Currently I am trying to solve the following equality for $|p|<1$:

$$p\left(\frac{-\ln(a)-\ln(1+\gamma p+p^2)+\ln(1-p+ap+\gamma p+p^2)}{1-p+\gamma p+p^2}\right)=a-1,$$

where $0<a<1$ and $\gamma > 0$.

I am doing this because I have a probability generating function which is given as a quotient, where the numerator is a difficult integral. In order to say something about the numerator, I am trying to find the root of the denominator. This gave me the equality shown above.

Honestly, I have no idea how to handle this. Does anybody have any idea?

Thank you in advance.


If you make the substitution $$ \zeta := 1-p+\gamma p + p^2$$ your equation becomes $$ \ln{\dfrac{\zeta+ap}{a(\zeta-p)}} = \dfrac{a-1}{p}\zeta. $$ using $\ln(a b) = \ln(a)\ln(b),\,\, \ln{\dfrac{a}{b}} = \ln(a)-\ln(b)$. Now let us substitute further $$ \eta:= \dfrac{\zeta+ap}{a(\zeta-p)} \Leftrightarrow \zeta = \dfrac{ap(1+\eta)}{a\eta-1} $$ for $\zeta \neq p, 1 \neq a\eta$. If you substitute this in the second equation you get independent (explicitly) of $p$ equation: $$ \ln{\eta} = a(a-1)\dfrac{1+\eta}{a\eta-1}\cdot $$ Now this equation can be solved approximately and you can get idea about the solutions graphically for different values of $a$ very easily. In the picture you see the graph for $a = 0.12$

If $\eta = 1$ then $\zeta = 0$ and you can calculate $p$ from the quadratic equation $$p^2 + p(\gamma-1) + 1 =\zeta$$ depending on $\gamma$.

Simulating over $a \in (0,1)$ seems to give that after $a \simeq 0.18$ there are no roots to this equation. Have also in mind that the case $\zeta = p$ should be considered separately.

P.S. The Wolfram Mathematica code that produced the manipulation is

Manipulate[ Plot[{Log[b], a (a - 1) (b + 1)/(a b - 1)}, {b, -10, 20}], {a, 0.01, 0.99, 0.01}]

  • $\begingroup$ Thanks for the response. I redid your steps, but I think there is something wrong. When computing $\zeta$ I find $\frac{ap(\eta +1)}{1-a \eta}$. Would there still be an 'easy' root then? $\endgroup$ – Zdenek Rovnez May 17 '17 at 21:13
  • 1
    $\begingroup$ @ZdenekRovnez Yes, you are right. It is even $\dfrac{ap(\eta+1)}{a\eta-1}$ :) In this way I don't see a nice root. Moreover, as you can see from my answer after a certain point there is no solution to the equation.. $\endgroup$ – Veliko May 17 '17 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.