing x[1,] in R shows) a person in a NEW person in a Vocational program and math score of 41.

It is for a NEW person because, as you remember from your reviewing, we do not need probability models to tell us what happened. We just look!

What if we wanted to compare the probability for a person with the same math score but in an Academic vocation? Like this, which is a real pain in the keister:

w = x[12,] # where I hunted for an x that had Academic
w$math = 41
p = MCMCpoisson.pred(fit,w)
plot(p,xlab='num_awards', ylab='Pr(num_awards|D,M)')

I get probability of 0.77 for 0 awards, 0.20 for 1, 0.03 for 2, and near-enough-0 for the rest.

But why can't we do this?

w = x[1,]
w$prog = 'Academic'
p = MCMCpoisson.pred(fit,w)

Because R strips the class of factor from w$prog and turns it into a character. Recall whatever we pass into the prediction functions has to be the same kind of data.frame that went in. The variable prog went in as a factor with 3 levels, and not a character. You might try this:

w = x[1,]
w$prog = 'Academic'
levels(w$prog) = levels(x$prog)

But that turns w$prog to General! I do not know why R strips the class from a factor variable when there is only one row upon substitutions, but it does. For instance, we can do this:

w = x[1:2,]
w$prog[1] = 'Academic'
w = w[-2,]

That works fine: summary(w) shows w$prog is still a three level factor.

So here is the solution:

w = x[1,]
w$prog[1] = 'Academic'
p = MCMCpoisson.pred(fit,w)

That [1] on the second line does the trick! Like I said, pain in the keister.

Okay, let's do our old trick of using all the old data as scenarios. First let's get the probabilities.


# more complicated, because y is open ended; a guess has to be made in 
# number of cols of q; if subscript out of bounds, increase size
q = matrix(0,nrow(x),16)
for(i in 1:nrow(x)){
  a = MCMCpoisson.pred(fit,x[i,])
  q[i,1:length(a)]=a
}

We do not know the limit of the number of awards, but we have to store them. See that "16"? Try putting 8. It will break because some scenarios generate positive (defined as above machine limit) probability for number of awards larger than 7 (recall 0 awards takes a column in q). Of course, you could always put some huge number in place of 16 and never have a problem. Until you want to make pretty plots. As you see next.


par(mfrow=c(2,2))
for (i in levels(x$prog)){
  j = which(x$prog == i)
  plot(0:(dim(q)[2]-1),q[1,],xlab='num_awards', ylab='Pr(num_awards|D,M)', main=i,ylim=c(min(q),max(q)), xlim=c(0,8), type='l')
  for(k in 2:nrow(x)){
    lines(0:(dim(q)[2]-1),q[j[k],])
  }  
}

You can see we limit here the number of awards probabilities to stop at 8 (in the xlim). That obviously has to be changed depending on the application.

Every scenario results in a distribution, probabilities for number of awards from 0 to 8 (max shown). In the scenarios everybody has a program, which are only three, hence three plots. But everybody also has a math score, which obviously varies. That variability does not do too much to change the probability of awards for General or Vocational programs NEW people. But it must for Academic.

The next trick would be to make a plot of only Academic NEW people, where perhaps you color the distributions so that higher math scores are a brighter and brighter red (or whatever color scheme you like).

Math scores goes from 33 to 75. Another fun thing would be to put these scores into discrete decision buckets (which you recall from your review we must always do), and then make the same plots for each of these decision buckets. How? Use the cut function. Do maybe


x$math.cut = cut(x$math, breaks = c(0,45, 52, 59, 80))

which chunks it into the observed quartiles, Do this only if these make sensible decision points. Then you need only change the code above to reflect the new plot, e.g.

...
for (i in levels(x$math.cut)){
j = which(x$math.cut == i)
...

I did it and it worked fine, and showed something interesting. That's your homework!

Next time tobit regression.

August 13, 2018 | 6 Comments

How To Do Predictive Statistics: Part V New (Free) Software Multinomial Regression

Previous post in the series (or click above on Class). REVIEW!

Download the code: mcmc.pred.R, mcmc.pred.examples.R. If you downloaded before, download again. This is version 0.22! Only the example code changed since last time.

For an explanation of the theory behind all this, which is markedly different than classical views, get this book: Uncertainty.

