Generating function from recurrence relation of binomial distribution
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Hello i have given recurrence like this :
$$p_{n,k}=(1-q)p_{n-1,k-1}+qp_{n-1,k}$$
my question is how to get (step by step) generating function from this recurrence?
we know that it's some king of distribution and from how it looks we can say it's binomial distribution.
recurrence-relations generating-functions recursive-algorithms
asked Nov 28 at 9:58
Gokuruto
123
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Hello i have given recurrence like this :
$$p_{n,k}=(1-q)p_{n-1,k-1}+qp_{n-1,k}$$
my question is how to get (step by step) generating function from this recurrence?
we know that it's some king of distribution and from how it looks we can say it's binomial distribution.
recurrence-relations generating-functions recursive-algorithms
asked Nov 28 at 9:58
Gokuruto
123
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 at 10:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Hello i have given recurrence like this :
$$p_{n,k}=(1-q)p_{n-1,k-1}+qp_{n-1,k}$$
my question is how to get (step by step) generating function from this recurrence?
we know that it's some king of distribution and from how it looks we can say it's binomial distribution.
recurrence-relations generating-functions recursive-algorithms
asked Nov 28 at 9:58
Gokuruto
123
Hello i have given recurrence like this :
$$p_{n,k}=(1-q)p_{n-1,k-1}+qp_{n-1,k}$$
my question is how to get (step by step) generating function from this recurrence?
we know that it's some king of distribution and from how it looks we can say it's binomial distribution.
recurrence-relations generating-functions recursive-algorithms
recurrence-relations generating-functions recursive-algorithms
asked Nov 28 at 9:58
Gokuruto
123
asked Nov 28 at 9:58
Gokuruto
123
asked Nov 28 at 9:58
Gokuruto
123
asked Nov 28 at 9:58
Gokuruto
123
asked Nov 28 at 9:58
Gokuruto
123
123
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 at 10:15
add a comment |
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 at 10:15
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 at 10:02
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 at 10:15
i think Probability would be the best one
– Gokuruto
Nov 28 at 10:15
add a comment |
1 Answer
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The probability generating function is $G_n(t):=sum_{k=0}^n p_{n,,k}t^k$, so $G_0=1$ and for $nge 1$ we have $$G_n(t)=q^n+sum_{k=1}^n ((1-q)p_{n-1,,k-1}+qp_{n-1,,k})t^k\=q^n+(1-q)tG_{n-1}(t)+q(G_{n-1}(t)-q^{n-1})\=(q+(1-q)t)G_{n-1}(t).$$Hence $G_n(t)=(q+(1-q)t)^n$ for all $n$.
answered Nov 28 at 10:28
J.G.
21.2k21933
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The probability generating function is $G_n(t):=sum_{k=0}^n p_{n,,k}t^k$, so $G_0=1$ and for $nge 1$ we have $$G_n(t)=q^n+sum_{k=1}^n ((1-q)p_{n-1,,k-1}+qp_{n-1,,k})t^k\=q^n+(1-q)tG_{n-1}(t)+q(G_{n-1}(t)-q^{n-1})\=(q+(1-q)t)G_{n-1}(t).$$Hence $G_n(t)=(q+(1-q)t)^n$ for all $n$.
answered Nov 28 at 10:28
J.G.
21.2k21933
add a comment |
up vote
0
down vote
accepted
The probability generating function is $G_n(t):=sum_{k=0}^n p_{n,,k}t^k$, so $G_0=1$ and for $nge 1$ we have $$G_n(t)=q^n+sum_{k=1}^n ((1-q)p_{n-1,,k-1}+qp_{n-1,,k})t^k\=q^n+(1-q)tG_{n-1}(t)+q(G_{n-1}(t)-q^{n-1})\=(q+(1-q)t)G_{n-1}(t).$$Hence $G_n(t)=(q+(1-q)t)^n$ for all $n$.
answered Nov 28 at 10:28
J.G.
21.2k21933
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The probability generating function is $G_n(t):=sum_{k=0}^n p_{n,,k}t^k$, so $G_0=1$ and for $nge 1$ we have $$G_n(t)=q^n+sum_{k=1}^n ((1-q)p_{n-1,,k-1}+qp_{n-1,,k})t^k\=q^n+(1-q)tG_{n-1}(t)+q(G_{n-1}(t)-q^{n-1})\=(q+(1-q)t)G_{n-1}(t).$$Hence $G_n(t)=(q+(1-q)t)^n$ for all $n$.
answered Nov 28 at 10:28
J.G.
21.2k21933
The probability generating function is $G_n(t):=sum_{k=0}^n p_{n,,k}t^k$, so $G_0=1$ and for $nge 1$ we have $$G_n(t)=q^n+sum_{k=1}^n ((1-q)p_{n-1,,k-1}+qp_{n-1,,k})t^k\=q^n+(1-q)tG_{n-1}(t)+q(G_{n-1}(t)-q^{n-1})\=(q+(1-q)t)G_{n-1}(t).$$Hence $G_n(t)=(q+(1-q)t)^n$ for all $n$.
answered Nov 28 at 10:28
J.G.
21.2k21933
answered Nov 28 at 10:28
J.G.
21.2k21933
answered Nov 28 at 10:28
J.G.
21.2k21933
answered Nov 28 at 10:28
J.G.
21.2k21933
21.2k21933
add a comment |
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Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 at 10:15