Bivariate generating function on the unit disk
$begingroup$
Suppose $sum_{m,n}^{infty}p(x,y)=1$ where $p(x,y)$ is the stationary distribution of some two dimensional random walk and let its bivariate generating function be defined as:
$P(x,y)= sum_{m}^{infty}sum_{n}^{infty}p(x,y)x^{m}y^{n}$.
Can someone please show why $P(x,y)$ is analytic in the open domain
${{{(x,y) in mathbb{C}^{2}: |x|<1, |y|<1}}}$
and continuous in the closed domain
${{{(x,y) in mathbb{C}^{2}: |x|le 1, |y|le 1}}}$.
complex-analysis
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
$endgroup$
add a comment |
$begingroup$
Suppose $sum_{m,n}^{infty}p(x,y)=1$ where $p(x,y)$ is the stationary distribution of some two dimensional random walk and let its bivariate generating function be defined as:
$P(x,y)= sum_{m}^{infty}sum_{n}^{infty}p(x,y)x^{m}y^{n}$.
Can someone please show why $P(x,y)$ is analytic in the open domain
${{{(x,y) in mathbb{C}^{2}: |x|<1, |y|<1}}}$
and continuous in the closed domain
${{{(x,y) in mathbb{C}^{2}: |x|le 1, |y|le 1}}}$.
complex-analysis
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
$endgroup$
add a comment |
$begingroup$
Suppose $sum_{m,n}^{infty}p(x,y)=1$ where $p(x,y)$ is the stationary distribution of some two dimensional random walk and let its bivariate generating function be defined as:
$P(x,y)= sum_{m}^{infty}sum_{n}^{infty}p(x,y)x^{m}y^{n}$.
Can someone please show why $P(x,y)$ is analytic in the open domain
${{{(x,y) in mathbb{C}^{2}: |x|<1, |y|<1}}}$
and continuous in the closed domain
${{{(x,y) in mathbb{C}^{2}: |x|le 1, |y|le 1}}}$.
complex-analysis
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
$endgroup$
Suppose $sum_{m,n}^{infty}p(x,y)=1$ where $p(x,y)$ is the stationary distribution of some two dimensional random walk and let its bivariate generating function be defined as:
$P(x,y)= sum_{m}^{infty}sum_{n}^{infty}p(x,y)x^{m}y^{n}$.
Can someone please show why $P(x,y)$ is analytic in the open domain
${{{(x,y) in mathbb{C}^{2}: |x|<1, |y|<1}}}$
and continuous in the closed domain
${{{(x,y) in mathbb{C}^{2}: |x|le 1, |y|le 1}}}$.
complex-analysis
complex-analysis
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
edited Dec 2 '18 at 21:36
Live Free or π Hard
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
asked Nov 16 '14 at 22:11
Live Free or π HardLive Free or π Hard
479213
479213
add a comment |
add a comment |
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