Circumventing Potentially Illegal Substitutions in Generating Functions
$begingroup$
This question comes from trying to solve the following problem from Probability: An Introduction by Grimmett and Welsh.
(b) In a two-dimensional random walk, a particle can be at any of the points $(x,y)$ which have integer coordinates. The particle starts at $(0,0)$ and at discrete intervals of time, takes a step of unit size. The steps are independent and equally likely to be any of the four nearest points. Show that the probability generating function of the time taken to reach the line $x+y=m$ is
$$left{frac{1 - sqrt{1-s^2}}sright}^mqquadtext{for }|s|leq 1.$$
(c) Let $(X,Y)$ be the random point on the line $x+y=m$ which is reached first. What is the probability generating function of $X-Y$?
So far I have deduced (hopefully correctly!) that if
$$F(s) = frac{1-sqrt{1-s^2}}s = sum_{ngeq 0}a_ns^n$$
is the generating function for the $m=1$ case in part (b), then the answer to part (c) for the $m=1$ case is
$$sum_{ngeq 0}a_nleft(frac{s+s^{-1}}2right)^n = Fleft(frac{s+s^{-1}}2right).$$
(The basic idea is to note that at each step of the process we may either increase or decrease $X-Y$ by $1$ with probability $tfrac12$ each; the left hand side is then the partition theorem when conditioning on the number of steps taken.) This would be fine and dandy, but the issue is that
$$sqrt{1 - left(frac{s+s^{-1}}2right)^2} = sqrt{1 - frac{s^2 + 2 + s^{-2}}4} = sqrt{-left(frac{s-s^{-1}}2right)^2},$$ and so I can't just plug in $tfrac{s+s^{-1}}2$ into the definition of $F$. (In particular, the PGF in part (b) assumes $|s|leq 1$, but as a real number $tfrac{s+s^{-1}}2$ is always $geq 1$ in absolute value.)
Is there a way to make sense of this "negative square root" in the context of Laurent series, or must I go about simplifying the generating function in a different way?
probability-theory generating-functions random-walk laurent-series
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
$endgroup$
add a comment |
$begingroup$
This question comes from trying to solve the following problem from Probability: An Introduction by Grimmett and Welsh.
(b) In a two-dimensional random walk, a particle can be at any of the points $(x,y)$ which have integer coordinates. The particle starts at $(0,0)$ and at discrete intervals of time, takes a step of unit size. The steps are independent and equally likely to be any of the four nearest points. Show that the probability generating function of the time taken to reach the line $x+y=m$ is
$$left{frac{1 - sqrt{1-s^2}}sright}^mqquadtext{for }|s|leq 1.$$
(c) Let $(X,Y)$ be the random point on the line $x+y=m$ which is reached first. What is the probability generating function of $X-Y$?
So far I have deduced (hopefully correctly!) that if
$$F(s) = frac{1-sqrt{1-s^2}}s = sum_{ngeq 0}a_ns^n$$
is the generating function for the $m=1$ case in part (b), then the answer to part (c) for the $m=1$ case is
$$sum_{ngeq 0}a_nleft(frac{s+s^{-1}}2right)^n = Fleft(frac{s+s^{-1}}2right).$$
(The basic idea is to note that at each step of the process we may either increase or decrease $X-Y$ by $1$ with probability $tfrac12$ each; the left hand side is then the partition theorem when conditioning on the number of steps taken.) This would be fine and dandy, but the issue is that
$$sqrt{1 - left(frac{s+s^{-1}}2right)^2} = sqrt{1 - frac{s^2 + 2 + s^{-2}}4} = sqrt{-left(frac{s-s^{-1}}2right)^2},$$ and so I can't just plug in $tfrac{s+s^{-1}}2$ into the definition of $F$. (In particular, the PGF in part (b) assumes $|s|leq 1$, but as a real number $tfrac{s+s^{-1}}2$ is always $geq 1$ in absolute value.)
Is there a way to make sense of this "negative square root" in the context of Laurent series, or must I go about simplifying the generating function in a different way?
probability-theory generating-functions random-walk laurent-series
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
$endgroup$
add a comment |
$begingroup$
This question comes from trying to solve the following problem from Probability: An Introduction by Grimmett and Welsh.
