Example of a Distribution that is not symmetric, but has a skewness of zero. In my probably class we saw that if a distribution is symmetric then the skewness will be zero. Intuitively this makes sense to me since if a data set is symmetric than for each point that is distance 'd' above the mean there will be a point that is distance 'd' below the mean (Although, in practice this is probably just very close to zero and actually zero).
However, it was implied in class that the converse is not true. That is, there may exist a distribution which is not symmetric, or skewed, but the skewness is zero.
Can anyone give me an actual example of a distribution that is skewed, but the values of the skewness equals zero?
Cheers
 A: There are infinite possible asymmetric distributions with zero skewness. To visualize a classical example, we can imagine a distribution where one tail is "long" but the other tail is "fat", so that the asymmetries even out. More precisely, all distributions for which the third standardized moment
$$\gamma_1=E [\frac {\mu_3}{\sigma^3}] $$
where $\mu_3$ is the third central moment and $\sigma$ is the SD, have skewness equal to zero. You can find a nice practical example of asymmetric distribution with zero skewness here.
A: We can construct a simple example such that $E(X)=0$ and $E(X^3)=0$, which implies that the third central moment is 0 and thus zero skewness. This gives us a hint. For instance, we only need take $X\in\{-3,-1,2\}$ and $P(X=-3)=a, P(X=-1)=b, P(X=2)=1-a-b$. Now, we have two unknows and two equations,
\begin{equation}
(-3)*a+(-1)*b+2*(1-a-b)=0~~\textrm{and}~~(-3)^3*a+(-1)^3*b+2^3(1-a-b)=0
\end{equation}
By solving $a=1/10$ and $b=1/2$, we get an asymmetric distribution and the skewness is 0. It is asymmetric by construction.
A: Here is a discrete example which has $\mathbb{E}[X]=\mathbb{E}[X^3]=0$, though there are many more on a similar theme.
$$\mathbb{P}(X=n) = \left\{
     \begin{array}{ll}
       0.03 &  n=-3 \\
       0.04 &  n=-2 \\
       0.25 &  n=-1 \\
       0.40 &  n=0 \\
       0.15 &  n=1 \\
       0.12 &  n=2 \\
       0.01 &  n=3  
     \end{array}
   \right.$$

A simpler, but less pretty, version could be 
$$\mathbb{P}(Y=m) = \left\{
     \begin{array}{ll}
       0.1 &  m=-3 \\
       0.5 &  m=-1 \\
       0.4 &  m=2   
     \end{array}
   \right.$$
though, despite again having $\mathbb{E}[X]=\mathbb{E}[X^3]=0$, it would have "median skewness" 
