# Show that $Y_1+Y_2$ have distribution skew-normal

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 $Z=Y_1+Y_2$

$$f_Z(z)=\int_{-\infty}^{\infty}2\phi(y_1|\mu_1,\sigma_1)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\phi(z-y_1|\mu_2,\sigma_2^2)\,\text{d}y_1$$

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

$$f_Z(z)=\int_{-\infty}^{\infty}2\frac{1}{\sqrt{2\pi\sigma_1}}\frac{1}{\sqrt{2\pi\sigma_2}}exp\Big(-\frac{1}{2\sigma_1^2}(y_1-\mu)^2-\frac{1}{2\sigma_2^2}((z-y_1)^2-\mu)^2\Big)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1$$

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

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.

• 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? Sep 19, 2016 at 13:05
• 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) Sep 19, 2016 at 13:09

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 $$M_{\mu + \sigma X}(t)=Ee^{(\mu+\sigma X)t} = e^{\mu t}M_X(\sigma t) = 2e^{\mu t+\sigma^2 t^2/2}\Phi(\frac\lambda{\sqrt{1+\lambda^2}}\sigma t).$$
• I don't understood how you get this $\delta'=\delta\frac{\sigma_1}\sigma$ can you give me more details?
• 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. Sep 24, 2016 at 21:23