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Let $X\sim SN(0,1,\lambda)$ a skew-normal distribution, where $\lambda$ is the shape parameter. Show that $-X\sim SN(0,1,-\lambda)$.

From the notes that I have if $X\sim SN(0,1,\lambda)$ the density of $X$ is $$f(x|\lambda)=2\phi(x)\Phi(\lambda x)\qquad x\in\mathbb{R}$$

where $\phi(.)$ is the density of standard normal and $\Phi(.)$ is the accumulated distribution function of standard normal.

$$P(-X\leq x)=1-P(X\leq -x)=1-\int_{-\infty}^{-x}2\phi(z)\Phi(\lambda z)dz$$

$$f(-x|\lambda)=\frac{\partial}{\partial x}P(-X\leq x)=\frac{\partial}{\partial x}\Big(1-\int_{-\infty}^{-x}2\phi(z)\Phi(\lambda z)dz\Big)$$

but I'm stucked, any tips?

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You are thinking way too hard. If $X$ has density $$f_X(x \mid \lambda) = 2\phi(x) \Phi(\lambda x),$$ then $Y = g(X) = -X$ has density $$f_Y(y \mid \lambda) = f_X(g^{-1}(y) \mid \lambda) \left|\frac{dg^{-1}}{dy}\right| = f_X(-y \mid \lambda) = 2\phi(-y) \Phi(\lambda(-y)).$$ Since $\phi(y) = \phi(-y)$ because the standard normal is symmetric, we immediately obtain $$f_Y(y \mid \lambda) = 2\phi(y) \Phi((-\lambda)y),$$ which is the density for a skew normal distribution with parameters $0$, $1$, and $-\lambda$.

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