2
$\begingroup$

Given $c>0$ and define a function $f:(0,\infty)\mapsto \mathbb{R}$ by \begin{equation} f(\sigma)=\frac1{\sqrt {2\pi}\sigma}\int_{-\infty}^\infty\max\{-c,\min\{x,c\}\}^2e^{-\frac{x^2}{2\sigma^2}}. \end{equation} We can see $f$ as the second moment of $X_c=\max\{-c,\min\{X,c\}\}$ where $X\sim N(0,\sigma^2)$. I could prove that $f$ is increasing using derivative which is very tedious. I'd like to see $f$ is increasing using expectation approach. Could anyone help? Thanks in advance

$\endgroup$

1 Answer 1

2
$\begingroup$

Let the random variable $X$ have any distribution $F$. We can remove the infinite endpoints in the integral, showing there are no worries about whether moments of $F$ exist:

$$\eqalign{ e(F,c) &= \mathbb{E}_F\left(\max\{-c, \min\{X,c\}\}^2\right) \\ &=\mathbb{E}_F\left(c^2 + (X^2 - c^2) I_{|X|\le c}\right) \\ &=c^2 - \mathbb{E}_F\left((c^2 - X^2) I_{|X|\le c}\right). }$$

(The notation "$I_{|X| \le c}$" is the indicator function for the set $|X| \le c$, equal to $1$ on the set and $0$ elsewhere.)

Now rescale $X$ by $\sigma \gt 0$; that is, suppose that $Y = X/\sigma$ has the distribution $F$, and consider the right hand term:

$$\eqalign{ \mathbb{E}_F\left(\left(c^2 - \left(\sigma Y\right)^2 \right) I_{|\sigma Y |\le c}\right). }$$

Because this expectation occurs with a negative sign in $e(F,c),$ it follows that $e(F,c)$ decreases as $X$ is rescaled by ever larger values of $\sigma.$

The function inside the expectation is non-negative: it is the part of a parabola rising above the axis. Its apex is always at height $c$ when $Y=0$, but it is being horizontally shrunk by a factor of $\sigma$.

Figure: three parabolae

As $\sigma$ increases (values of $\sigma=1/2, 1, 2$ are shown here), the parabola shrinks. The expectations are the probability-weighted integrals of these functions. The question asks to show that these expectations decrease as the parabola shrinks.

We can easily show this algebraically (but without the same geometric insight) by noting that whenever $\sigma^\prime \gt \sigma,$

$$c^2 - \left(\sigma Y\right)^2 \gt c^2 - \left(\sigma^\prime Y\right)^2.$$

Consequently, because $I_{|\sigma Y |\le c} \ge I_{|\sigma^\prime Y |\le c}$, the expression inside the expectation is non-decreasing (and is strictly decreasing whenever $|Y| \lt c/\sigma$). Therefore the expectation itself cannot increase and in fact must decrease whenever $F$ assigns positive probability to the interval $(-c/\sigma, c/\sigma)$.

The setting of the question is the case where $F$ is a standard normal distribution. Because it assigns positive probability density to all numbers, a fortiori it assigns positive probabilities to all intervals $(-c/\sigma, c/\sigma),$ whence the desired conclusion follows.

$\endgroup$
1
  • $\begingroup$ Awesome. Thank you very much for your help @whuber. $\endgroup$
    – Jlamprong
    Feb 14, 2014 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.