Mandatory rant

We’ll use another built-in dataset, the Netherlands voting data. Accessed like this:

data(Nethvote)
x = Nethvote

As before, assigning it to x is for simplicity sake. Find out all about the data with ?Nethvote. Essentially, voters could pick one of several parties. And, supposing those parties last until the NEXT election, and given some demographic information, we want the probability that

     Pr(Y | new X, old X&Y, Model & Assumptions)

Which—surprise!—is identical to the same probability we want in every predictive model! The emphasis on the NEXT election cannot be stressed too highly. Why? Glad you asked. Let me shout the answer:

There is NEVER a need to model what happened, only what might happen.

We do not need a probability model to tell us what we saw. We need only use our eyes. If we want to know if more religious people voted for vote (i.e. party) = CDA WE JUST LOOK. There is no need to do a “hypothesis test”, which is insane. Either more religious would have voted for CDA, or they wouldn’t have. AND THAT IT IS.

The classical idea, frequentist or Bayes, p-value of Bayes factor, of ascertaining whether more religious “really” voted more frequently for CDA is nuts. About the future? Well, that’s what model is for. To quantify the probability more religious will vote CDA accepting as an assumption religion is probative. It is our decision whether we choose religion as probative or not; two people looking at the same data, and even same model, can come to different conclusions.

I shout, because these ideas are central to the (old, ancient) predictive approach. They are foreign to the hypothesis testing classical methods, which aim to uncover occult forces in data. We will have none of that. Our concern is only observables and measures.

On to the data!

The real beginning

Because of a limitation (out of my control) of MCMCmnl, we have to keep track of the model formula. So we call the method a little differently than when we did ordinary or logistic regression.

form = formula('vote ~relig + class + income + educ + age * urban')

lv = levels(x[, as.character(form[[2]]) ])

fit = MCMCmnl(form, mcmc.method="IndMH", B0=0, mcmc=5000, thin=10, tune=0.5, baseline='D66', data=x)

Notice form is a standard R formula. This one was chosen to match the one in the native help function. Fool around with others. The object lv holds the levels of the “y” observable. It’s written in a generic way, so that it works with any data set. We could have, of course, just wrote lv = levels(x$vote), but that works only on data frames with vote as an outcome. Notice, too, that we can change the baseline. We don’t have to: it will default to the normal R base level. We keep track of the levels because you’re allowed to change them, and MCMCmnl doesn’t save the model formula. Ah, well.

Predictions are somewhat different than before, too. We have pass in the model formula and levels of the y. We also need, as ever and as core of the predictive method, a scenario. How about this one? Mixing code and output, and ignoring the ‘dist’ measures, which we don’t use.


x[1,]

vote distD66 distPvdA distVVD distCDA relig class income educ age urban
PvdA 2.669695 2.335121 4.109881 6.45008 0 0 1 2 5 1

Then

p = MCMCmnl.pred(fit,form,x[1,],lv)
p

I get


> p
     D66      CDA     PvdA      VVD 
0.076580 0.067476 0.822900 0.033044 

So, given non-region, class of 0, and so on, the probability a NEW voter will go D66 is about 8%. Your results will vary a bit, since as ever this is a numerical approximation. But they’ll be close. The most likely vote will be cast at 82% is for PvdA for NEW voters of this sort, and the least likely is VVD at 3%. I don’t know Dutch politics, so I offer no opinions on what this means.

The idea, if it isn’t clear, is that you get a probability for each possible category, because why? Because that’s what we wanted!

The form and lv ensure everything is labeled correctly at the end. Pain in the keister. But as yet there are no wrappers for any of these methods to make things easier.

How about all the scenarios in the data? You bet:


p = MCMCmnl.pred(fit,form,x[1,],lv)
for(i in 1:nrow(x)){
  # this preserves the proper names for p's columns
  if(i>1) p=rbind(p,MCMCmnl.pred(fit,form,x[i,],lv))
}
p = as.data.frame(p, row.names=FALSE)

par(mfrow=c(2,2))
for (i in 1:4){
    plot(x$class,p[,i],main=names(p)[i], ylab='Pr(Vote|D,M)',col=x$relig+1)
}

Notice we stacked the answers one on top of the other, and turned p into a data.frame. The plot is for each category or level of vote, as a function of class (which really does have all those odd values; probably the output of some other model). For fun, I colored the points by religion yes/no.

