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Let $X\sim SN(0,1,\lambda)$ (skew-normal) with density given by $$f(x|\lambda)=2\phi(x)\Phi(\lambda x)$$ Show that

a) $\lim_{\lambda\rightarrow > \infty}f(x|\lambda)=2\phi(x)I_{[0,\infty]}(x) \qquad \forall x\neq 0$

b)$X^2\sim \chi_1^2$ distribution

Since $\phi(.)$ and $\Phi(.)$ is density function and accumulated distribution function of a standard normal, I did

$$\lim_{\lambda\rightarrow \infty}f(x|\lambda)=2\phi(x)\lim_{\lambda\rightarrow \infty}\int_{-\infty}^{\lambda x}\phi(z)dz=2\phi(x)=2\phi(x)\int_{-\infty}^\infty \phi(z)dz=2\phi(x)$$

Is this wrong ? Why the indicator function in $[0,\infty]$?

For b) I tried

$$P(X^2\leq x)=P(X\leq \sqrt{x})=\int_{-\infty}^\sqrt{x}2\phi(z)\Phi(z)dz$$ $$=\int_{-\infty}^\sqrt{x}\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\Big(\int_{-\infty}^{\lambda z} \frac{1}{\sqrt{2\pi}}e^{-u^2/2}du\Big)dz$$

But that gives me a product $0\times \infty$.

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  • $\begingroup$ The presence of indicator function in $[0,\infty]$ is because, if you take $x<0$, then the integral converges to $0$ as $\lambda\rightarrow\infty$. $\endgroup$ – Sungjin Kim Sep 17 '16 at 1:09
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The solution is explained in here.

(a) is as I commented, we proceed (b).

The idea is exploiting the symmetry of $\phi(x)$. We will achieve (b) by proving that $|X|$ and $|Z|$ have identical distributions where $Z\sim N(0,1^2)$ is the standard normal distribution. Then $X^2$ will be identically distributed as $Z^2$ which is $\chi_1^2$.

$$ \begin{align} P(|X|\leq x) &= \int_{-x}^x 2 \phi(u) \Phi(\lambda u) du\\ &=\int_0^x 2\phi(u) \Phi(\lambda u ) du + \int_{-x}^0 2\phi(u)\Phi(\lambda u) du\\ &=\int_0^x 2\phi(u) \Phi(\lambda u ) du - \int_x^0 2\phi(-u)\Phi(-\lambda u) du \\ &=\int_0^x 2\phi(u) \Phi(\lambda u ) du+ \int_0^x 2\phi(u) \Phi(-\lambda u) du \\ &=\int_0^x 2\phi(u) \left(\Phi(\lambda u ) + \Phi(-\lambda u) \right) du\\ &=\int_0^x 2\phi(u) du = P(|Z|\leq x). \end{align} $$

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