# 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?

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.

• 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 :) Commented Mar 31 at 23:04
• @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).
– Dave
Commented Mar 31 at 23:23
• "$\mathbb E[X] = \underset{\mu}{\arg\min}\left\{\left\vert X-\mu\right\vert^2\right\}$" -- Wrong. The correct statement is $\mathbb E[X] = \underset{\mu}{\arg\min}\color{red}{E}\left\{\left\vert X-\mu\right\vert^2\right\}$. Commented Apr 1 at 0:44
• 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. Commented Apr 1 at 2:02
• @picky_porpoise I think a proof of that is exactly what I want!
– Dave
Commented Apr 1 at 13:59

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]).