# Determining the probability density function from an equation

I have the following (for me quite interesting) densities for which I am completely stuck. I only hope that you can provide me some help.

Let me introduce my problem. I have two probability densities:

$$g_0(y)=f(\delta(y),\mu_0,\lambda_0)f_0(y)$$ and $$g_1(y)=f(1-\delta(y),\mu_1,\lambda_1)f_1(y)$$

where $f_{0,1}$ and $g_{0,1}$ are some continuous probability denstiy functions on $\mathbb{R,}$ $\mu_{0,1}$ and $\lambda_{0,1}$ are some variables to be determined such that $g_{0,1}$ are some densities, i.e., $\int dG=1$ and $\delta(y)\in[0,1].$

When I write the densities explicitely, they are

$$g_0(y,\delta)=\exp\left(-\frac{-W\left(\exp\left((-\delta(y)+\mu_0+\lambda_0)/\lambda_0\right)\right)\lambda_0-\delta(y)+\mu_0+\lambda_0}{\lambda_0}\right)f_0(y)$$

$$g_1(y,\delta)=\exp\left(-\frac{-W\left(\exp\left((-1+\delta(y)+\mu_1+\lambda_1)/\lambda_1\right)\right)\lambda_1-1+\delta(y)+\mu_1+\lambda_1}{\lambda_1}\right)f_1(y)$$ where $W$ is the Lambert $W$ function, i.e., given $y=xe^x\rightarrow x=W(y)$.

Now I have $$g_1(y,\delta)-g_0(y,\delta)=0$$ for which I have a solution $\delta_s(y)=f_s(\mu_0,\mu_1,\lambda_0,\lambda_1,f_0(y),f_1(y))$.

Question:

What is

$$g_0(y,\delta_s)$$ or equivalently $$g_1(y,\delta_s)$$

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What is the first $f$ ? –  leonbloy Sep 28 '12 at 21:11
@leonbloy you mean $g_0(y)=f(..)$ this $f$? –  Seyhmus Güngören Sep 28 '12 at 21:15