I have a cumulative distribution function: $$G(x) = -ae^{-xb} - ce^{-xd}+h$$ The associated probability density function is: $$g(x) = abe^{-xb} + cde^{-xd}$$ My problem concerns $x\ge 0, X \in R$.
I know that the mean (expected value) of $x$ can be computed by: \begin{align} E[x] &= \int_0^{\infty} x g(x)~dx \\ &=\int_0^{\infty}xabe^{-xb}~dx~+~\int_0^{\infty}xcde^{-xd}~dx\\ &=\frac{a}{b} + \frac{c}{d} \end{align}
The median is when $G(x) = 0.5$. This requires finding the roots of $$ 0 = -ae^{xb} - ce^{xd}+h-0.5 $$ Based on the Abel–Ruffini theorem I know that there are no general solutions to this problem.
My question relates to the component exponential decay equations contained in $G(x)$: \begin{align} F(x) &= ae^{xb}\\ J(x) &= ce^{xd} \end{align}
The mean of $F(x)$ is $\frac{a}{b}$ and median $\frac{a\ln2}{b}$. The mean of $J(x)$ is $\frac{c}{d}$ and median $\frac{c\ln2}{d}$.
MY QUESTION: The ratio of the mean of $g(x)$ to the mean of $f(x)$ is $$\frac{a}{b} + \frac{c}{d} \div \frac{a}{b}$$ But is the ratio of the MEDIAN of $g(x)$ TO THE MEDIAN of $f(x)$ the same? I am assuming not, because there is no generic formula to solve for the median of $g(x)$. But I don't know how to prove this, so am not sure.
Your intuition is correct, the ratio of mean to median of a random variable $X$ with density of shape $abe^{-ax}+cde^{-cx}$ is not always the same as the ratio of mean to median of an exponentially distributed random variable. (The latter ratio, as your post pointed out, is $\frac{1}{\ln 2}$.)
To show this, it is enough to give an example. Let $X$ have density function $g(x)=\frac{1}{2}\left(e^{-x}+2e^{-2x}\right)$ (for $x\gt 0$). The mean of $X$ is $\frac{3}{4}$. To compute the median, we solve the equation $$\frac{1}{2}(2-e^{-m}-e^{-2m})=\frac{1}{2}.$$ This is one of the rare cases where we can compute explicitly: $m=\ln\left(\frac{1+\sqrt{5}}{2}\right)$. The golden section strikes again.