# Why do we need $\gamma>2$ in SCAD penalty?

The SCAD penalty $$p(x | \lambda, \gamma)$$ from https://myweb.uiowa.edu/pbreheny/7600/s16/notes/2-29.pdf or the paper "Variable Selection via Nonconcave Penalized Likelihood and its Oracle Properties" is in the form of $$p(x | \lambda, \gamma) = \begin{cases} \lambda |x|\ \ \ \ \mathrm{if}\ |x|\leq \lambda,\\ \frac{2 \gamma \lambda |x| - x^2 - \lambda^2}{2(\gamma-1)}\ \ \ \ \mathrm{if}\ \lambda<|x|<\gamma\lambda,\\ \frac{\lambda^2(\gamma + 1)}{2}\ \ \ \ \mathrm{if}\ |x|\geq \gamma\lambda, \end{cases}$$ for $$\lambda>0$$ and $$\gamma>2$$. I am curious why we need $$\gamma>2$$ in this definition.

• have you tried plotting the penalty with $\gamma>2$ and $\gamma \leq 2$?
– Ben
Commented Jul 7, 2023 at 17:47
• Since the function $p(x|\lambda,\gamma)$ is even, we can only focus on $x \geq 0$. I think $p(x|\lambda,\gamma) = \lambda x$ if $x \in [0, \lambda]$. Then it increases as a concave function in $x \in (\lambda, \gamma \lambda)$. And after $x \geq \lambda \gamma$, it is a flat line. The whole function of $x \geq 0$ is continuous with continuous derivative. If we just check the existence of this function, I think $\gamma > 1$ is enough. Commented Jul 7, 2023 at 19:25