Let $Y_1\sim SN(\mu_1,\sigma_1^2,\lambda)$ and $Y_2\sim N(\mu_2,\sigma_2^2)$ independents. Show that $Y_1+Y_2$ have a skew-normal distribution and find the parameters of this distribution.

Since the random variables are independent I tried to use convolution. Let $Z=Y_1+Y_2$


Here $\phi()$ and $\Phi()$ are the standard normal pdf and cdf, respectively.


For simplified notations, let $k=2\frac{1}{\sqrt{2\pi\sigma_1}}\frac{1}{\sqrt{2\pi\sigma_2}}$

\begin{align*}f_Z(z)&=k\int_{-\infty}^{\infty}\exp\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_1^2(y_1-\mu_1)^2+\sigma_2^2((z-y_1)-\mu_2)^2\Big)\Big)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1\\ &=k\int_{-\infty}^{\infty}\exp\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_2^2(y_1^2-2y_1\mu_1+\mu_1)+\sigma_1^2((z-y_1)^2-2(z-y_1)\mu_2+\mu_2^2)\Big)\Big)\\&\quad\times\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1=k\int_{-\infty}^{\infty} \exp\\&\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_2^2(y_1^2-2y_1\mu_1+\mu_1)+\sigma_1^2(z^2-2zy_1+y_1^2-2z\mu_2+2y_1\mu_2+\mu_2^2)\Big)\Big)\\&\quad\times\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1 \end{align*}

But I'm stuck at this point.

EDIT: Following the suggestions in the comments, taking $\mu_1=\mu_2=0$ and $\sigma_1^2=\sigma_2^2=1$ \begin{align*} &\int_{-\infty}^\infty 2\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\exp\Big(-\frac{1}{2}[y_1^2+z^2-2zy_1+y_1^2]\Big)\Phi(\lambda y_1)dy_1 \\&\int_{-\infty}^\infty 2\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\exp\Big(-\frac{1}{2}y_1^2\Big)\Phi(\lambda y_1) \exp\Big(-\frac{1}{2}(z-y_1)^2\Big)dy_1\end{align*}

is skew-normal.

  • 2
    $\begingroup$ Trying a simpler case of $\mu_1 = \mu_2 = 0$, $\sigma_1 = \sigma_2 = 1$ will reduce the clutter quite a bit and make you see the forest instead of the trees? $\endgroup$ Sep 19, 2016 at 13:05
  • 1
    $\begingroup$ I think Dilip's suggestion is a good one, but you might want to check your expansion of the first quadratic term carefully. (It won't fix your immediate problem but it will matter in the end) $\endgroup$
    – Glen_b
    Sep 19, 2016 at 13:09

1 Answer 1


Reparameterizing the skew in terms of $\delta=\lambda/\sqrt{1+\lambda^2}$ and using the mgf of the skew normal (see below), since $Y_1$ and $Y_2$ are independent, $Z=Y_1 + Y_2$ has mgf \begin{align} M_Z(t) &= M_{Y_1}(t)M_{Y_2}(t) \\ &= 2e^{\mu_1 t +\sigma_1^2 t^2/2}\Phi(\sigma_1\delta t)e^{\mu_2t +\sigma_2^2 t^2/2} \\ &= 2e^{(\mu_1+\mu_2)t + (\sigma_1^2+\sigma_2^2)t^2/2}\Phi(\sigma_1 \delta t) \\ &= 2e^{\mu t + \sigma^2 t^2/2}\Phi(\sigma \delta' t), \end{align} that is, the mgf of a skew normal with parameters $\mu=\mu_1+\mu_2$, $\sigma^2=\sigma_1^2+\sigma_2^2$ and $\sigma\delta'=\sigma_1\delta$ where $\delta'$ is the new skew parameter. Hence,
$$ \delta'=\delta\frac{\sigma_1}\sigma=\delta\frac{\sigma_1}{\sqrt{\sigma_1^2+\sigma_2^2}}. $$ In the other parameterization, the new skew parameter $\lambda'$ can be written, after some algebra, e.g. as $$ \lambda' = \frac{\delta'}{\sqrt{1-\delta'^2}}=\frac{\lambda}{\sqrt{1 + \frac{\sigma_2^2}{\sigma_1^2}(1+\lambda^2)}}. $$

The mgf of a standard skew normal can be derived as follows: \begin{align} M_X(t)&=Ee^{tX} \\ &=\int_{-\infty}^\infty e^{xt}2\frac1{\sqrt{2\pi}}e^{-x^2/2}\Phi(\lambda x)dx \\ &=2\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{-\frac12(x^2-2tx)}\Phi(\lambda x)dx\\ &=2\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{-\frac12((x-t)^2-t^2)}\Phi(\lambda x)dx \\ &=2e^{t^2/2} \int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{-\frac12(x-t)^2}P(Z\le \lambda x)dx, & \text{where }Z \sim N(0,1) \\ &=2e^{t^2/2} P(Z\le \lambda U), & \text{where }U \sim N(t,1)\\ &=2e^{t^2/2} P(Z - \lambda U \le 0) \\ &=2e^{t^2/2} P(\frac{Z - \lambda U +\lambda t}{\sqrt{1+\lambda^2}} \le \frac{\lambda t}{\sqrt{1+\lambda^2}}) \\ &=2e^{t^2/2}\Phi(\frac\lambda{\sqrt{1+\lambda^2}}t). \end{align} The mgf of a skew normal with location and scale parameters $\mu$ and $\sigma$ is then $$ M_{\mu + \sigma X}(t)=Ee^{(\mu+\sigma X)t} = e^{\mu t}M_X(\sigma t) = 2e^{\mu t+\sigma^2 t^2/2}\Phi(\frac\lambda{\sqrt{1+\lambda^2}}\sigma t). $$

  • $\begingroup$ I don't understood how you get this $\delta'=\delta\frac{\sigma_1}\sigma$ can you give me more details? $\endgroup$
    – user72621
    Sep 24, 2016 at 21:07
  • $\begingroup$ You just equate the quantities appearing before $t$ and $t^2$ in the exponential and in the argument of the $\Phi$-function to find the new parameters. $\endgroup$ Sep 24, 2016 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.