Let $X\sim F(x)$ be a (univariate) random variable defined by distribution function $F$. If the expected value exists, it is equal to $ \mathbb E[X] = \underset{\mu}{\arg\min}\left\{\mathbb E\left\vert X-\mu\right\vert^2\right\} $.

However, the expected value does not have to exist.

What conditions are there on the exponent $p$ such that $ \underset{\mu}{\arg\min}\left\{\mathbb E\left\vert X-\mu\right\vert^p\right\} $ must exist? How low do we have to go before it must exist? We trivially get the $\arg\min$ equal to $\mathbb R$ if $p=0$ and get the median as the $\arg\min$ for $p=1$. What about $p=1.5$ or $1.9$ or $1.99$?

I wonder if $2$ is the supremum, but not in the set, of $p$ that guarantee an $\arg\min$, and that any $p<2$ must have a defined $\arg\min$. If there is interesting commentary about uniqueness, I am on board with going down that rabbit hole.

My inspiration came from here, and I wonder if a reasonable answer to the linked question is to lower $p$ until the $\arg\min$ must exist.

  • $\begingroup$ Can you say a little more about what you mean when you say that $\textrm{argmin}...$ exists? I imagine you mean that the set of minimizers has one and only one member (for example: it is given by $\mathbb{R}$ if $p=0$)? And also: are you sure you don't want to take an absolute value before exponentiation, i.e. $|X-\mu|^p$? relevant for e.g. $p=3$. Anyways, awesome question and I can't wait to see the answers :) $\endgroup$ Commented Mar 31 at 23:04
  • $\begingroup$ @JohnMadden I have made some edits that I hope clarify my question. I’m really interested in lowering $p$ below $2$ until the $\arg\min$ must exist. I have my doubts that any $p>2$ requires the $\arg\min$ to exist (but am open to learning why that is wrong if it is). $\endgroup$
    – Dave
    Commented Mar 31 at 23:23
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    $\begingroup$ "$\mathbb E[X] = \underset{\mu}{\arg\min}\left\{\left\vert X-\mu\right\vert^2\right\}$" -- Wrong. The correct statement is $\mathbb E[X] = \underset{\mu}{\arg\min}\color{red}{E}\left\{\left\vert X-\mu\right\vert^2\right\}$. $\endgroup$
    – Zhanxiong
    Commented Apr 1 at 0:44
  • 2
    $\begingroup$ For the symmetric stable distributions, $p=1$ corresponds to the Cauchy, which leads to the median, as you've noted, but for $p<1$, say, $1/2$, does the expectation $|X-\mu|^{1/2}$ always exist? I doubt it, but am not in a position to work out an example ATM. $\endgroup$
    – jbowman
    Commented Apr 1 at 2:02
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    $\begingroup$ @picky_porpoise I think a proof of that is exactly what I want! $\endgroup$
    – Dave
    Commented Apr 1 at 13:59

2 Answers 2


The choice $p=0$ is the only one such that the arg min in question always exists. For any $p>0$ we can find a distribution $F$ such that $\mathbb{E} \vert X - \mu \vert^p$ does not exist for any $\mu \in \mathbb{R}$ and thus the arg min is empty.

The simplest way to see this, is to take the Pareto distribution with parameter $\alpha > 0$ on $[1, \infty)$ for $F$. Its density is given by $$ f(x) = \frac{\alpha}{x^{\alpha + 1}} $$ for $x > 1$. From basic analysis we can see that the $p$-th moment $\mathbb{E}\vert X \vert^p$ of an $\alpha$-Pareto distribution exists only if $\alpha > p$ and as pointed out in the answer by jacques, the existence of $\mathbb{E} \vert X - \mu \vert^p$ is equivalent to the existence of $\mathbb{E} \vert X \vert^p$. Hence, $\underset{\mu}{\arg\min} \, \mathbb{E}\left\vert X-\mu\right\vert^p $ is empty for a sufficiently heavy-tailed Pareto distribution.


Generally, for $E[\vert X-\mu \vert^p]$ to be finite, it is necessary and sufficient that $E[\vert X \vert^p] < \infty$, i.e. that $X \in L^p$ (I think you forgot the expectation in your question), because you can estimate $\vert X - \mu\vert^p \leq 2^p (\vert X \vert^p + \vert\mu\vert^p)$ (and for the other direction use $X = (X - \mu ) + \mu$). In that case, also the arg-min will be well-defined.

Generally, for finite measure spaces, and in particular for probability spaces (where the total mass is 1), $L^p \subset L^q$ for $p > q$ (this is a consequence of Jensen's inequality applied to the function $\varphi(x)=\vert x \vert^{p/q}$... although it can also be proven using Hölder's inequality applied to $f = \vert X \vert^q$ and $g = 1$). In general, the characterization of the median/mean of a random variable in terms of its minimizing $E \vert X - \mu \vert $ and $E[(X - \mu)^2]$ is not valid in general (e.g. mean exists if $X \in L^1$, but the criterion is well-defined if $X \in L^2$ [example: Pareto distribution for $\alpha \in (1, 2)$ has finite mean but infinite variance, hence the criterion is not well-defined]... moreover, the median always exists, but the criterion is only well-defined if $X \in L^1$ [example: Cauchy distribution]).

  • $\begingroup$ What do you mean in that last comment about the median? $\endgroup$
    – Dave
    Commented Apr 1 at 2:41
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    $\begingroup$ @Dave The median exists for all distributions, but only for the ones with finite mean can it be represented as the arg min of the expected absolute distance. $\endgroup$ Commented Apr 1 at 12:47

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