What can we say about population mean from a sample size of 1?

Question

I am wondering what we can say, if anything, about the population mean, $\mu$ when all I have is one measurement, $y_1$ (sample size of 1). Obviously, we'd love to have more measurements, but we can't get them.

It seems to me that since the sample mean, $\bar{y}$, is trivially equal to $y_1$, then $E[\bar{y}]=E[y_1]=\mu$. However, with a sample size of 1, the sample variance is undefined, and thus our confidence in using $\bar{y}$ as an estimator of $\mu$ is also undefined, correct? Would there be any way to constrain our estimate of $\mu$ at all?

Yes, a confidence interval on $\mu$ can be constructed under certain assumptions. If no one posts it, I will track it down. — soakley, Jun 18 '15 at 15:43
See http://stats.stackexchange.com/questions/1807 for another version of the same question (the average of a sample is available, but not its sample size, so effectively the average is a *single* observation from the unknown sampling distribution) and http://stats.stackexchange.com/questions/20300 for a related discussion. — whuber, Jun 18 '15 at 16:49
a recent article discussing optimality of these estimators in the normal case: http://www.tandfonline.com/doi/full/10.1080/00031305.2017.1360796 — user795305, Sep 19 '17 at 12:59

score 42 · Answer 1 · edited Jun 19 '15 at 17:35

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If the population is known to be normal, a 95% confidence interval based on a single observation $x$ is given by $$x \pm 9.68 \left| x \right| $$

This is discussed in the article "An Effective Confidence Interval for the Mean With Samples of Size One and Two," by Wall, Boen, and Tweedie, The American Statistician, May 2001, Vol. 55, No.2. (pdf)

edited Jun 19 '15 at 17:35

Tim

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answered Jun 18 '15 at 16:16

soakley

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I hate to sound stupid but.... surely not. This depends on units and doesn't behave properly at all (by properly I mean scalar multiplication....) – Alec Teal Jun 19 '15 at 05:48
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Doesn't sounds stupid to me -- good question. I briefly looked at the paper and indeed, it does seem to suggest exactly what @soakley describes. But is the method actually giving a CI that has a "minimum coverage probability" of 95%? See p. 102, right column. – Wolfgang Jun 19 '15 at 12:18
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@Alec Just because a procedure depends on units of measurement (that is, it is not invariant) does not mean it is automatically invalid or even bad. This one is valid: read the article and do the math. Many will grant that it is a little *disturbing*, though. Even more surprisingly, you don't even have to assume the underlying distribution is Normal: a similar result holds for any unimodal distribution (but 9.68 has to be increased to about 19 or so): see the links I provided in a comment to this question. – whuber Jun 19 '15 at 13:29
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A later issue of the journal had three letters to the editor, one of which brought up Alec Teal's point about the units. The reply from Wall says this: "The confidence interval is not equivariant (i.e., its coverage probability depends on the ratio of ${{\left| \mu \right| } \over {\sigma}}$...)" Later she says "The confidence interval is not based on a pivotal quantity..." It's an unusual approach and result, no doubt! – soakley Jun 19 '15 at 13:35
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Just to save y'all a bit of work: the [letters to the editor & reply](http://amstat.tandfonline.com/doi/abs/10.1198/000313002753631420) @soakley notes appeared in [*The American Statistician*, vol. 56, no. 1 (2002)](http://amstat.tandfonline.com/toc/utas20/56/1). – Stephan Kolassa Jun 19 '15 at 18:54
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This seems to give confidence intervals covering the mean with probability about $95\%$ when $\sigma \approx | \mu | \gt 0$ but with much higher probabilities otherwise. If $\mu = 0$ then clearly the probability is $100\%$ as the confidence intervals always contain $0$. – Henry Jun 19 '15 at 22:05

score 28 · Answer 2 · answered Jun 18 '15 at 15:42

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Sure there is. Use a Bayesian paradigm. Chances are you have at least some idea of what $\mu$ could possibly be - for instance, that it physically cannot be negative, or that it obviously cannot be larger than 100 (maybe you are measuring the height of your local high school football team members in feet). Put a prior on that, update it with your lone observation, and you have a wonderful posterior.

answered Jun 18 '15 at 15:42

Stephan Kolassa

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(+1) One observation will be overwhelmed by the prior, so it would seem that what you get out of the posterior will not be much more than what you put into the prior. – whuber Jun 18 '15 at 16:51
What if we combined such a prior with the kind of likelihood implied by that wretched $x \pm 9.68 \left|x\right|$? – Simon Kuang Jun 19 '15 at 20:25
@SimonKuang: one conceptual problem is that we can only use the $x\pm 9.68|x|$ interval *after* we have observed $x$, so this can't enter the *prior*. – Stephan Kolassa Jun 20 '15 at 05:21
@StephanKolassa No, this interval (and the associated distribution) forms the likelihood. Our prior is separate. – Simon Kuang Jun 20 '15 at 06:58
@SimonKuang: yes, you are right, my mistake. Unfortunately, I don't have the time to go through this at this time, but if you do this, please post what you find! – Stephan Kolassa Jun 20 '15 at 14:06

score 14 · Answer 3 · edited Jun 21 '15 at 12:27

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A small simulation exercise to illustrate whether the answer by @soakley works:

