I want to generate three random variable X, Y, and Z that have a specific coskewness. I am defining coskewness between three random variable instead of between two random variable which seems to be more common:
$\displaystyle S(X,Y,Z)={\frac{E[(X-E[X])(Y-E[Y])(Z-E[Z])]}{\sigma_{X}\sigma_{Y}\sigma_{Z}}}$
As I understand it, mean and std are Sufficient statistic for a gaussian distribution so higher order statistics are not unique and can be calculated using only those two; Am I correct? and if so does that mean for my case, I need to look into other pdf families?
I have seen this question link but not sure how I can use it for three r.v.
Suppose we want a coskewness of $\xi$. If $\xi=0$, this is easy. Just take $X,Y,Z\sim N(0,1)$ independent.
For $\alpha> 0$, there exists a random variable $A$ with mean zero, variance $1$, and $\mathbb{E}(A^4)=\alpha^2+1$. To see this, define $f(t)=\frac{t^3+1}{t+1}$ for $t\geqslant 1$. Note that $f(1)=1$, $\lim_{t\to \infty}f(t)=\infty$, and $f$ is continuous. Therefore for any $0<\alpha<\infty$, since $f(1)=1<\alpha^2+1<\infty$, there exists $t>1$ such that $$\alpha^2+1 = \frac{t^3+1}{t+1}.$$ Let $A$ be a random variable taking the values $-1$, $0$, and $t$ according to $$\mathbb{P}(A=a)=\left\{\begin{array}{ll} \frac{t}{t^2+t} & : a = -1 \\ \frac{1}{t^2+t} & : a = t \\ \frac{t^2-1}{t^2+t} & : a = 0.\end{array}\right.$$
Then $$\mathbb{E}(A)=t\cdot \frac{1}{t^2+t} -1\cdot \frac{t}{t^2+t} =0,$$ $$\mathbb{E}(A^2) = t^2\cdot \frac{1}{t^2+t}+1\cdot \frac{t}{t^2+t}=1,$$ $$\mathbb{E}(A^4) = t^4\cdot \frac{1}{t^2+t}+1\cdot \frac{t}{t^2+t}=\frac{t^3+1}{t+1}=\alpha^2+1.$$
Let $X=A$, $Y=A$, and $$Z=A^2-1.$$ Then $$\mathbb{E}(X)=\mathbb{E}(Y)=\mathbb{E}(Z)=0.$$ Moreover, $\sigma_X=\sigma_Y=1$ and $\sigma_Z=\alpha$. This is because $$\text{Var}(Z)=\text{Var}(A^2)=\mathbb{E}(A^4)-\mathbb{E}(A^2)^2=\alpha^2+1-1=\alpha^2,$$ and $\sigma_Z=\sqrt{\alpha^2}=\alpha$.
Note that $$XYZ=A^4-A^2,$$ which has mean $\alpha^2+1-1=\alpha^2$. Since $X,Y,Z$ are mean zero, the coskewness is $\alpha^2/\alpha=\alpha$.
Fix $\xi\in\mathbb{R}$. If $\xi=0$, let $X,Y,Z$ be independent $N(0,1)$ random variables, which gives coskewnss $0=\xi$.
If $\xi\neq 0$, let $\alpha=|\xi|$ and let $X,Y,Z$ be as above with this $\alpha$. Then if $\xi>0$, $X,Y,Z$ have coskewness $\alpha=\xi$. If $\xi<0$, $-X,Y,Z$ have coskewness $-\alpha=\xi$.