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Suppose a theory claims that a random variable $R$ (of unknown distribution $F$) must satisfy a certian upper bound $R < c$ (where $c$ is known constant). Suppose I perform a set of measurements $X_1,...,X_n$ which are iid~$F$, but because of errors, I observe $X_1+E_1,...,X_n+E_n$ where $E_1,...,E_n$ are independent Gaussian with $E_i$ having standard deviation $\sigma_i$.

Is there some statistical method of testing whether the upper bound is indeed an upper bound based on the measurements?

Now, relax the fact that we know the distributions of $E_i$ to be Normal. My current problem is that I am given the values of $X_i + E_i$ with 68% confidence intervals for what $X_i$ should have been, and there is no additional information given (imagine someone handed me the data and perished afterwards; more realistically, it is too difficult to handle but somehow a software churned out the 68% confidence intervals). Is there still some statistical technique I can consider?

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  • $\begingroup$ Is anything known about the error term? Is $X$ known to be positive? What about $X + \text{error}$? Is it known to be positive? $\endgroup$ – Glen_b Jun 17 '14 at 11:30
  • $\begingroup$ I've edited my question to answer your queries. Nothing is known/constrained to be positive. $\endgroup$ – renrenthehamster Jun 17 '14 at 13:23
  • $\begingroup$ Presumably these "68% confidence intervals" are in fact $\pm$ 1 standard deviation intervals based on a normal assumption. Knowing the $\sigma_i$ values is critical. Your question now seems to have the potential to be answered. $\endgroup$ – Glen_b Jun 18 '14 at 1:05
  • $\begingroup$ The answer should be a very definite no. Here's why. Observe $Y_i=X_i+E_i$. Let $Y$ be the largest of them. $(Y_i-Y+c)=(X_i-Y+c)+E_i$ is in the desired form: it looks like some draw $(X_i-Y+c)$ from an unknown distribution plus iid Gaussian error. Moreover, every one of them is less than $c$ by construction, which is perfectly consistent with the hypothesis $R\lt c,$ so you cannot rule it out. This demonstrates that you need to add more assumptions. The first one I would consider adding is that the $E_i$ have common means of $0$! $\endgroup$ – whuber Jun 21 '14 at 12:52
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A simple way: Not having a distribution for $X$ makes it a bit tricky, but at least in the case of normal errors, I think this can be done.

I think there's no useful upper limit on $c$ provided by the data in the scenario you describe, since the unobserved values (the ones without error) might have any distribution - as such very large values of $c$ are consistent with any data no matter how small, because it may just be that $F$ just has a very tiny tail that has almost no chance to produce any sample values near the upper limit.

However, I think it's perhaps possible to get a useful lower bound on $c$ that could be used to exclude values that aren't really consistent with the assumptions.

[Excuse me but I am going to move to a slightly more conventional notation. Hopefully it will still be clear.]

Let $Y_i=X_i+\epsilon_i$

"Underlying" values $X_i\stackrel{_\text{iid}}{\sim}F$

Observed values are $Y_i=X_i+\epsilon_i\,$, where $\epsilon_i/\sigma_i\stackrel{_\text{iid}}{\sim}N(0,1)\,$ and $\sigma_i, i=1,...,n$ are known.

Method: We could compute for each $Y_i$ the probability that $X_i<c$. If the product of that probability over the observations is low, we reject the hypothesis that the supposed value $c$ ($c_0$) is an upper bound on the values $X$ can take.

In particular, the set of probabilities computed for each $X_i$ individually would correspond to the case of testing whether $X_i$ is no lower than $c_0$, which is equivalent to testing for $X_i=c_0$, $i=1,2,\ldots,n$.

enter image description here

In the example, the 4 sets of probabilities are where all the X's are right at the hypothesized upper bound (black), below the upper bound (red), above the upper bound (blue, and what we want to reject) and randomly distributed below the bound in such a way that although the hypothesized $c_0$ is the least upper bound on the distribution, none of the X's are likely to get close to it (green).

We can combine the individual probabilities in various ways (I earlier said "take the product", and that works - as long as you compare the product with the right thing). A common way to combine these probabilities into a single test statistic would be via Fisher's method. This takes $-2\times$ the sum of the logs, so it's monotonic in the product I mentioned before. That's then compared with a $\chi^2_{2n}$.

If we do that for the data in the above example, we get p-values of:

Black:  0.3077957
Red:    1
Blue:   1.882208e-09   <--- we reject the case where the bound was below the X's
Green:  1

The green one was a beta distribution on (5,25) and $c_0$ was 25 (i.e. an exact upper bound on the distribution), but it was a very right skew distribution (most X values were close to 5, and the largest observed value was about 13). The test won't have good power in a case like the green one if all the $X$s are below $c_0$, even though $c_0$ is below the largest possible $X$. That really can't be helped.

It should have reasonably good power against $c_0<c$ in cases where $c$ is a sharp least upper bound on the X's (like a uniform, or a beta(10,1) say), which means you should get some observations very close to $c$ in reasonable sized samples.


