Hi. I have an update on a question that I asked a couple of weeks ago but it has been tagged as a duplicate

I edited the question but I do not know if doing so will bump it up or anything, or if I should just post a new question

This is the question I am referring to: stats.stackexchange.com/questions/286239/…

@Glen_b Thanks for the pointers. I've fixed up the grammar and expanded some unclear sections. Please let me know if there's still something murky. stats.stackexchange.com/questions/282634/…

@user164144 An edit should put it in a review queue which is looked at by any users with enough reputation to see the queue (it requires a number of votes). It should also bump it to the top of the front page at the time of the edit.

Oh, actually that's marked as duplicate, right?

Oh, yeah, that one.

Uh, well, it's not a duplicate of that other post any more ... but it's not not clear to me what you're asking about. The point of the exercise was to estimate the area of a circle,

Because the distribution was uniform (in the square) the proportion of points inside the circle is an estimate of the proportion of the total area that's within the circle.

Note that you don't necessarily even know what the value of $\pi$ is for this exercise

Anyway, you'll need to define $\theta$ in your question and show more of what you did for it to be clear what's happening.

The more usual presentation of this exercise is on a circle of radius 1, where the aim of the exercise is to get an estimate of $\pi$.

@Glen_b Have you seen this Q stats.stackexchange.com/questions/289410 ? It has fair amount of upvotes and there is an answer there that also has several upvotes but I believe is wrong (will hopefully get edited and fixed but hasn't been yet). The Q looks like something you should be able to answer. Perhaps it's even a duplicate.

Thank you @Glen_b. My question is really just whether $100\frac{\sum z_i}{10}$ or $25\pi\frac{\sum z_i}{10}$ is the correct estimator. I got the first one with the assumption that the $\z_i's$ are $Bern(\theta/100)$ but I wonder why I should not use the second one because it seems to be more intuitive. Using the second one requires me to assume that the $Z_i's$ are $Bern(\theta/25\pi)$ which I cannot seem to justify.

I do not think that the value of \pi is unknown in the problem because part of the problem was to describe the parameter space for $\theta$ which I described as $[0,25\pi]$. I don't think the point of the problem is to estimate $\pi$, because there is a part 2 to the problem where another method is used (not monte carlo) to estimate the area of the circle and the last objective is to compare the MSEs of the estimators from the two methods.

@user164144 1. Estimator of what? 2. What's theta? 3. If it's intuitive, can you explain in simple terms how that intuition leads you to that particular estimate? I didn't get it. 4. If you can't justify it why do you think it's intuitive?

...

Part of the difficulty with your question is you need to be clearer what you're actually trying to answer. The present question dives in without making what you're trying to find out clear.

@amoeba I had read it but I couldn't figure out what the situation even was. I didn't have a clear sense of what was going on in the first few paragraphs there, so I couldn't follow what the code was trying to do (and the algorithm wasn't explained beforehand -- I have an issue with that because the problem with trying to infer an algorithm from code is you can't tell whether the code does what they intended or not -- is something that looks wrong an error in the algorithm, ...

an error in the code or an error in my understanding of one or the other?

it's intuitive since if my estimator is 25pi\sumZ_i/10, I am estimating by getting the proportion of points out of 10 that fall in the circle and multiplying by the area of the biggest possible circle instead of by 100 which is much larger than that area. So, for example, if all my points happen to fall in the circle, my estimate using 25pi is 25 pi while my estimate using 100 is 100 which is not even in the parameter space of \theta.

@user164144

Please start by explaining clearly

what you're estimating

what theta is

Ok. The problem is to estimate the area of a circle (\theta) in a square of known dimension 10x10

The procedure to do this is to randomly generate 10 points (x_i,y_i)

theta is the area of the circle? Wow. I really can't tell that from the question.

I assumed it was a proportion

sorry about that

So you're estimating theta, which is the area of the circle. Cool.

Yup, I just looked at the problem as I edited it. I, yeah, you're right it's not stated that area=\theta

Now what circle are we talking about? I assumed it was already a maximal circle. Is it just any circle?

yes any circle. not the maximal

Anywhere in the square?

yes

Okay now, what's the definition of $Z$?

