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$\delta M=\beta(x-y)M_y+\mu x(y-1)M_x+\delta y$, where $M(x,y)=\sum_{n=0}^{\infty}{\sum_{k=0}^{\infty}{s_{n,k}x^ny^k}}$ is the generating function for a certain probability distribution $\{s_{n,k}\}$ (the exact formula for $s_{n,k}$ is unknown), and $\delta$, $\beta$, $\mu$ are all constants.

The problem comes out of a probability model, and data show that the distribution should have a finite mean and infinite second moment ($M_{xx}(1,1)=\infty$). My question is that is there any way (or any theory on it) to get the asymptotic of $M_{xx}(x,1)$ when $x\rightarrow 1^{-}$? (by asymptotic I mean something like $M_{xx}(x,1)\sim C(1-x)^{-\zeta}$)

The background for this problem is here: Asymptotic behaviour of a two-dimensional recurrence relation

Thank you!

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For $\delta=0$ the general solution has an explicit form: $M(x,y)=c_1(x)(\beta(x-y))^{-\frac\delta\beta}+\frac{\beta x+\delta y}{\delta(\beta+\delta)}$, where $c_1$ is an arbitrary function. Putting $y=1$ gives $M(x,1)=c_1(x) (\beta (1-x))^{-\frac{\delta }\beta }+\frac{\delta +\beta x}{\beta+\delta}\;$. So the asymptotic depends upon behavior of $c_1$ as $x\to1-\;$. May be inserting initial conditions could allow to obtain $c_1$ to see if power asymptotic holds in this case. Also it is possible to obtain an explicit solution $r(x,y)=c_1(y)(\mu x)^{\frac{\delta}{\mu-\mu y}}+y\ $ for $\beta=0$. –  Andrew Aug 15 '11 at 11:09
    
Thank you for your reply. But in my model, though $\delta$ should be small (it is supposed to be the rate of exit of firms), it has to be a positive number. And $\beta$ (supposed to be job-filling rate) should be very large. –  epsilon Aug 15 '11 at 21:16

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