My understanding of what an estimator and an estimate is: Estimator: A rule to calculate an estimate Estimate: The value calculated from a set of data based on the estimator

Between these two terms, if I am asked to point out the random variable, I would say the estimate is the random variable since it's value will change randomly based on the samples in the dataset. But the answer I was given is that the Estimator is the random variable and the estimate is not a random variable. Why is that ?


3 Answers 3


Somewhat loosely -- I have a coin in front of me. The value of the next toss of the coin (let's take {Head=1, Tail=0} say) is a random variable.

It has some probability of taking the value $1$ ($\frac12$ if the experiment is "fair").

But once I have tossed it and observed the outcome, it's an observation, and that observation doesn't vary, I know what it is.

Consider now I will toss the coin twice ($X_1, X_2$). Both of these are random variables and so is their sum (the total number of heads in two tosses). So is their average (the proportion of head in two tosses) and their difference, and so forth.

That is, functions of random variables are in turn random variables.

So an estimator -- which is a function of random variables -- is itself a random variable.

But once you observe that random variable -- like when you observe a coin toss or any other random variable -- the observed value is just a number. It doesn't vary -- you know what it is. So an estimate -- the value you have calculated based on a sample is an observation on a random variable (the estimator) rather than a random variable itself.

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    $\begingroup$ +1, the thread worth mentioning is:stats.stackexchange.com/questions/7581/… $\endgroup$
    – Tim
    Commented Dec 7, 2017 at 9:58
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    $\begingroup$ but once we observe, why is it an estimate at all? there is nothing to be estimated after observation? $\endgroup$ Commented Nov 18, 2018 at 16:44
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    $\begingroup$ It's an estimate of an unobserved population parameter. For example in the coin tossing experiment where you don't know the coin to be fair, the observed average number of heads in $n$ tosses is a suitable estimate of the probability of a head. $\endgroup$
    – Glen_b
    Commented Feb 15, 2019 at 0:06
  • $\begingroup$ I'm really confused now because @Tim linked a thread that explicitly said an estimator is not a random variable $\endgroup$ Commented Mar 11, 2019 at 21:15
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    $\begingroup$ If you have a function (say with vector argument), $g$, then $g$ is just a function, but the value of that function when $g$ is applied to a collection of variates ($X=(X_1,X_2,...,X_n)$) whose components are random variables (peerhaps corresponding to some random sampling procedure on some population), then $T=g(X)$ will be a random variable. If you were to define $g$ as the estimator then $g$ is just a function. But if you called $T$ the estimator then $T$ is a random variable. Strictly this latter usage (as I have above) is rather loose (but quite common). ... ctd $\endgroup$
    – Glen_b
    Commented Mar 11, 2019 at 22:10

My understandings:

  1. An estimator is not only a function, which input is some random variable and output another random variable, but also a random variable, which is just the output of the function. Something like $y=y(x)$, when we talk about $y$, we mean both the function $y()$, and the result $y$.
  2. Example:an estimator $\overline X=\mu(X_1,X_2,X_3)=\frac{X_1+X_2+X_3}{3}$, we mean both $\mu()$ ,which is a function,and its result $\overline X$,which is random variable.
  3. The difference between estimator and estimate is about before observing or after observing.
  4. Actually, similar to an estimator, an estimate is both a function and a value(the function output) too. But the estimate is in the context of after observing, and by contrast, the estimator is in the context of before observing.

A picture illustrates the idea above:enter image description here

I have researched this question during my weekend, after reading lots of material from the internet, i am still confused. Although I am not completely sure that my answer is right, it seems like, to me, it's the only way to let everything make sense.

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    $\begingroup$ +1 You are making some good distinctions. Given your interest and dedication, might I recommend consulting a good textbook rather than relying entirely on the Internet? Textbooks can go deeply into a subject in a consistent manner, whereas depth and consistency are very hard to find online. $\endgroup$
    – whuber
    Commented May 12, 2019 at 16:11
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    $\begingroup$ hi whuber, I highly recommend this newonlinecourses.science.psu.edu/stat414 as an undergraduate level learning material of probability and statistic, and All of Statistics by Larry is also a good book for the beginner. Almost all my stat teachers recommend mathematical statistics by j. shao as a graduate level Textbook. I do agree with you that consistency and depth are very important for learning, I think textbooks and courses are for consistency while wiki and StackExchange are for depth. $\endgroup$
    – dawen
    Commented May 13, 2019 at 10:13
  • $\begingroup$ Link is gone, but web.archive.org seems to have it: web.archive.org/web/20190621221736/https://… $\endgroup$
    – Glen_b
    Commented Jun 14, 2023 at 1:27

Let $\left(\mathcal X, \mathcal A, \left(\mathbb P_\vartheta\right)_{\vartheta \in \Theta}\right)$ be a statistical model consisting of

  • a set $\mathcal X$ (the sample space),
  • a $\sigma$-algebra $\mathcal A$ on $\mathcal X$,
  • a family of probability measures $\left(\mathbb P_\vartheta\right)_{\vartheta \in \Theta}$ on $\mathcal A$ with
  • index set $\Theta$ (the parameter space) of cardinality bigger than one.

Let $(\Sigma, \mathcal S)$ be another event space, i.e., a measurable space.

Let $\tau: \Theta \to \Sigma$ be any function.

Then, an estimator $T$ of $\tau$ is defined to be any random variable from $(\mathcal X, \mathcal A)$ to $(\Sigma, \mathcal S)$.

In this answer I describe the relation between estimands, estimators, and estimates.


Georgii, H.-O. (2013). Stochastics: Introduction to probability and Statistics (E. Baake & M. Ortgiese, Trans.). Walter de Gruyter.

  • $\begingroup$ Straight to the point. +1. $\endgroup$ Commented Oct 16, 2023 at 1:23

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