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I have a random variable which I know is uniformly distributed, and I expect it to be distributed between the range $[0,1]$. Then, I generate (simulate) 100 realizations of the variable.

The question: Which is the best way to find the probability that the underlying "actual" range of the variable is in fact $[0,1]$ from the realizations?

Its clear that if the 100 realizations are within the range $[0.51, 0.69]$, the probability is very low, while if they are within the range $[0.01, 0.99]$, it should be high.

I tried simply to compute the actual vs expected range ratio $$ ((0.69-0.51)/(1-0)) = 0.08, $$ and assumed that it was the probability $p$ of each realization. Then computed the probability of it happening 100 times as $p^{100}$.

The problem is that, for a correct case, in which the original range was in fact $[0,1]$, the realization range turned out to be $[0.02,0.93]$ and the computed probability is very low $0.91^{100} = 0.0012$, which does not makes sense.

So, what is it that I am getting completely wrong?

EDIT. Clarifications thanks to @whuber (1) I am attempting to check that the range of the underlying variable is in fact [0,1] (for validation purposes) (2) I just want to check the range and know how likely is that it is in fact [0,1]. (3) realizations are independent $rand(1,100)$ in Matlab

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migrated from mathoverflow.net Apr 14 '16 at 14:03

This question came from our site for professional mathematicians.

  • $\begingroup$ Not much. You are testing $H_0: X \sim u(0,1) $ vs $H_a: X \sim U(a,b)$ with $0 \le a,b \le 1$ and not equality in both cases. You propose the perfectly reasonable test statistic Z = Max - min, where you reject if $Z < \lambda$ and determine $\lambda$ by $P_0(Z < \lambda) = .05$, say . This calc. is a little off as you ignore the min, also it's odd you get exactly 0 as the min, which shouldn't happen, but I doubt the slight differences make much difference. $\endgroup$ – michael Apr 14 '16 at 11:33
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    $\begingroup$ What do you mean by 'random process' here? Are the realizations actually single random numbers (drawn independently) or is there some time series? $\endgroup$ – Juho Kokkala Apr 14 '16 at 14:30
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    $\begingroup$ @whuber (1) I am attempting to check that the range of the underlying process is in fact [0,1] (for validation purposes) (2) I just want to check the range and know how likely is that it is in fact [0,1]. (3) realizations are independent $rand(1,100)$ in Matlab. (4) none, probably I am misusing the concept "random process" $\endgroup$ – Jose Ospina Apr 14 '16 at 16:13
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    $\begingroup$ That comment substantially clarifies your question. Would you mind editing your original post to incorporate those refinements? $\endgroup$ – whuber Apr 14 '16 at 16:37
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    $\begingroup$ The expected range is not one for a finite sample from a uniform distribution so your ratio formula is off. This looks like what might be called a two-sided German tank problem to me. $\endgroup$ – soakley Apr 14 '16 at 18:04
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Although it is meaningless to find a probability (unless you first specify a prior distribution of the endpoints), you can find the relative likelihood. A good basis for comparison would be the alternative hypothesis that the numbers are drawn from a uniform distribution between a lower bound $L$ and upper bound $U$.

Sufficient statistics are the minimum $X$ and maximum $Y$ of all the data (assuming each number is obtained independently). It doesn't matter whether you draw the data in batches or not. When drawn from the interval $[0,1]$, the joint distribution of $(X, Y)$ is continuous and has density

$$\eqalign{f(x,y) &= \binom{n}{1,n-2,1}(y-x)^{n-2}\mathcal{I}(0\le x\le y\le 1) \\ &= n(n-1)(y-x)^{n-2}\mathcal{I}(0\le x\le y\le 1).}$$

When scaled by $U-L$ and shifted by $L$, this density becomes

$$f_{(L,U)}(x,y) = (U-L)^{-n} n(n-1)(y-x)^{n-2}\mathcal{I}(L\le x\le y\le U).$$

Obviously this is greatest when $L = x$ and $U=y$.

