6

In this question, OP is interested, apparently, in a loss function with the ambiguous name "normal loss function" which is defined as $$ L(z) =\phi(z) - z \left(1 - \Phi(z)\right). $$

Based on this quick plot, the function is nonnegative and essentially linear for sufficiently small values $z$.

enter image description here It's not obvious to me why this must be non-negative, nor is it obvious why it must be essentially linear for small $z$. Is there an intuitive explanation for this behavior?

Sycorax
  • 90,934
  • 3
    This result is contained in the extensive results reported by Cardinal at https://stats.stackexchange.com/questions/7200/evaluate-definite-interval-of-normal-distribution/7206#7206. – whuber Apr 03 '18 at 16:18

1 Answers1

7

Since on the interval $x \in [z, \infty)$ it is clear that $x \ge z,$ use the fact that $\phi^\prime(x) = -x \phi(x)$ to conclude

$$\phi(z) = \int_z^\infty (-\phi^\prime(x))dx = \int_{z}^\infty x \phi(x) dx \ge z \int_z^\infty \phi(x) dx = z(1-\Phi(z)),$$

QED.

Concerning the second question, observe that $$L^\prime(z) = \phi^\prime(z) + z\phi(z) - (1-\Phi(z)) = 1-\Phi(z),$$ yielding $L$ as an integral $$L(z)=\int_{\infty}^z (1-\Phi(x))dx \gt 0,$$ which provides another solution to the first question. Since $1-\Phi(z) \approx 1$ for $z\ll 0,$ $L$ approaches an asymptote with slope $-1$ as $z\to-\infty.$

whuber
  • 322,774