well?

Not your personal homework army.

>>9082350
Can two people have the same height?

Just think about all of the orders they could arrive in.

>>9082768
There is no unique answer unless we at least assume everyone has a different height.

>>9082350

you would have to have some information about the distribution of heights

File: fugen16.png (56KB, 1044x647px) Image search: [Google]

56KB, 1044x647px

>>9082798
The prompt is actually right, you don't need to know anything about the distribution. The distribution of record breaking events is fairly stable no matter which input distribution you use, and they all have the same peak number of about 7.

# draw 1000 random numbers from various distributions, a bunch of times
x <- replicate(100,runif(1000,70,150))
y <- replicate(100,rbeta(1000,0.5,0.5))
z <- replicate(100,rcauchy(1000))
a <- replicate(100,rgamma(1000,2,3))
b <- replicate(100,rnbinom(1000,100,0.2))
d <- replicate(100,rnorm(1000,mean=100,sd=10))

# function for finding number of record break events
findRecords <- function(x){
currentMax <- 0
maxCount <- 0
for (each in x){
if (each>currentMax) {
currentMax <- each
maxCount <- maxCount+1
}
}
return(maxCount)
}

# function for making histogram of max numbers across matrix
maxHist <- function(x) {
maxes <- rep(0,100)
for (each in 1:100) {
maxes[each] <- findRecords(x[,each])
}
hist(maxes)
}

# graph the data
maxHist(x)
dev.new()
maxHist(y)
dev.new()
maxHist(z)
dev.new()
maxHist(a)
dev.new()
maxHist(b)
dev.new()
maxHist(d)

>>9082916
wait no they don't all have hte same peak number but it's fairly similar. one has a peak about 4, one about 9, another about 6.

it's certainly much more robust ot the underlying distribution than I expected, and I threw in some weirdos like beta(0.5,0.5).

I smell Central Limit Theorem, somebody solve this please, my stats knowledge isn't good enough

>>9082350
>worst case:
only one record b/c tallest person came first

>best case
1000 records b/c they came in order of height

Therefore the answer is the average of 1 and 1000, 500.5

>>9082350
Metric or Imperial? Answer is different based on limit of precision of measuring tape.

I will show by induction that for N people, the expected number of records is H_n = 1 + 1/2 + ... + 1/n.

Base case:
1 person, 1 record. H_1 = 1.

Inductive case:
Assume you can expect H_n records with n people. Now add another person at the end. The probability that the new person is the tallest of all n + 1 people is 1/(n+1), so the expected number of records is H_n + 1/(n+1) = H_(n+1).

So we expect H_1000 = 1 + 1/2 + ... + 1/1000, which is approximately 7.485

>>9082350
this topic is also interesting
https://en.wikipedia.org/wiki/Order_statistic

>>9082350
It does depend on the probaility distribution.

If you have the distribution where P(X=1) = 1 then there is always only one record.

If you instead have the distribution with P(X=1) = 1/2 and P(X=2) = 1/2 then there are slightly more than 1.5 records on average.

>>9082350
I'll admit that my first intuition was binary log 1000, but after thinking about it I realized it's just regular old log with base e 1000

Solution is basically 1 + 1/2 + 1/3 + ... + 1/1000

>>9082916
>Iz true u dun needs to know notin about the distribution
>E-except they r not same height
Can't solve the problem without a specific distribution, maybe you can get a range of values by trying every possibility but nothing more than that

>>9084727

If you look away from people with the same height (as in even extremely minute differences will be noted) then it is enough that there exists a total order, no more info is necessary

>>9084729
What do you mean no more is necessary? Not even your simulation converged to one value so there has to be a range based on all possible distributions of the height, so more is necessary to find an expected value of records rather than its range

>>9082798
Central Limit Theorem, motherfucker. Or some variation/application thereof.

>>9084772
Yeah a normal with unknown parameters because the question is too ambiguous

>>9084764
For a person in spot m there is a 1/m chance he is the tallest in the series 1..m, so you just sum the resulting series.

To the people who think distribution is necessary, assume normal then poisson with whatever parameters and show what the difference would be

>>9084790
>the tallest
That assumes that an unique tallest person exist. If several people have the same height then this doesn't work.

>>9084764
that person you're replying to isn't the one who did the simulation, i am

my gut feeling is that there probably isn't a singular expected value, but that it's good enough to pin a range to the wall

>>9084669
>The probability that the new person is the tallest of all n + 1 people is 1/(n+1)
isn't that only true if heights are uniformly distributed?

>>9084794
A deterministic distribution trivially has only one record.

It's 1 + 1/2 + ... + 1/1000, it's a standard result on record statistics:

https://books.google.com/books?id=dAS3CgAAQBAJ&lpg=PA229&ots=6qaFPfmTof&dq=order%20statistics%20harmonic%20numbers&pg=PA229#v=onepage&q=order%20statistics%20harmonic%20numbers&f=false

>>9084799
All that is needed is a total order.

>>9084800
that's my point f4m

>>9084795
If you assume that then the whole question becomes pointless. Arguing semantics gets old, it's clearly intended that there are no people equally tall...

>>9084795
that's a question of measurement precision, not statistics. there is no reason to expect there are truly identical heights in this population

>>9084806
>>9084806

DESU you need that for every real number, P(size=0), (maybe you even need that the diwtribution of a person's size has a density with respect to the Lebesgue's measure).

If the sizes are integers almost surely the above reasoning is wrong (because of ties).

File: 1479033455721.jpg (109KB, 1080x1349px) Image search: [Google]

109KB, 1080x1349px

I'd guess it's [math]\sum\limits_{i = 1}^{1000} \frac{1}{i} = H_i[/math] which is somewhere between 7 and 8.

>>9084836
If you allow for ties, then the question statement is just wrong about the answer being independent of the distribution. The answer of 1 + 1/2 + ... + 1/1000 is right if there are no ties, but if you imagine a world where everyone is the same height then it obviously won't work there. (You'll get either 1 or 1000 instead depending on whether ties count as new records or not)