0
$\begingroup$

Is there anything we can say about $\lim_{x\rightarrow \infty}\frac{F(x)}{f(x)}$ where $F(x)$ is the CDF of the normal distribution and $f(x)$ is the pdf? I know this function is increasing, but what can we say about the limit?

More specifically, consider two random variables $v_b \sim N(\mu_b,\sigma_b)$ and $v_s \sim N(\mu_s,\sigma_s)$ with correlation $\rho$. I am trying to understand the existence and uniqueness of positive solutions to the following: $$g(\pi):=\mu_b-\pi-\frac{\rho\sigma_b\mu_s}{\sigma_s^2}+\frac{\rho\sigma_b}{\sigma_s^2}\pi-\frac{F_{v_s}(\pi)}{f_{v_s}(\pi)}=0$$

This is not an easy task, but I thought understanding the limiting behavior of the Mill's ratio may help.

Improve this question
$\endgroup$
7
  • $\begingroup$ This equation doesn't seem to make physical sense, because it mixes quantities of various units. E.g., if $v_b$ is a length and $v_s$ a mass, then $\mu_b$ is a length, $\rho\sigma_b\mu_s/\sigma_s^2$ is a length per unit mass, and $F_{v_s}(\pi)/f_{v_s}(\pi)$ is a mass. $\endgroup$
    – whuber ♦
    Commented Mar 2, 2021 at 15:03
  • $\begingroup$ Why are you interpreting $v_b$ as a length and $v_s$ as a mass? This has nothing to do with physics. $\endgroup$
    – Dmlawton
    Commented Mar 2, 2021 at 17:17
  • $\begingroup$ I am not interpreting them as such: "E.g." means for example. This is called the units calculus or quantity calculus and is useful for detecting basic algebraic errors in formulas. The inconsistency raises red flags and makes me wonder whether this formula is really the one you want to be applying to whatever your original problem might be. $\endgroup$
    – whuber ♦
    Commented Mar 2, 2021 at 17:18
  • $\begingroup$ Got it, so does this imply there was a mistake somewhere? As in, this equation doesn't make intuitive sense at all? $\endgroup$
    – Dmlawton
    Commented Mar 2, 2021 at 17:46
  • $\begingroup$ Yes, that's my concern, but I can't advise you because I don't know where this equation came from or what it is intended to mean. $\endgroup$
    – whuber ♦
    Commented Mar 2, 2021 at 18:43

1 Answer 1

Reset to default
1
$\begingroup$

I may be missing something, but, I think the answer here is straightforward.

as $x \rightarrow \infty$; $F(x) \rightarrow 1$ and $f(x)\rightarrow 0$. Thus the Mills ratio $\frac{F(X)}{f(x)}$ will go to infinity.

The other case is a bit more interesting. when $x \rightarrow -\infty$, $F(x) \rightarrow 0$ and $f(x)\rightarrow 0$. So you need l'hopital's rule.

$$\frac{F(X)}{f(x)}|_{x\rightarrow -\infty} = \frac{F'(x)}{f'(x)}=\frac{f(x)}{-x*f(x)}=\frac{1}{-x}$$

So $$\frac{F(X)}{f(x)}|_{x\rightarrow -\infty}\rightarrow 0$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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