This is only one possible plot of many. Other obvious ones will suggest themselves to you. Do them as homework.

Everything is more complex because the model itself is more complex. There isn’t any real or general way to make this easy, either. Nor should there be!

“But, Briggs, can’t I do an average probability for each class level, using all the old scenarios? That way I can tell the impact of ”

Sure you can. But why would say impact when you meant influence? Second, it would be fooling yourself. Because your model included all those other things, you have to state probability only with regard to and conditional on all those other things. Otherwise you’re talking weird.

If you want to discuss only class, then build a model with only class.


form = formula('vote ~ class')
lv = levels(x[, as.character(form[[2]]) ])
fit = MCMCmnl(form, mcmc.method="IndMH", B0=0, mcmc=5000, thin=10, tune=0.5, baseline='D66',data=x)

Then you can say what you want about class considered only by itself. Or whatever.

The key lesson is that you specified a model with all those measures, so you can only speak of the model with all those measures. If you don’t want to speak of them, remain model-silent of them.

Mini rant

We are done with multinomial. But not really. It should be used in place of ordinary regression almost always. Why? Because all measures are discrete and finite, thus all Y are, thus all Y are better approximated by multinomials. Now, all Y are approximated by continuity, which is an ENORMOUS assumption, and untrue. No measure can be continuous, and none infinite in actuality.

All data should be transformed into the units of decision. We talked about this before with regard to CGPA data. If you are a dean only interested in counting numbers of students scoring 3 or hihger in CGPA (or whatever), then you have naturally created an analysis were the Y is dichotomous. Or maybe you want 3 or above, which naturally implies under 3s are of interest, and then 4s (to be given special recognition, say). Then we have a trichotom. Multinomial can handle this, ordinary regression cannot.

Two people can have the same data and come to different conclusions about it, as happens all the time in real life. People have different decisions to make, and different consequences to face about those decisions. Therefore, every analysis, i.e. model, should be tailored to the decision at hand. Since every decision, like every measure, is discrete and finite in act, then so should by every model.

“But Briggs, if I quash the data into buckets like you say, then I lose information. I won’t know the difference, in this case, between a CGPA of 2.876543 and 2.876544. I’m losing power or whatever. Besides, I’ve heard discretizing data is bad.”

You heard wrong. I remind you that there is no difference between 2.876543 and 2.876544—not one bit! nor between 0 and 2, or 0 and 2.9—when any decision you make recognizes no difference between these CGPAs! If you are going to make different decisions, then you will have different buckets, and thus a different model, and different results.

This is not a bug, it is a feature. Just like the conditionality of all probability.

Next is Poisson regression.

August 8, 2018 | 2 Comments

How To Do Predictive Statistics: Part IV New (Free) Software Logistic Regression

Previous post in the series (or click above on Class).

Download the code: mcmc.pred.R, mcmc.pred.examples.R. If you downloaded before, download again. This is version 0.21! Only the example code changed since last time.

For an explanation of the theory behind all this, which is markedly different than classical views, get this book: Uncertainty.

We’ve done ordinary regression to death. So it’s time we move to unordinary but still humble regression, i.e. logistic regression, which, like all models, aims for this:

     Pr(Y | new X, old X&Y, Model & assumptions)

where in this case Y is a yes/no, up/down, 0/1 proposition. Like whether Y = “A baby is born with (what doctors say is) low birth weight.”

R has a built-in dataset to play with, invoked with


data(birthwt)
x=birthwt
x$race = as.factor(x$race)

The first line brings it up, the second assigns it to x only for the sake of consistency of all the other code we have been writing, and the last turns the numbers 1, 2, 3 into a factor. You can read all about the dataset by typing ?birthwt.

As an ever-important reminder, the model we build will not be causal. It will be probabilistic. We are not saying, for instance, smoking causes low birth weight, but we do aim to check the conditional probability of low birth weight given a mother smokes or not. Conditional on, as the equation above says, the old data, new guesses of the data, the model itself, and all the other assumptions that went into the mix. Assuming all that is true, we get the probability.

