How to find a specific inverse function.

Question

I am developing a new probability distribution. The probability density function is as follows:

$$f(x) \, = \, \frac{x^{r+1}}{(r+1)^{x}}\frac{ln(r+1)^{r+2}}{ (r+1)!}$$ with $0<x$ and $r = 1, \, 2, \, 3,\, ... \,$ . The distribution function is as follows:

$$F(x) = 1 - \sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$

Define $G(x) = 1 - F(x)$:

$$G(x) = \sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$

I am searching for the inverse of $G(x)$. Does a solution exist?

Answer 1

For $r=1$,

$$G(x)= 2^{-x}\left(1+ x \ln2+ \frac12(x \ln2)^2\right)$$

is of the form $2^{-x}P(x)$ where $P$ is a polynomial. Such functions are not invertible analytically. The same holds for $r>1$.

Answer 2

May I start remarking that it looks for me like some Poisson distribution: $$G(x) = \sum_{i=0}^{r+1} (1/i!)(r+1)^{-x} x^i ln((r+1))^i$$

Now, to start set $a:= ln(r+1) >0 $ then $$G(x) = \sum_{k=0}^{r+1} e^{-ax} \frac{(ax)^k}{k!}= e^{-ax} \sum_{k=0}^{r+1} \frac{(ax)^k}{k!} $$ This shows (which is needed later), that $G$ can be expressed as a product of two power series and thus can be written as a power series itself.

Now let us get the derivative of $G$ and find out if it is monotonic: $$G'(x) = -a e^{-ax} \sum_{k=0}^{r+1} \frac{(ax)^k}{k!} + e^{-ax} \sum_{k=1}^{r+1} a\frac{(ax)^{k-1}}{(k-1)!} = \\ -a e^{-ax} \sum_{k=0}^{r+1} \frac{(ax)^k}{k!} + ae^{-ax} \sum_{k=0}^{r} \frac{(ax)^k}{k!} =\\ =-a e^{-ax}\frac{(ax)^{r+1}}{(r+1)!} $$ Then if $r>0$ we have $a =ln(r+1) > 0$ and hence the the last line gets negative for positive $x>0$ . Under this condition for $r$ and $x$ then $G'$ is negative on $\mathbb{R}^+$, thus it is monotonic (decreasing) and therefore invertable.

Furthermore as has been noted previously that $G$ can be expressed as a power series, we could apply the Lagrange–Bürmann formula to find the coefficients of the inverse function's power series coefficients.

Addendum: Please note that
$$G(x) = e^{-ax} \sum_{k=0}^{r+1} \frac{(ax)^k}{k!} = \frac{\Gamma(r+2,ax)}{ \Gamma(r+2)}=Q(r+2,ax)$$ (see here)where $\Gamma(.,..)$ is the incomplete Gamma function, $\Gamma(.)$ is the well known Gamma function, and $Q$ is the regulized Gamma function. All these functions are available by many math libraries, and their inverses are so as well. Just in case you need explicit expressions take a look here.