How can I show that $b$ in this proportional hazard model $y=1-x^{e^{bz}}$ is the percent change in $y$ with a unit change in $z$?
For brevity, I have neglected to show the functional form of $x$, but $x$ is always less than or equal to one.
I've estimated the model and $b$ is the percent change in $y$ with a unit change in $z$, but I am unsure how to show this mathmatically.
This is what I've come up with so far:
- $y=1-x^{e^{bz}}$
- $=1-x^{e^{b(z+1)}}$
- $=1-x^{e^{bz+b}}$
- $=1-x^{e^{bz}e^b}$
I get lost at this step. I've tried substituting $1-y$ for $x^{e^{bz}}$, but don't quite know where to go with it. I've also tried the approach of taking logs of both sides and I get $ln(y)=ln(y)e^b$, but when I convert back I get $y=ye^{eb}$.
Just to note, it's easy to show that for $y=Ae^{bx}$ that y increases at a constant relative rate of b.
- $=Ae^{b(x+1)}$
- $=Ae^{bx+b}$ -> $Ae^{bx}e^b$
- $=ye^b$
Denote $y_0, z_0$ be the original value. Continue from your calculation,
$$y = 1 - x^{e^{bz_0}e^b} = 1 - \left(x^{e^{bz_0}}\right)^{e^b} = 1 - (1 - y_0)^{e^b}$$
So I do not see $b$ is simply the percentage change of $y$. Unless $y_0$ is small enough such that you can take a first order approximation:
$$ y \approx 1 - (1 - e^by_0) = e^by_0 \Rightarrow b = \ln\frac {y} {y_0}$$
and thus $b$ is the log rate of change (log-return in financial sense). Or another way to interpret is
$$ b = \ln \frac {1 - y} {1 - y_0}$$
This is exact but not very easy to interpret either.