I think it is easiest if I describe my question using an example:


$$ \mathbf{x} \triangleq [ x_1,x_2,x_3] $$ and for sake of explanation let's assume that $$ \{x_i \in [0,3] \}_{i=1}^{3} $$ which is to say that each element in the array $\mathbf{x}$ can take one of four values: 0,1,2 or 3. All combinations are allowed including repetitions e.g. $\mathbf{x}=[1,1,1]$.

Let us further say that $\mathbf{x}$ follows the distribution $P(\mathbf{X}\mid \boldsymbol \theta)$ and a sample is drawn thus: $\mathbf{x} \sim P(\mathbf{X}\mid \boldsymbol \theta)$.

The expected value of $\mathbf{X}$ is simply $$ \mathbb{E}[\mathbf{X}] = \sum_i \mathbf{x}_i \cdot P(\mathbf{x}_i) $$

But (here starts the real question) suppose that I now apply an operation to $\mathbf{X}$ e.g. concatenation of a reversed copy of itself i.e. $$ f(\mathbf{x}) = \mathbf{x} \cup \texttt{reverse}(\mathbf{x}) $$ Where if we have drawn a sample e.g. $\mathbf{x} = [1,2,3]$ then $f(\mathbf{x}) = [1,2,3,3,2,1]$.

What can we say about the expectation of $f(\mathbf{X})$? I believe I am onto the Law of the unconscious statistician (https://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician) but I do not understand if the law holds for my above function (i.e. concatenation which is the operation of interest here).

I.e. what can we say about the expected value of $\mathbf{X}$ is simply $$ \mathbb{E}[f(\mathbf{X})] = \sum_i f(\mathbf{x}_i) \cdot P(f(\mathbf{x}_i)) $$ in terms of the original distribution over $\mathbf{x}$, if anything at all?

An explanation with an example would be most helpful (or even an explained proof) since I do not currently understand how this can possibly work (for my function $f(\cdot)$ since it is neither differentiable nor the inverse monotonic).

  • 1
    $\begingroup$ This notation is inconsistent (and somewhat nonsensical in places, although the general intent is evident), which makes it difficult to determine what you're really trying to ask. Your definition of $\mathbf x$ in particular is that of a vector valued random variable. Thus, $E[\mathbf x]$ also is a vector. Furthermore -- I hope this is easy to see -- your function $f:\mathbb R^3\to\mathbb R^6$ is linear. All you have to do is apply linearity of expectation to conclude $E[f(\mathbf x)] = f(E[\mathbf x]).$ If you differ with this interpretation, please clarify. $\endgroup$
    – whuber
    May 25 at 23:42
  • $\begingroup$ Please explain where it is inconsistent and I shall endeavour to fix it so that it makes more sense. Your definition of $f$ is further incorrect since $f$ is a discrete map and not continuous (but I may have misunderstood you?) I do not think we mean the same thing? But again, please tell me where I can improve on the notation. I struggle to see how the expectation of $\mathbf{x} \in \mathbb{Z}^3$ -- but I may misunderstand here, is a concatenation of a reverse copy a linear operation? $\endgroup$
    – Astrid
    May 25 at 23:49
  • $\begingroup$ If you insist on that nit, $f$ has a cpntinuous extension to all of $\mathbb R^3.$ It is represented by the matrix $$\pmatrix{1&0&0&0&0&1\\0&1&0&0&1&0\\0&0&1&1&0&0}.$$ As far as notation goes, the first formula is idiosyncratic and not likely to be understood by many; the second uses set brackets in an unusual way; and the text suggests you are thinking of an expectation as a number rather than a vector, indicating other potential problems with understanding the notation. $\endgroup$
    – whuber
    May 26 at 11:24
  • $\begingroup$ I’m sorry I don’t agree. The first formula is just an array definition. Standard fare. The second is domain assignment of each element. Also standard. And the standard definition of the expectation does give a number for one random variable, and an array for arrays. I’m not insisting on f being discrete, it is fundamental to my question. Imposing another form imposes another question not of interest. $\endgroup$
    – Astrid
    May 26 at 13:32
  • $\begingroup$ Re "standard fare:" the notation of computing does not have a well-understood mathematical or statistical meaning. By using it, you both obscure the issues and, apparently, confuse yourself. $\endgroup$
    – whuber
    May 26 at 13:48

1 Answer 1


I think this is a lot simpler than you think it is :)

We have a random vector $\mathbf X=(X_1,X_2, \cdots, X_n)$, with expectation $E[\mathbf X] = (E[X_1], E[X_2], \cdots, E[X_n])$.

The expectation of a vector is just a vector of each element's expectation.

Then, the reverse concatenation $(\mathbf X, \mathbf X_\text{rev})=(X_1,\cdots, X_n, X_n, \cdots, X_1)$ has expectation:

$$ \begin{aligned} E[(\mathbf X, \mathbf X_\text{rev})] &= (E[X_1], \cdots, E[X_n], E[X_n],\cdots, E[X_1]) \end{aligned} $$

  • $\begingroup$ Interesting and thank you! My confusion persists a bit. I want to treat the whole vector as the sample rather than the individual bits. Is there a world in which that is done? So we would have sufficient statistics over vectors and not elements as shown in your answer? $\endgroup$
    – Astrid
    May 26 at 2:44
  • $\begingroup$ Hi @Astrid. But the whole vector isn't the sample - only the first half (i.e. $\mathbf X$) is the sample (under how you've defined your problem). I'm not sure I understand what you're asking. $\endgroup$
    – Alex J
    May 26 at 3:03
  • $\begingroup$ No you’re right apologies; same question but regarding only the first half. So treating the first half as the sample and then providing statistics over that. $\endgroup$
    – Astrid
    May 26 at 4:34
  • $\begingroup$ Still not sure I understand - what about your question is not answered by my second sentence (starting from "We have a random vector [...]")? If I replaced the words "random vector" with "random sample", my argument still holds. $\endgroup$
    – Alex J
    May 26 at 4:39
  • $\begingroup$ You’ve explained it very well. Thank you again. I’ll have to collect my thoughts to see if I can pose a better question. $\endgroup$
    – Astrid
    May 26 at 4:41

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