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Solution of generating function does not make sense

1

$begingroup$

Consider the generating function
$$G(x,t) = sum_n=0^N P_n(t) x^n,$$
with $G(1,t) = 1$ and $G(x,0) = x^m$. From a master equation, I obtained the following partial differential equation for $G$:
$$fracpartial Gpartial t = (x-1) left[2 a N G - (2a x + b) fracpartial Gpartial x right].$$
This seems ripe for the method of characteristics and if we consider $G(x,t) = G(x(t),t)$ we obtain the characteristics
$$C e^(2a + b)t = fracx-12a x + b,$$
along which $G$ evolves
$$G(x,t) = K[1 - 2 a C e^(2a + b)t]^-N.$$
The boundary condition $G(1,t)$ implies that along that characteristic, we must have $C = 0$ and so $K = 1$ (since $G(1,t) = 1$). At this point, I am a little confused since
$$G(x,t) = [1 - 2 a C e^(2a+b)t]^-N$$
Implies with the characteristic that
$$G(x,t) = G(x) = left(frac2ax+b2a+bright)^N.$$
This is the generating function for the binomial distribution with $N$ trials and $p = frac2a2a+b$, however there seems to be no time dependence on this solution (e.g., if I solve the PDE at steady state, I would have arrived at the same result), how is this possible?

probability generating-functions

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asked Apr 6 at 20:12

GregoryGregory

1,351412

$endgroup$

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1

$begingroup$

Consider the generating function
$$G(x,t) = sum_n=0^N P_n(t) x^n,$$
with $G(1,t) = 1$ and $G(x,0) = x^m$. From a master equation, I obtained the following partial differential equation for $G$:
$$fracpartial Gpartial t = (x-1) left[2 a N G - (2a x + b) fracpartial Gpartial x right].$$
This seems ripe for the method of characteristics and if we consider $G(x,t) = G(x(t),t)$ we obtain the characteristics
$$C e^(2a + b)t = fracx-12a x + b,$$
along which $G$ evolves
$$G(x,t) = K[1 - 2 a C e^(2a + b)t]^-N.$$
The boundary condition $G(1,t)$ implies that along that characteristic, we must have $C = 0$ and so $K = 1$ (since $G(1,t) = 1$). At this point, I am a little confused since
$$G(x,t) = [1 - 2 a C e^(2a+b)t]^-N$$
Implies with the characteristic that
$$G(x,t) = G(x) = left(frac2ax+b2a+bright)^N.$$
This is the generating function for the binomial distribution with $N$ trials and $p = frac2a2a+b$, however there seems to be no time dependence on this solution (e.g., if I solve the PDE at steady state, I would have arrived at the same result), how is this possible?

probability generating-functions

share|cite|improve this question

asked Apr 6 at 20:12

GregoryGregory

1,351412

$endgroup$

add a comment | 

$begingroup$

Consider the generating function
$$G(x,t) = sum_n=0^N P_n(t) x^n,$$
with $G(1,t) = 1$ and $G(x,0) = x^m$. From a master equation, I obtained the following partial differential equation for $G$:
$$fracpartial Gpartial t = (x-1) left[2 a N G - (2a x + b) fracpartial Gpartial x right].$$
This seems ripe for the method of characteristics and if we consider $G(x,t) = G(x(t),t)$ we obtain the characteristics
$$C e^(2a + b)t = fracx-12a x + b,$$
along which $G$ evolves
$$G(x,t) = K[1 - 2 a C e^(2a + b)t]^-N.$$
The boundary condition $G(1,t)$ implies that along that characteristic, we must have $C = 0$ and so $K = 1$ (since $G(1,t) = 1$). At this point, I am a little confused since
$$G(x,t) = [1 - 2 a C e^(2a+b)t]^-N$$
Implies with the characteristic that
$$G(x,t) = G(x) = left(frac2ax+b2a+bright)^N.$$
This is the generating function for the binomial distribution with $N$ trials and $p = frac2a2a+b$, however there seems to be no time dependence on this solution (e.g., if I solve the PDE at steady state, I would have arrived at the same result), how is this possible?

probability generating-functions

share|cite|improve this question

asked Apr 6 at 20:12

GregoryGregory

1,351412

$endgroup$

share|cite|improve this question

asked Apr 6 at 20:12

GregoryGregory

1,351412

share|cite|improve this question

asked Apr 6 at 20:12

GregoryGregory

1,351412

asked Apr 6 at 20:12

GregoryGregory

1,351412

asked Apr 6 at 20:12

GregoryGregory

1,351412