The relative likelihood is their ratio, best expressed as a logarithm:

$$\Lambda(X,Y) = \log\left(\frac{f_{(X,Y)}(X,Y)}{f(X,Y)}\right) = -n\log(Y-X).$$

A small value of this is evidence for the hypothesis $(L,U)=(0,1)$; larger values are evidence against it. Of course if $X \lt 0$ or $Y \gt 1$ the hypothesis is controverted. But when the hypothesis is true, for large $n$ (greater than $20$ or so), $2\Lambda(X,Y)$ will have approximately a $\chi^2(4)$ distribution. Assuming $X \ge 0$ and $Y \le 1$, this enables you to reject the hypothesis when the chance of a $\chi^2(4)$ variable exceeding $2\Lambda(X,Y)$ becomes so small you can no longer suppose the large value can be attributed to chance alone.

I will not attempt to prove that the $\chi^2(4)$ distribution is the one to use; I will merely show that it works by simulating a large number of independent values of $2\Lambda(X,Y)$ when the hypothesis is true. Since you have the ability to generate large values of $n$, let's take $n=500$ as an example.

Figure

$100,000$ results are shown for $n=500$. The red curve graphs the density of a $\chi^2(4)$ variable. It closely agrees with the histogram.


As a worked example consider the situation posed in the question where $n=100$, $X= 0.51$, and $Y=0.69$. Now

$$-2\Lambda(0.51, 0.69) = -2(100\log(0.69 - 0.51)) = 343.$$

The corresponding $\chi^2(4)$ probability is less than $10^{-72}$: although we would never trust the accuracy of the $\chi^2$ approximation this far out into the tail (even with $n=100$ observations), this value is so small that certainly these data were not obtained from $100$ independent uniform$(0,1)$ variables!

In the second situation where $X=0.01$ and $Y=0.99$,

$$-2\Lambda(0.01, 0.99) = -2(100\log(0.99 - 0.01)) = 4.04.$$

Now the $\chi^2(4)$ probability is $0.40 = 40\%$, quite consistent with the hypothesis that $(L,U)=(0,1)$.


BTW, here's R code to perform simulations. I have reset it to just $10,000$ iterations so that it will take less than one second to complete.

n <- 500 # Sample size
N <- 1e4 # Number of simulation trials
lambda <- apply(matrix(runif(n*N), nrow=n), 2, function(x) -2 * n * log(diff(range(x))))
#
# Plot the results.
#
hist(lambda, freq=FALSE, breaks=seq(0, ceiling(max(lambda)), 1/4), border="#00000040", 
     main="Histogram", xlab="2*Lambda")
curve(dchisq(x, 4), add=TRUE, col="Red", lwd=2)
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It's not clear exactly what you need, but let's first take a look at how to estimate the lower and upper boundaries. It is claimed on Wikipedia that the uniformly minimum variance unbiased estimators for the continuous uniform distribution on $(a,b)$ are the maximum spacing estimators $$ \hat{a}={{nx_{(1)}-x_{(n)}} \over {n-1}},\ \ \hat{b}={{nx_{(n)}-x_{(1)}} \over {n-1}}, $$ where $x_{(1)}$ is the minimum observed value, $x_{(n)}$ is the maximum observed value, and $n$ is the sample size. Note that $\hat{a}$ can be negative and $\hat{b}$ can be greater than one when the population is $U(0,1)$.

I don't know if this is enough for you or not. It seems you really want to test whether your output is truly $U(0,1).$ If that is the case, why not just conduct a K-S test since your desired distribution is fully specified? In this case, if the null hypothesis is true, your p-values should also have a $U(0,1)$ distribution.

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    $\begingroup$ What I need is a test on the data that passes when the underlying distribution is in fact $U(0,1)$ and fails when is not (for example if it is $U(0,0.5)$. That's all. Thanks for the suggestion, I don't know what a K-S test is. $\endgroup$ – Jose Ospina Apr 15 '16 at 18:54
  • $\begingroup$ Kolmogorov-Smirnov goodness-of-fit test. $\endgroup$ – soakley Apr 15 '16 at 18:59

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