# Set the number of trials, M
M=10^6
# Set the true mean for each trial
mu=rep(0,M)
# Set the true standard deviation for each trial
sd=rep(1,M)
# Set counter to zero
count=0
for(i in 1:M){
 # Control the random number generation so that the experiment is replicable 
 set.seed(i)
 # Generate one draw of a normal random variable with a given mean and standard deviation
 x=rnorm(n=1,mean=mu[i],sd=sd[i])
 # Estimate the lower confidence bound for the population mean
 lower=x-9.68*abs(x)
 # Estimate the upper confidence bound for the population mean
 upper=x+9.68*abs(x)
 # If the true mean is within the confidence interval, count it in
 if( (lower<mu[i]) && (mu[i]<upper) ) count=count+1
}
# Obtain the percentage of cases when the true mean is within the confidence interval
count_pct=count/M
# Print the result
print(count_pct)
[1] 1

Out of one million random trials, the confidence interval includes the true mean one million times, that is, always. That should not happen in case the confidence interval was a 95% confidence interval.

So the formula does not seem to work... Or have I made a coding mistake?

Edit: the same empirical result holds when using $(\mu, \sigma)=(1000,1)$;
however, it is $0.950097 \approx 0.95$ for $(\mu, \sigma)=(1000,1000)$ -- thus pretty close to the 95% confidence interval.

edited Jun 21 '15 at 12:27

Community

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answered Jun 19 '15 at 10:14

Richard Hardy

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Code is fine -- but you really don't need the loop (just ``rnorm(M, mean=mu, sd=sd)`` directly and then vectorize the rest; this will also be much faster). And one really doesn't need to run a simulation to show that $x \pm f|x|$ will *always* capture 0 when $f \ge 1$. – Wolfgang Jun 19 '15 at 12:15
Totally agree with you on coding/computation efficiency! I did it on purpose, though, so as to replicate the way one thinks (not the way computer likes it best). But I do not quite get your latter point. It is correct, but how is it relevant? Here $x$'s are random, and it is about the confidence interval capturing the population mean which need not be zero. – Richard Hardy Jun 19 '15 at 12:42
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Indeed, for $\mu$ not equal to 0, this is useful (and +1 for providing the code in the first place!). I just meant that for $\mu = 0$, it's a foregone conclusion that 0 will always be captured. – Wolfgang Jun 19 '15 at 13:13
Oh, I got the point now. In my simulation, I indeed used $\mu=0$. I should try some other value instead to assess sensitivity. – Richard Hardy Jun 19 '15 at 13:20
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(@Wolfgang) This is not the way to test a confidence interval. The definition does not require that the $\alpha$-level CI covers the mean $1-\alpha$ of the time *in every case*: it requires only that (a) has *at least* that much coverage in every case and (b) it approximates that coverage in some cases. Thus, for your approach to be valid and convincing you would have to search a large number of possibilities. Try `sim – whuber Jun 19 '15 at 13:25
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I understand the point you are trying to make, but I strongly disagree with the statement that "this is not the way to test a confidence interval". In the definition/construction of a CI, the parameter is a fixed constant. In your simulation, $\mu$ keeps changing. For fixed $\mu$, if the method truly gives a 95% CI, then it should cover $\mu$ in 95% of the cases. It doesn't. Also, even with your construction, ``sim(0.1)`` yields a coverage very close to 1 (of course, now we are getting closer again to $\mu$ being fixed at 0). – Wolfgang Jun 19 '15 at 13:37
Of course, one can construct an interval that has *at least* 95% coverage. But that's not a 95% CI in the strict sense (and one needs to clarify then that the method is conservative). – Wolfgang Jun 19 '15 at 13:42
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@Wolfgang check the definition used by paper quoted, it is: $P(X - \zeta |X| \leq \mu \leq X + \zeta |X|) \geq 1 - \alpha $, i.e. the probability that $\mu$ is in the interval is **at least** 0.95. – Tim Jun 19 '15 at 17:45
@Tim Ah, got it, thanks. But then I wouldn't call it a 95% CI -- or at least strongly emphasize that the coverage could be higher and under some conditions, it could be essentially 1. Otherwise, let me suggest another 95% CI: $(-\infty, +\infty)$ (regardless of data) ;) – Wolfgang Jun 19 '15 at 18:54
@Wolfgang choosing appropriate $\zeta$ value makes it possible for you to find the interval that encompasses $\mu$ with probability of at least 0.95. With lower $\zeta$ you have less confidence in it. Picking $\zeta = \infty$ provides that the CI will for certain encompass $\mu$, but also does not give you much clue what $\mu$ can be, while $\zeta = 9.68$ provides you some information, given the extremely limited data, with 0.95 probability. It makes sense. – Tim Jun 19 '15 at 19:05
This is not the programming language I am most familiar with currently, but ... You seem to have always drawn from $Normal(0,1)$. The stated CI will always include zero, so this simulation merely tests that $0$ is in the CI. A "real" test would also vary the mean and standard deviation. (That is, the paper discusses single samples drawn from arbitrary normal distributions. This simulation does not appear to draw from arbitrary normal distributions, so one should be careful expecting identical results.) – Eric Towers Jun 20 '15 at 03:31
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@EricTowers, you point is well accepted. However, I just needed a counterexample to show that the formula does not work universally. soakley gave the formula with no warnings and no conditions (except for normality), so I just showed that some are needed, otherwise it may fail. – Richard Hardy Jun 20 '15 at 05:09
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Again, $\mu$ is a constant. So, it's perfectly fine to simulate with $\mu = 0$. Of course, then coverage must be 1. The method provides a CI that has *at least* 95% coverage and the example shows (whether by simulation or via reasoning) that in some conditions, coverage could be as high as 100%. So, it's not a 95% CI. It's still a pretty clever method for drawing some kind of inference from so little information. – Wolfgang Jun 20 '15 at 07:13
@RichardHardy: It is the case that the 95% CI is the given interval when the specified experiment is performed -- single sample from arbitrary normal distributions. If you choose to perform a different experiment on some subset of the space of arbitrary normal distributions, it is likely that the CI could be shrunk or that the degree of confidence in the same interval is increased. This in no way invalidates the claim about CI confidence in the non-restricted experiment. – Eric Towers Jun 20 '15 at 07:27
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@EricTowers, my idea was very simple. To disprove a general statement, one counterexample is enough. I provided such an example. (Meanwhile, to prove a general statement, all possible cases should be examined, and all of them should satisfy the statement.) – Richard Hardy Jun 20 '15 at 07:45
@RichardHardy: You have assumed more than is in the words "If the population is known to be normal, a 95% confidence interval based on a single observation" by taking more than one observation. You have not performed the described experiment. – Eric Towers Jun 20 '15 at 22:33
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I have performed it – a million times. Exactly the same experiment (the same population $N(\mu,\sigma^2)$, the same sample size of 1, random sampling) every time. After that, I looked at the outcomes, and here is where I got that the coverage is one million out of one million. Had I done the experiment only once instead, I would not have been able to get the information I needed. – Richard Hardy Jun 21 '15 at 04:49