Don’t forget to re-source the code, if you don’t have it in memory (with relevant path, if necessary).

source('mcmc.pred.R')

Fit the classic parameterized model:

fit = MCMClogit(low~age+race+smoke, data=x)

Why just these measures and not the others also included? No reason at all, except that the help for MCMClogit had just these. By all means, try out the others!

We next form a scenario, the “new X”, so that we can calculate the probability. A minor limitation is that the scenario has to be an R data.frame matching the input measures. A limitation is NOT that it has to have all input measures. You (in this case I) said all those measures must be specified to get the probability, so every time you want the probability, you have to specify values of all measures.

Perplexed which scenario to try, since few of us are low birth weight experts? The old data can be used as if it were new. For instance (mixing code and output):

w = x[1,]
w
low age lwt race smoke ptl ht ui ftv bwt
85 0 19 182 2 0 0 0 1 0 2523

Ignore the “85”, it is a line number produced by whoever sorted the original data so that all low birth weights came first. It is assigned to w because of the limitation I mentioned of passing in the kind of data.frame, which includes knowledge of all factor levels of the relevant measures. We could also have passed in x[1,]. If we use this scenario we want to calculate

     Pr(low | age = 19, race=2, smoke=0, D, M & A)

where D = the old data. The answer is (drumroll please):

p = MCMClogit.pred(fit,w)
p

I get 0.34. Since this is an approximation to a complicated integral, your answer might differ slightly.

What if we wanted the model-old-data-conditional probability of low birth weight given a 19-year-old mother of race 2 but who smoked? This:

w$smoke = 1
p = MCMClogit.pred(fit,w)
p

I get 0.61, about double. Yours may differ slightly. So, given everything we punched into the model (the priors etc. included), the chance a 19-y-o race = 2 woman smokes doubles, in our estimation, of having a low birth weight baby. Any individual woman like that either will or won’t have a low birth weight baby, a state caused by various things. We are not measuring cause, only probability.

What about 20-y-o’s? 21-y-o’s? Higher? What about race = 1 and race = 3? The only way to find out is to do the calculations.

Again we can create lots of scenarios using the old data, which has the virtue of containing scenarios actually found in nature. But it has the disadvantage of only having scenarios actually measured. But we can use it to get some idea of what to expect. For instance:


# all the old x
p = NA
for(i in 1:nrow(x)){
  p[i] = MCMClogit.pred(fit,x[i,])
}

plot(x$age,p,col=x$race,ylab='Pr(low wt|old data,M)', pch=x$smoke+1)
grid()
legend('topright',c('r1','r2','r3','s0','s1'), col=c(1,2,3,1,1), pch = c(3,3,3,1,2), bty='n')

The legend is a hack because of author laziness, but you get the idea. If we wanted pretty pictures we’d be using ggplot2, but then I’d have to spend too much time explaining that code, and we’d miss the important bits.

The idea is that each old measure was taken as a new scenario. The model-etc.-conditional probabilities were estimated, and the result plotted by the measures.

This is the model-estimated-probability! That’s why everything looks so smooth. It says, more or less, that race = 1 non-smokers have the lowest probability, race = 1 smokers and race = 2, 3 non-smokers are the same in probability, and race = 2,3 smokers have the highest probability. And that regardless of race or smoking older women have smaller chances.

For people who are like the gathered data.

About those people I know nothing except what is printed in the help file. Whether these women are like women all over the world in all countries in all time periods of all economic and spiritual backgrounds and so on and so forth, I have no clue. But I doubt it.

Given all those necessary and hard-to-over-emphasize caveats, it is well to recall these are the probabilities of low birth weight babies given whatever conditions we specify. For individual mothers. What if you were a hospital administrator making a guess of the number of low birth weight babies in the coming year, these (perhaps) costing more for whatever reasons? Then you’d have to have another layer of modeling on top of our simple one, which includes patient-type scenarios. How many race = 1, 2, 3 and how many smokers, what ages. And how many patients!

We could do that easily for single-measure patients. Like when w = x[1,]. Then the model is binomial-like. After specifying the number of patients. If you don’t know that, you have to model it, too, as just mentioned.

I’ll leave that for homework. Meanwhile, we are done with logistic regression. It’s as easy as it looks. Next time we start multinomial regression!