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How does the expected value of a continuous random variable relate to its arithmetic mean, median, etc. in a non-normal distribution (eg. skew-normal)? I'm interested in any common/interesting distributions (eg. log-normal, simple bi/multimodal distributions, anything else weird and wonderful).

I'm looking mostly for qualitative answers, but any quantitative or formulaic answers are also welcome. I'd particularly like to see any visual representations that make it clearer.

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  • $\begingroup$ Can you be a bit clearer? The arithmetic mean and median are functions we apply to data, not anything intrinsic to particular distributions... for example, data doesn't have to be Normal in order for you to calculate the sample mean. $\endgroup$ – guest Feb 28 '12 at 7:59
  • $\begingroup$ Ok, so the question should technically be "how does the expected value relate to the mean, median etc. of data drawn randomly from a particular probability distribution?" I'm looking for simple, intuitive understandings, similar to the way you can intuitively say that when a distribution is more skewed, the median and the mean are further apart, and the median may give a better indication of where the data lies. $\endgroup$ – naught101 Feb 28 '12 at 10:40
  • $\begingroup$ Heh. Thanks Marco. I've clearly been reading things wrong. May as well write that as an answer, I'll chose it at he best answer. $\endgroup$ – naught101 Mar 2 '12 at 5:02
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(partially converted from my now-deleted comment above)

The expected value and the arithmetic mean are the exact same thing. The median is related to the mean in a non-trivial way but you can say a few things about their relation:

  • when a distribution is symmetric, the mean and the median are the same

  • when a distribution is negatively skewed, the median is usually greater than the mean

  • when a distribution is positively skewed, the median is usually less than the mean

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  • $\begingroup$ Interesting. What examples are there of the unusual behaviour of a negatively skewed distribution where the mean is greater than the median? $\endgroup$ – naught101 Mar 4 '12 at 6:58
  • $\begingroup$ @naught101: is this a typo? A negatively skewed distribution is one in which the left-of-centre outcomes occur more frequently than the right-of-centre outcomes, and therefore the "tail" of low frequency outcomes goes out to the right. In such a situation, the hump on the left will always pull the (arithmetic) mean left of centre, while the tail on the right will keep the median greater than the mean. $\endgroup$ – Assad Ebrahim Oct 23 '14 at 17:50
  • $\begingroup$ @AssadEbrahim: No, it was a reference to Macro's comment "the median is usually greater than the mean" - I was asking for counter examples. $\endgroup$ – naught101 Oct 24 '14 at 0:34
  • $\begingroup$ @naught101: The counter-examples in the case of a unimodal distribution are his next line: when the hump is to the right then the tail to the left pulls the median below the mean. The longer the tail, the greater the gap between median and mean. $\endgroup$ – Assad Ebrahim Oct 24 '14 at 8:49
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    $\begingroup$ What are the practical circumstances in which one would use a median over a mean or vice versa? For example in survival analysis where lifetimes follow an exponential distribution, should I use the median (so half the things last longer, half last less) or the mean (the "expected" lifetime) if I had to predict life/death as a binary outcome? $\endgroup$ – drevicko Jun 19 '15 at 5:33
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There is a nice relationship between the harmonic, the geometric, and the arithmetic mean of a log-normally distributed random variable $X \sim \mathcal{LN}\left( \mu,\sigma^2 \right)$. Let

  • $\mathrm{HM}(X) = \mathrm{e}^{\mu - \frac{1}{2}\sigma^2}$ (harmonic mean),
  • $\mathrm{GM}(X) = \mathrm{e}^{\mu}$ (geometric mean),
  • $\mathrm{AM}(X) = \mathrm{e}^{\mu + \frac{1}{2}\sigma^2}$ (arithmetic mean).

It is not difficult to see that the product of the harmonic and the arithmetic mean yields the square of the geometric mean, i.e.

$$ \mathrm{HM}(X) \cdot \mathrm{AM}(X) = \mathrm{GM}^2(X). $$

Since all values are positive, we can take the squre root and find that the geometric mean of $X$ is the geometric mean of the harmonic mean of $X$ and the arithmetic mean of $X$, i.e.

$$ \mathrm{GM}(X) = \sqrt{ \mathrm{HM}(X) \cdot \mathrm{AM}(X) }. $$

Furthermore, the well-known HM-GM-AM inequality

$$ \mathrm{HM}(X) \leq \mathrm{GM}(X) \leq \mathrm{AM}(X) $$

can be expressed as

$$ \mathrm{HM}(X) \cdot \sqrt{\mathrm{GVar}(X)} = \mathrm{GM}(X) = \dfrac{\mathrm{AM}(X)}{\sqrt{\mathrm{GVar}(X)}}, $$

where $\mathrm{GVar}(X) = \mathrm{e}^{\sigma^2}$ is the geometric variance.

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For completeness, there are also distributions for which the mean is not well defined. A classic example is the Cauchy distribution (this answer has a nice explanation of why). Another important example is the Pareto distribution with exponent less than 2.

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    $\begingroup$ Several iff's. A power law is not a distribution, but a Pareto distribution is a power law. This relates to the non-integrability of a log-convex power function at $x=0$. For a power law, you mean less than 2, not greater than 2. $\endgroup$ – Carl Apr 22 '18 at 7:10
  • $\begingroup$ @Carl good points - I edited the answer accordingly. Many thx (: $\endgroup$ – drevicko Apr 25 '18 at 2:33

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