score 9 · Accepted Answer · answered Sep 02 '15 at 08:24

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Here is a brand-new article on this question for the Poisson case, taking a nice pedagogical approach:

Andersson. Per Gösta (2015). A Classroom Approach to the Construction of an Approximate Confidence Interval of a Poisson Mean Using One Observation. The American Statistician, 69(3), 160-164, DOI: 10.1080/00031305.2015.1056830.

answered Sep 02 '15 at 08:24

Stephan Kolassa

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...unfortunately behind a paywall. – Tim Sep 02 '15 at 08:35
@Tim: that is so. Then again, an [ASA membership](http://www.amstat.org/membership/index.cfm) is not terribly expensive, and you get access to *The American Statistician*, *JASA* and quite a few other journals at a very reasonable price, which I personally very happily pay out of my own pocket. I really think you get your money's worth here. YMMV, of course. – Stephan Kolassa Sep 02 '15 at 08:39
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+1 but the Poisson case is radically different from the normal case because the variance has to equal the mean. The Poisson result is pretty straightforward whereas the $x\pm 9.68 |x|$ result for the normal case is counter-intuitive and mysterious. – amoeba Feb 05 '16 at 12:50
@amoeba: quite correct, but the OP did not specify any restrictions on the distribution. – Stephan Kolassa Feb 05 '16 at 13:36
This is so brief that it would better serve as a comment. But since it is the accepted answer, you probably will not wish to convert it to a comment. Could you then perhaps summarize the main points of the article? – Richard Hardy Oct 16 '19 at 19:40

score 0 · Answer 5 · answered Oct 16 '19 at 18:16

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See Edelman, D (1990) 'A confidence interval for the center of an unknown unimodal distribution based on a sample size one' The American Statistician, Vol 44, no 4. Article covers the Normal and Nonparametric cases.

answered Oct 16 '19 at 18:16

David Edelman

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Welcome to Stats.SE. Can you please edit you answer to expand it, in order to include the main points of the book you cite? It will be more helpful both for the original poster and for other people searching in this site. By the way, take the opportunity to take the [Tour], if you haven't done it already. See also some tips on [answer], on [formatting help](https://stats.stackexchange.com/help/formatting) and on writing down equations using [LaTeX / MathJax](https://math.meta.stackexchange.com/q/5020). – Ertxiem - reinstate Monica Oct 16 '19 at 18:23
Welcome to our site, David. Your contribution, as author of that article (which I believe has been cited in several threads here), is greatly appreciated, so any perspective or comments you can provide in this answer would be most welcome. – whuber Oct 16 '19 at 21:36

What can we say about population mean from a sample size of 1?

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