Doubt on a recurrence equation

Question

I have the following recursive equation for a pmf $P(s)$: $$P(s+1)=2 \lambda (1-\lambda) P(s)+ \lambda^2 \sum_{s'=0}^s P(s') P(s-s')$$ Here $s$ is a natural number and I have two initial conditions $P(0)=0$ and $P(1)=(1-\lambda)^2$. I tried to solve it using a generating function $G(x)=\sum_{s=0}^\infty x^s P(s)$. Using it I get: $$\frac{1}{x} G(x)=2 \lambda (1-\lambda) G(x) + \lambda^2 G^2(x)$$ but I think something is wrong (i.e. it does not satisfy $G(1)=1$). Any suggestions?

Answer 1

As the recurrence does not hold for $s=0$ we should have

\begin{align}G(x)&=\sum_{n=0}^\infty P(n)x^n\\&=(1-\lambda)^2x+\sum_{n=1}^\infty P(n+1)x^{n+1}\\&=(1-\lambda)^2x+\sum_{n=1}^\infty[2\lambda(1-\lambda)P(n)+\lambda^2\sum_{k=0}^nP(k)P(n-k)]x^{n+1}\\&=(1-\lambda)^2x+2\lambda(1-\lambda)xG(x)+\lambda^2xG(x)^2\end{align}

which satisfies $G(1)=1$.

Answer 2

Cintuing Simply Beautiful Art's work.

$G(x)=(1-\lambda)^2x+2\lambda(1-\lambda)xG(x)+\lambda^2xG(x)^2 $

so, writing $c$ for $\lambda$,

$0 =c^2xG(x)^2+(2c(1-c)x-1)G(x)+(1-c)^2x $.

$\begin{array}\\ D &=(2c(1-c)x-1)^2-4c^2x(1-c)^2x\\ &=4c^2(1-c)^2x^2 -4c(1-c)x+1-4c^2x(1-c)^2x\\ &=-4c(1-c)x+1\\ &=1+(4c^2-4c)x\\ &=1+(4c^2-4c+1-1)x\\ &=1-x+(2c-1)^2x\\ \end{array} $

so

$\begin{array}\\ G(x) &=\dfrac{-(2c(1-c)x-1)\pm\sqrt{1-x+(2c-1)^2x}}{2c^2x}\\ &=\dfrac{2c^2x-2c+1\pm\sqrt{1-x+(2c-1)^2x}}{2c^2x}\\ &=1-\dfrac{2c-1\mp\sqrt{1-x+(2c-1)^2x}}{2c^2x}\\ G(1) &=1-\dfrac{2c-1\mp\sqrt{(2c-1)^2}}{2c^2}\\ &=1-\dfrac{2c-1\mp(2c-1)}{2c^2}\\ &=1-\dfrac{0, 4c-2}{2c^2}\\ &=1, 1-\dfrac{3c-1}{c^2}\\ \end{array} $

So $G(1) = 1$ for the positive square root.