Mixing
power series expansion in two variables
$begingroup$
Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
$$
g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
$$
Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
$$
sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
$$
How can I expand $g$, so I can determine $q$.
real-analysis calculus multivariable-calculus power-series
asked Jan 4 at 19:54
love_mathlove_math
1595
$endgroup$
add a comment |
$begingroup$
Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
$$
g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
$$
Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
$$
sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
$$
How can I expand $g$, so I can determine $q$.
real-analysis calculus multivariable-calculus power-series
asked Jan 4 at 19:54
love_mathlove_math
1595
$endgroup$
add a comment |
$begingroup$
Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
$$
g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
$$
Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
$$
sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
$$
How can I expand $g$, so I can determine $q$.
real-analysis calculus multivariable-calculus power-series
asked Jan 4 at 19:54
love_mathlove_math
1595
$endgroup$
Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
$$
g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
$$
Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
$$
sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
$$
How can I expand $g$, so I can determine $q$.
real-analysis calculus multivariable-calculus power-series
real-analysis calculus multivariable-calculus power-series
asked Jan 4 at 19:54
love_mathlove_math
1595
asked Jan 4 at 19:54
love_mathlove_math
1595
edited Jan 4 at 22:57
love_math
asked Jan 4 at 19:54
love_mathlove_math
1595
asked Jan 4 at 19:54
love_mathlove_math
1595
asked Jan 4 at 19:54
love_mathlove_math
1595
1595
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since the constraints that you give result into $$|pq x_1 x_2|<1$$
then you can apply the geometric series to the denominator and the expansion is simply
$$
eqalign{
& left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
& = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
& = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
& = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
& quad quad quad Downarrow cr
& cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
$endgroup$
$begingroup$
The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
$endgroup$
– love_math
Jan 4 at 22:55
$begingroup$
@love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
$endgroup$
– G Cab
Jan 5 at 1:35
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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$begingroup$
Since the constraints that you give result into $$|pq x_1 x_2|<1$$
then you can apply the geometric series to the denominator and the expansion is simply
$$
eqalign{
& left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
& = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
& = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
& = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
& quad quad quad Downarrow cr
& cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
$endgroup$
$begingroup$
The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
$endgroup$
– love_math
Jan 4 at 22:55
$begingroup$
@love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
$endgroup$
– G Cab
Jan 5 at 1:35
add a comment |
$begingroup$
Since the constraints that you give result into $$|pq x_1 x_2|<1$$
then you can apply the geometric series to the denominator and the expansion is simply
$$
eqalign{
& left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
& = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
& = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
& = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
& quad quad quad Downarrow cr
& cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
$endgroup$
$begingroup$
The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
$endgroup$
– love_math
Jan 4 at 22:55
$begingroup$
@love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
$endgroup$
– G Cab
Jan 5 at 1:35
add a comment |
$begingroup$
Since the constraints that you give result into $$|pq x_1 x_2|<1$$
then you can apply the geometric series to the denominator and the expansion is simply
$$
eqalign{
& left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
& = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
& = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
& = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
& quad quad quad Downarrow cr
& cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
$endgroup$
Since the constraints that you give result into $$|pq x_1 x_2|<1$$
then you can apply the geometric series to the denominator and the expansion is simply
$$
eqalign{
& left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
& = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
& = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
& = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
& quad quad quad Downarrow cr
& cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
edited Jan 5 at 1:33
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
answered Jan 4 at 20:26
G CabG Cab
18.5k31237
18.5k31237
$begingroup$
The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
$endgroup$
– love_math
Jan 4 at 22:55
$begingroup$
@love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
$endgroup$
– G Cab
Jan 5 at 1:35
add a comment |
$begingroup$
The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
$endgroup$
– love_math
Jan 4 at 22:55
$begingroup$
@love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
$endgroup$
– G Cab
Jan 5 at 1:35
$begingroup$
The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
$endgroup$
– love_math
Jan 4 at 22:55
$begingroup$
The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
$endgroup$
– love_math
Jan 4 at 22:55
$begingroup$
@love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
$endgroup$
– G Cab
Jan 5 at 1:35
$begingroup$
@love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
$endgroup$
– G Cab
Jan 5 at 1:35
add a comment |
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