(I know thats already there, but look at it)

Z_i's are indicator functions for when (x_i,y_i) fall in the circle

Now when is Z equal to 0?

when the point is not in the circle

When it's in the part of the square that's not in the target circle.

so i defined Z_i's as Bern(\theta/100) where theta is in [0, 25pi]

yes

Imagine for a moment that the target circle just happened to be the largest possible circle.

and everything sort of falls into place from there except that intuitively, I think that considering Z_i's to be Bern(\theta/25pi) seems better

ok

WHat proportion of the Z's would be 0 then?

You have to focus on what the Z's are here.

25pi/100

25pi/100 of the z's would be 0 in that case, right?

Well, that's teh proportion of z's that are 1 not 0, but okay, let's discuss that. What does your second estimator say about the area of the circle then?

it would be 25pi

versus 100 using the 1st estimator

That's where you went wrong.

my 1st estimator is 100(SumZi)/10 and my second is 25pi(SumZi)/10

If the expected proportion of points in the circle (i.e. E(sum Z)/10) is 25pi/100, and the estimator is 25 pi times that, then your estimate of the area is 25 pi x 25 pi /100.

... You are instead somehow thinking that 25 pi x 25 pi /100 is 25 pi.

I think I get that part, but what i'm confused about is, if I do use the first estimator in practice 100SumZi/10 and I happen to have all my points fall in the circle, then my estimate would be 100, would it not?

Sure.

And if none of your points fall in the circle your estimate would be 0, would it not?

But if I use 25piSumZi/10 my estimate would be 25pi

yes, but that's the case whether I use estimator 1 or 2

so from what I am seeing there, why should 100SumZ/10 be used over 25piSumZ/10 when the former will overestimate beyond parameter bounds while the latter will not?

of course, the second will be biased but that is if we assume that the Zi's are bern (theta/100)

if we assume that the Z_i's are Bern(\theta/25pi) then it will not be biased.

but I cannot justify doing that because of how Z_i's is defined

So your argument for using an estimator that has a wildly wrong average value is that it's right in a situation that happens quite rarely

no, i know that an estimator like that is a bad estimator

Why do you say "if we assume the Zi's are bernoulli(theta/100)"? The described situation guarantees it.

yes exactly. But I've been thinking if the parameterization of theta affects it in some way. Since we know theta is [0,25pi]. I was thinking if this somehow leads to the conclusion that Z_i are Bern(\theta/25pi)

Like, I tried thinking about it as what if Z_i's are indicator functions of when (c_i,y_i) is in the circle given that the maximum area of the circle is 25pi

sorry that should be (x_i,y_i)

Note that 1. the target circle may not be in the maximal circle at all. It could be up in one corner for example 2. you have to use a different indicator if you want to change the estimator. That is, you have to use not Zi but something else (say Vi) and carefully define it.

so i thought maybe P(x_i,Y_i in the circle| max area=25pi) = (\theta/100)/(25pi/100) which will give me what I want, but that seems wrong too because that would mean that if the circle is the biggest circle, the probablity that a point would fall in it is 1 which is wrong.

Your concern is that you have an estimator that has the right expectation but may occasionally give an estimate which is obvious nonsense, but your attempts to fix that problem are far worse than the problem you worry about.

Much of what you've discussed (the motivation for your second estimator for example) would be useful background to your quesiton

Yes. I guess I'm just not seeing the big picture. If I did simulations on all possible outcomes then it would probably become clear that the 1st estimator is better than my attempt to fix it

I guess I also just wanted to ask that question to make sure I wasn't missing anything. That is, that the answer really was 25piSumZ/10 and I just could not see how to justify it. Your inputs make it clear that it is not. Thank you.

I now understand your question well enough to post an answer to it but it would be better if it was clarified; in particular, it could be clearer that you were trying to fix a problem where the first estimator could yield an impossibly large value

ok. I will edit it.

You could improve the first estimator in a mean square error sense (reducing the variance at the cost of some bias) by stopping the estimator from being impossibly large.

or maybe just hedge it by making the estimate 25pi if it gets impossibly large?