I want to find the solution to:
$x = (1-F(x))/f(x)$ where $F()$ is the distribution and $f()$ the density of a gaussian random variable.
Notice that it is also equivalent to solving $x = 1/h(x)$ where $h()$ is the hazard function.
In other words (in the gaussian case), I want to find the solution to:
$x = \frac{\Phi(-(x-\mu)/\sigma)}{\frac{1}{\sigma}\phi((x-\mu)/\sigma)} $
My first question would be: is there any closed form solution?
I supposed there was no closed form solution to it, so I just solved it numerically. However it seems like I found a weird property:
if $X \sim N(\mu, \sigma)$ (with distrib F and density f) and $Y = aX \sim N(a\mu, a\sigma)$ (with distrib $F_a$ and density $f_a$), then what happens numerically is that I get a linear relationship if I plot the solution $x^*$ with respect to $a$.
More precisely it would seem that $x^{**}$ solution to $x^{**} = (1-F_a(x^{**}))/f_a(x^{**})$ is actually equal to: $x^{**} = a x^*$ (where $x^*$ is solution to $x^* = (1-F(x^*))/f(x^*)$.
i.e., if we write $x^* \equiv c(\mu, \sigma)$ (where $c$ stands for constant), then $x^{**} \equiv c(a \mu, a \sigma) = a \ c(\mu, \sigma)$.
What I was wondering is (i) whether or not it is true (seems to be numerically but we never know), (ii) if so, how to prove it? (so far I'm only able to show that if the solution is linear, then indeed it is consistently linear via the equation, but it's completely tautological).
Ideally I would like to be able to determine a closed form solution to $x = (1-F(x))/f(x)$ in the gaussian case but I don't think it's possible.
Notice that this linearity property seems to also hold for lognormal distributions. (in which case $Y = aX \sim LN(\mu + log(a), \sigma)$ if $X \sim LN(\mu, \sigma)$)
If $x^*$ satisfy the equation
$$ x^* = \frac {\displaystyle \Phi\left(-\frac {x^*-\mu} {\sigma}\right)} {\displaystyle \frac {1} {\sigma}\phi\left(\frac {x^*-\mu} {\sigma}\right)} = \frac {1} {h(x^*;\mu,\sigma)}$$
Then we also have $$ ax^* = \frac {\displaystyle \Phi\left(-\frac {(ax^*)-(a\mu)} {(a\sigma)}\right)} {\displaystyle \frac {1} {(a\sigma)}\phi\left(\frac {(ax^*)-(a\mu)} {(a\sigma)}\right)} = \frac {1} {h(ax^*;a\mu, a\sigma)}$$
So the result is not surprising.