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I am looking for a random variable $b$ which has a specific Co Skewnes $S$ to another random variable $b$.

$S(a,a,b) =\frac{\mathbf{E}[(a-\mu_a)(a-\mu_a)(b-\mu_b)]}{\sigma_a \sigma_a \sigma_b} = c$ where c is a real number and $b$ is skewed normally distributed.

Furthermore:

$\mu_a = 0\\ \mu_b=0 \\ {sd}_a = 1 \\ {sd}_b = 1 \\ {skew}_a = f \\ Cov(a,b) = 0$

$\mu$ referring to the mean; $sd$ referring to standard deviation; where f is any real number; $skew$ referring to skewness; $Cov$ refering to Covariance

I know how to find $a$ which fullfills $\mu_a = 0$; ${sd}_a = 1$; ${skew}_a = f$ through a skewnormal, but I have no clue how to find $b$. If possible $b$ should have some relation to the normal distribution.

Due to my restrictions it simplifies to $S(a,a,b) =\mathbf{E}[(a^2b)] = c$

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If I understand correctly:

You know the distribution of $a$, which is skew-normal with known mean $0$, standard deviation $1$, and skewness $f$.

You seek a distribution for $b$, with known mean $0$ and standard deviation $1$, such that $$E[a b]=0$$ (equivalent to cov$(a,b)=0$ when $a$ and $b$ have mean $0$) and $$E[a^2 b]=c$$

Assuming $P(a,b)=P(a) P(b)$ $$E[a^2 b]=\int a^2 P(a) da\underbrace{\int b P(b) db}_{=\mu_b=0}=c$$ So $a$ and $b$ cannot be independent.

Assuming instead that $b$ is a function of $a$, $b=f(a)$. The covariance of $a$ and $b$ must be $0$ $$\int a f(a) P(a) da=0$$ The mean of $b$ must be $0$: $$\int f(a) P(a) da=0$$ The standard deviation of $b$ must be $1$: $$\int f(a)^2 P(a) da=1$$ The co-skewness condition is $$\int a^2 f(a) P(a) da=c$$

These are four conditions. Let $f(a)$ be a third-degree polynomial with four coefficients $f(a)=f_0+f_1 a + f_2 a^2 + f_3 a^3$. We should be able to solve the four conditions for the four coefficients. Indeed three of these conditions are linear in the $f_i$, only one is quadratic.

Let's start with the three linear conditions $$\int a f(a) P(a) da=\sum_{i=0}^3 f_i \int a^{i+1} P(a) da=0$$ $$\int f(a) P(a) da=\sum_{i=0}^3 f_i \int a^i P(a) da=0$$ $$\int a^2 f(a) P(a) da=\sum_{i=0}^3 f_i \int a^{i+2} P(a) da=c$$ In matrix form $$\left[\begin{matrix}\int P(a) da & \int a P(a) da & \int a^2 P(a) da & \int a^3 P(a) da \\ \int a P(a) da & \int a^2 P(a) da & \int a^3 P(a) da & \int a^4 P(a) da \\ \int a^2 P(a) da & \int a^3 P(a) da & \int a^4 P(a) da & \int a^5 P(a) da \end{matrix}\right]\left[\begin{matrix}f_0 \\ f_1 \\ f_2 \\ f_3\end{matrix}\right]=\left[\begin{matrix}0 \\ 0 \\ c\end{matrix}\right]$$ Simplifying $$\left[\begin{matrix}1 & 0 & 1 & f \\ 0 & 1 & f & \int a^4 P(a) da \\ 1 & f & \int a^4 P(a) da & \int a^5 P(a) da \end{matrix}\right]\left[\begin{matrix}f_0 \\ f_1 \\ f_2 \\ f_3\end{matrix}\right]=\left[\begin{matrix}0 \\ 0 \\ c\end{matrix}\right]$$ (it may be possible to write down explicit expressions for the remaining two integrals as well, since you know $P(a)$). If this linear system of equations is consistent, it can be solved for $f_0,f_1,f_2$ as a function of $f_3$, which can then be substituted in $$\int f(a)^2 P(a) da=1$$ to get a quadratic equation in $f_3$.

As an example, let $f=0$ and $c=\sqrt{2}$. In this case the transformation is $b=f(a)=\frac{a^2-1}{\sqrt{2}}$ and the transformed distribution is $$P(b)=\frac{\sqrt[4]{2} e^{-\frac{b}{\sqrt{2}}-\frac{1}{2}}}{\sqrt{\pi } \sqrt{2 b+\sqrt{2}}}$$ for $b>-1/\sqrt{2}$, and $0$ otherwise.

A distribution with prescribed co-skewness relative to the normal distribution

Note that if you don't demand $sd_b=1$, the problem simplifies greatly: the transformation can be quadratic rather than cubic and all three coefficients can be found from the linear system.

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  • $\begingroup$ Thanks a lot @Wouter. I haven't fully understood the answer, because I did not take enough time. It looks really detailed. Thanks a lot! I will take a longer look and mark it as correct as soon as I understood it thanks! $\endgroup$
    – PalimPalim
    Commented Sep 9, 2017 at 10:34
  • $\begingroup$ I do not understand what $P(a)$ stands for. Is it probability, i.e for a normal distribution with $mean=0$ and $sd=0$ $P(0)$ would be $0.5$? $\endgroup$
    – PalimPalim
    Commented Sep 18, 2017 at 10:53
  • $\begingroup$ It's the probability density function $\endgroup$
    – Wouter
    Commented Sep 18, 2017 at 12:51
  • $\begingroup$ Looking for example at the mean. I think it must be $\int f(a) P(f(a))df(a) = 0$. I don't think this is the same as $\int f(a) P(a)da = 0$. Here they write about transformations of r.v.s, but only for strictly increasing. How do you get from $P(f(a))$ to $P(a)$? $\endgroup$
    – PalimPalim
    Commented Sep 25, 2017 at 11:24
  • $\begingroup$ That's because it's easy for strictly increasing functions. You know the cumulative distribution function $F(x_0)=P(x\leq x_0)$, where $P$ is the peobability. You want to determine $P(f(x)\leq x_0)$, so you need to actually determine which $x$ obey $f(x)\leq x_0$. It is possible to solve this for cubics, but very unpleasant. (you'll note that I picked an example where the cubic becomes quadratic, for this reason) $\endgroup$
    – Wouter
    Commented Sep 25, 2017 at 13:24

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