Somewhat more complicated alternative, by simulation.

Since the distribution of the $\epsilon_i$ are completely specified, it's possible to simulate from their distribution; consequently/equivalently we can simulate from $(X_i|Y_i,\sigma_i)$ and compute the per-simulation maximum. Then if $c_0$ was too low in the tail of this distribution, we'd regard it as unlikely to have produced the sample.

I've tried a few examples and this seems to at least give plausible answers but I haven't yet explored its coverage under different distributions for $X$ (different $F$); you shouldn't rely on it without some degree of checking of its properties.

(if I get a chance to check it myself I will post an update)


On the non-normal error case, it's much trickier, because we know nothing other than a 68% interval. I think that interval is just based on a normal assumption (i.e. a $\pm$ 1 sd interval - and presumably where the $\sigma_i$ are based on something other than the data).

However, assuming the 68% interval were a valid 68% interval and assuming that the distribution of the $\epsilon_i$ are symmetric, we can still get somewhere.

The lower bound of the 68% interval for an observation (presumably $Y_i-\sigma_i$) would lie entirely above $c_0$ with probability $0.16$ ($1-0.68/2$).

Hence the number of points whose interval lower bounds lies above the hypothesized value for $c$ is $\text{binomial}(n,0.16)$. If too many lie above it to be consistent with that, we would reject the possibility that the $X$'s all lie below $c_0$.


Finally, even without symmetry, we could use $\text{binomial}(n,0.32)$ as a worst case.

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  • $\begingroup$ Thanks for the well-thought out answer. In the first method, should we consider some form of geometric mean/arithmetic mean for the p-value? Otherwise taking more samples will definitely decrease the p-value $\endgroup$ – renrenthehamster Jun 19 '14 at 11:30
  • $\begingroup$ You're right; it needs to be treated slightly differently (a bit more like my recent edits to the non-normal case at the end). I will edit to fix it in a few hours. $\endgroup$ – Glen_b Jun 19 '14 at 19:43
  • $\begingroup$ I've updated my answer to deal with the problem you pointed to. Taking the product wasn't the problem: it was treating that product as a p-value rather than a test statistic that was wrong. Once I fixed that, it's fine (Fisher's method sums the logs rather than takes the product, but that's effectively the same thing - it puts the same ordering on the sample space - except more convenient to work with, because double the sum of the logs can be compared with a chi-squared distribution) $\endgroup$ – Glen_b Jun 21 '14 at 4:14
  • $\begingroup$ Thanks for the detailed answered. I got the general gist. But I have a lingering query. Why is $P(X<c_0)$ the p-value (that is inputted into the Fisher's method)? I guess the equivalent question is: why is the hypothesis testing equivalent to testing $X_i = c_0$? $\endgroup$ – renrenthehamster Jun 23 '14 at 12:18
  • $\begingroup$ You could think of it like this: it treats each observation as its own (admittedly very low power) test, which you then combine into a single statistic. [But you could alternatively not think of those probabilities as p-values, and just use Fisher's method as a way of constructing an overall test statistic from them]... or were you asking 'why compare with $c_0$'? $\endgroup$ – Glen_b Jun 23 '14 at 12:23
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Use the distribution of bootstrapped sample maxima. The right-sided quantile of your specified level of confidence should not exceed the upper bound. You should be testing whether the upper bound is exceeded due to the measurement error.

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    $\begingroup$ Could you please give more details on exactly how to do what you suggested? For example, references, or actual concrete names and of steps that I should take? $\endgroup$ – renrenthehamster Jun 17 '14 at 10:05
  • $\begingroup$ "[...] the distribution of bootstrapped sample maxima" means to draw random samples from your sample, and calculate the maximum each time. Say your data is in a vector X=(x1,x2,...,xn) you: 1) Draw a new sample based on X, for example you might get X_sample_1=(x1,x2,x2,x3,x1,x3,...,x4) 2) calculate max(X_sample_1) 3) Repeat as many time as you deem necessary 4) Look at the distribution $\endgroup$ – pkofod Jun 17 '14 at 11:07
  • $\begingroup$ I see. But how about the measurement errors of x1,...,xn? Does that play a role somewhere? Intuitively, if I obtained some x1 that is 1.5c with error 0.1c, then I am more certain that the bound fails as compared to if the error is 1.0c $\endgroup$ – renrenthehamster Jun 17 '14 at 11:24
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    $\begingroup$ You should be careful; using the bootstrap with sample maxima doesn't always work (e.g. see here, also see this discussion -- if you can bear to wade through the failure of the author to spellcheck). As such you should take great care to make sure that your suggested bootstrap has desirable properties. $\endgroup$ – Glen_b Jun 19 '14 at 1:24
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    $\begingroup$ I'd suggest some simulation studies under a variety of different possibilities (e.g. a beta with the first parameter large and the second small, a beta with the second parameter large and the first small, with both small and with both large would be one place to start - seeing how meaningful the intervals can be in at least those cases). $\endgroup$ – Glen_b Jun 19 '14 at 1:25

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