Many tutorials demonstrate problems where the objective is to estimate a confidence interval of the mean for a distribution with known variance but unknown mean.

I have trouble understanding how the mean would be unknown when the variance is known since the formula for the variance assumes knowledge of the mean.

If this is possible, could you provide a real life example?

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    $\begingroup$ Are you sure you mean unknown, rather than undefined ? In a sample we can obviously always compute the mean, but there are distributions that have an undefined mean as a result of the intergral that defines it diverging. $\endgroup$ May 24 at 18:13
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    $\begingroup$ I meant unknown mean and ot undefined. $\endgroup$ May 24 at 18:14
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    $\begingroup$ You write as if there is "the" formula for the variance. There are many formulas and some of them do not refer to a mean at all. One is described at stats.stackexchange.com/a/18200/919. At stats.stackexchange.com/a/130772/919 I write (at length, but from an elementary standpoint) about a problem where the mean is known and inference about the variance is needed. $\endgroup$
    – whuber
    May 24 at 18:58
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    $\begingroup$ There are algorithms that estimate a mean (by sampling) when an upper bound on the distribution's variance is known (or at least assumed). $\endgroup$
    – Peter O.
    May 24 at 20:56
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    $\begingroup$ Just to add some context, these types of problems almost never occur in the real world and are just pedagogical tools to help you solidify basic statistical concepts before learning how to solve more realistic problems. (Almost) no one uses the one-sample z-test for the mean of a continuous variable, and (almost) no one tries to compute a confidence interval around a mean with known variance. Still, these basic concepts need to be taught, even if you'll never use them in their purest form. $\endgroup$
    – Noah
    May 25 at 6:40

A practical example: suppose I have a thermometer and want to build a picture of how accurate it is. I test it at a wide variety of different known temperatures, and empirically discover that if the true temperature is $T$ then the temperature that the thermometer displays is approximately normally distributed with mean $T$ and standard deviation 1 degree Kelvin (and that different readings at the same true temperature are independent). I then take a reading from the thermometer in a room in which the true temperature is unknown. It would be reasonable to model the distribution of the reading as normal with known standard deviation (1) and unknown mean.

Similar examples could arise with other measuring tools.

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    $\begingroup$ I'd still argue that this is a case where you are estimating the standard deviation rather than truly knowing it. $\endgroup$
    – Dave
    May 24 at 19:43
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    $\begingroup$ How about if you buy a fancy lab thermometer and the data sheet that comes with it tells you the standard deviation of error? In some sense it's presumably estimated there, too, but the estimate is what you've got and you trust the manufacturer to have done it right. $\endgroup$ May 24 at 21:04
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    $\begingroup$ @Dave you can design a machine to have a known standard deviation by adding noise from a well-parametrised source. But then, any time you interact with the physical world, you don't trully "know" anything, you just have estimations. $\endgroup$
    – Davidmh
    May 25 at 12:38

In the real world, these types of problems primarily happen in manufacturing. You usually see them when there are strong constraints on the behavior of a variable. For example, the normal distribution assumes that a value can take on any value over $(-\infty,\infty)$ but if you are building cars, it is never going to happen that a tire will be larger than the factory it is being constructed in, let alone of nearly infinite size. Indeed, barring a monumental equipment failure, the diameter will never be much outside some easily and well-defined maximum or minimum. There also will never be a penny-sized tire either. I am not sure what a negative diameter could mean.

A simple example would be a submarine hiding in the ocean. Because it is of fixed size and shape its variance is fixed. Its location is not fixed. Indeed, it is hiding.

You might have some way to collect data about the submarine's location. The data could be a location somewhere on the ship, for example, a point near the fantail. Maybe it could be from a variety of points around the vessel. If the data collected depends on the geometry of the ship, then the data generation function will have a fixed variance. However, as the mean is somewhere in the ocean we do not know what it is.

One other note, not all formulations of the variance assume that the mean is known or that a point estimate exists for it. Consider the posterior probability $\Pr(\mu;\sigma^2|X)$ where $X$ is the observed data, $\mu$ is the population mean, and $\sigma^2$ is the population variance. A point estimate of the variance can be obtained by first marginalizing out $\mu$ so that $$\Pr(\sigma^2|X)=\int_{-\infty}^{\infty}\Pr(\mu;\sigma^2|X)\mathrm{d}\mu.$$

From that point, a utility function could be imposed upon the distribution and a point found. While there is information about $\mu$, it is a probability distribution instead of a known point or an estimator. There is more than one way to estimate variance depending on the goals and the circumstances.


Suppose we have a random variable $X\sim N(\mu, \sigma^2)$ where $\mu$ is the mean and unknown but with $\sigma^2=1$ variance. Now to answer your question how this could be possible, we can consider one way to relate the mean to the variance and see if we can use this method to back out what the mean should be. I'll use the fact that $Var(X)=\mathbb{E}[(X-\mu)^2]$. So,

$$1=Var(X)=\mathbb{E}[(X-\mu)^2]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=(\mu^2 + 1)-\mu^2 = 1$$

Here I use the fact that the second moment of a normal is $\mathbb{E}[X^2]=\mu^2 + \sigma^2$. Notice how we just get back $1$ again which basically means we have 2 unknowns and 1 equation and so the mean $\mu$ is free to vary.

This is just one example that we can know the variance but not the mean. Intuitively, the mean is like the "location" of a random variable, and the variance is how much that random variable is "spaced out" around that location. But we could always move that location around and keep the same variance. Hope that helps!

Edit: I am interpreting this question as asking how we could have a random variable whose mean we do not know but whose variance is known (as in the example above). That is, given a random variable and it's variance can we somehow always back out the mean. Another interpretation could be how is possible for $\mathbb{E}[X]$ to be undefined. For this I would give the example of a Cauchy distribution.

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    $\begingroup$ However, the variance of a Cauchy, while known, is infinite for any mean. $\endgroup$
    – ttw
    May 25 at 3:11
  • $\begingroup$ @ttw Thanks for pointing that out! I was just looking for something with an undefined first moment there. $\endgroup$
    – Ariel
    May 30 at 18:52

Say that you want to measure a voltage $V(t)$ that is varying in time and has a certain additive noise $e(t)$ (a random variable), for example, $$ V(t) = \sin(ft) + e(t) $$ Often one assumes that $e(t)$ has a constant zero mean and constant positive variance. One is interested in $\sin(ft)$, and the variance of $e(t)$ is unknown, how should one proceed? One way is to jointly handle $\sin(ft)$ and the variance of $e(t)$ simultaneously, but that may turn out to be complicated. Another way is to measure the voltage with no signal, i.e., $$ V(t) = e(t) $$ from which you can estimate the variance of $e(t)$. Now you can go back to the original problem with the estimated variance of $e(t)$, where you now assume that you know it. This often simplifies the analysis of the original problem.


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