How does the expected value relate to mean, median, etc. in a non-normal distribution? 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.
 A: (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 
A: There is a nice relationship between the harmonic, the geometric, and the arithmetic mean of a log-normally distributed random variable $X \sim \mathcal{LN}\left( \mu,\sigma^2 \right)$. The parameters of the distribution are related to the different means in the following way:

*

*$\mathrm{HM}(X) = \mathrm{e}^{\mu - \frac{1}{2}\sigma^2}$ (harmonic mean),

*$\mathrm{GM}(X) = \mathrm{e}^{\mu}$ (geometric mean),

*$\mathrm{AM}(X) = \mathrm{e}^{\mu + \frac{1}{2}\sigma^2}$ (arithmetic mean).

Using these identities, 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 exactly 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.
A: 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.
A: While it is correct that mathematically mean and expectation value are defined identically, for a skewed distribution this naming convention becomes misleading.
Imagine you are asking a friend about the housing prices in her city because you really like it there and actually think about moving to that city.
If the distribution of housing prizes were unimodal and symmetric, then your friend can tell you the mean price of houses and indeed you can expect to find most houses on the market around that mean value.
However, if the distribution of housing prices is unimodal and skewed, for example right-skewed with most houses in the lower price range to the left and only some exorbitant houses on the right, then the mean will be "skewed" to high prices on the right.
For this unimodal, skewed house price distribution you can expect to find most houses on the market around the median. 