(b) In a two-dimensional random walk, a particle can be at any of the points $(x,y)$ which have integer coordinates. The particle starts at $(0,0)$ and at discrete intervals of time, takes a step of unit size. The steps are independent and equally likely to be any of the four nearest points. Show that the probability generating function of the time taken to reach the line $x+y=m$ is
$$left{frac{1 - sqrt{1-s^2}}sright}^mqquadtext{for }|s|leq 1.$$
(c) Let $(X,Y)$ be the random point on the line $x+y=m$ which is reached first. What is the probability generating function of $X-Y$?
So far I have deduced (hopefully correctly!) that if
$$F(s) = frac{1-sqrt{1-s^2}}s = sum_{ngeq 0}a_ns^n$$
is the generating function for the $m=1$ case in part (b), then the answer to part (c) for the $m=1$ case is
$$sum_{ngeq 0}a_nleft(frac{s+s^{-1}}2right)^n = Fleft(frac{s+s^{-1}}2right).$$
(The basic idea is to note that at each step of the process we may either increase or decrease $X-Y$ by $1$ with probability $tfrac12$ each; the left hand side is then the partition theorem when conditioning on the number of steps taken.) This would be fine and dandy, but the issue is that
$$sqrt{1 - left(frac{s+s^{-1}}2right)^2} = sqrt{1 - frac{s^2 + 2 + s^{-2}}4} = sqrt{-left(frac{s-s^{-1}}2right)^2},$$ and so I can't just plug in $tfrac{s+s^{-1}}2$ into the definition of $F$. (In particular, the PGF in part (b) assumes $|s|leq 1$, but as a real number $tfrac{s+s^{-1}}2$ is always $geq 1$ in absolute value.)
Is there a way to make sense of this "negative square root" in the context of Laurent series, or must I go about simplifying the generating function in a different way?
probability-theory generating-functions random-walk laurent-series
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
$endgroup$
This question comes from trying to solve the following problem from Probability: An Introduction by Grimmett and Welsh.
(b) In a two-dimensional random walk, a particle can be at any of the points $(x,y)$ which have integer coordinates. The particle starts at $(0,0)$ and at discrete intervals of time, takes a step of unit size. The steps are independent and equally likely to be any of the four nearest points. Show that the probability generating function of the time taken to reach the line $x+y=m$ is
$$left{frac{1 - sqrt{1-s^2}}sright}^mqquadtext{for }|s|leq 1.$$
(c) Let $(X,Y)$ be the random point on the line $x+y=m$ which is reached first. What is the probability generating function of $X-Y$?
So far I have deduced (hopefully correctly!) that if
$$F(s) = frac{1-sqrt{1-s^2}}s = sum_{ngeq 0}a_ns^n$$
is the generating function for the $m=1$ case in part (b), then the answer to part (c) for the $m=1$ case is
$$sum_{ngeq 0}a_nleft(frac{s+s^{-1}}2right)^n = Fleft(frac{s+s^{-1}}2right).$$
(The basic idea is to note that at each step of the process we may either increase or decrease $X-Y$ by $1$ with probability $tfrac12$ each; the left hand side is then the partition theorem when conditioning on the number of steps taken.) This would be fine and dandy, but the issue is that
$$sqrt{1 - left(frac{s+s^{-1}}2right)^2} = sqrt{1 - frac{s^2 + 2 + s^{-2}}4} = sqrt{-left(frac{s-s^{-1}}2right)^2},$$ and so I can't just plug in $tfrac{s+s^{-1}}2$ into the definition of $F$. (In particular, the PGF in part (b) assumes $|s|leq 1$, but as a real number $tfrac{s+s^{-1}}2$ is always $geq 1$ in absolute value.)
Is there a way to make sense of this "negative square root" in the context of Laurent series, or must I go about simplifying the generating function in a different way?
probability-theory generating-functions random-walk laurent-series
probability-theory generating-functions random-walk laurent-series
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
asked Dec 28 '18 at 21:46
David AltizioDavid Altizio
11
11
add a comment |
add a comment |
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