I'd like to prove that a random variable $X$ and it's linear transformation $Y = aX+b, a>0$ have the same skewness coefficient, where the skewness coefficient is, according to Pearson, given by:
$$\gamma_x = \frac{\kappa_3}{\kappa_2^{1.5}}=\frac{E[(X-E[X])^3]}{\sigma_X^3}$$
Where $\kappa_2 = \frac{d^2}{dt^2}k_x(t)$ at $t=0$ and $\kappa_3 = \frac{d^3}{dt^3}k_x(t)$ at $t=0$ and $\kappa_x(t):= log [m_x(t)]$.
Since the variance of a random variable is equal to $\kappa_2$ I determined that the denominator of the skewness coefficient of $Y$, $\gamma_Y$, is equal to $(a^2\sigma_X^2)^{1.5}=a^3\sigma_X^3$.
The numerator of $\gamma_Y$ is equal to $E[(Y-E[Y])^3] = E[(Y-(aE[X]+b))^3]$, any help with calculating this expectation is appreciated.
As $Y = aX + b$, we have $$ Y - E[Y] = aX + b - aE[X] - b = a(X - E[X]) $$ Hence, the numerator of $\gamma_Y$ is $$ E[(Y - E[Y])^3] = a^3 E[(X- E[X])^3]. $$