Generating function from recurrence relation of binomial distribution
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 '18 at 9:58
Gokuruto
123
add a comment |
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 '18 at 9:58
Gokuruto
123
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 '18 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 '18 at 10:15
add a comment |
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 '18 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 '18 at 9:58
Gokuruto
123
asked Nov 28 '18 at 9:58
Gokuruto
123
asked Nov 28 '18 at 9:58
Gokuruto
123
asked Nov 28 '18 at 9:58
Gokuruto
123
asked Nov 28 '18 at 9:58
Gokuruto
123
123
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 '18 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 '18 at 10:15
add a comment |
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 '18 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 '18 at 10:15
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 '18 at 10:02
Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 '18 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 '18 at 10:15
i think Probability would be the best one
– Gokuruto
Nov 28 '18 at 10:15
add a comment |
1 Answer
1
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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 '18 at 10:28
J.G.
23k22137
add a comment |
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1 Answer
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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 '18 at 10:28
J.G.
23k22137
add a comment |
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 '18 at 10:28
J.G.
23k22137
add a comment |
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 '18 at 10:28
J.G.
23k22137
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 '18 at 10:28
J.G.
23k22137
answered Nov 28 '18 at 10:28
J.G.
23k22137
answered Nov 28 '18 at 10:28
J.G.
23k22137
answered Nov 28 '18 at 10:28
J.G.
23k22137
23k22137
add a comment |
add a comment |
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Which kind of generating function do you want? Probability, moment, cumulant?
– J.G.
Nov 28 '18 at 10:02
i think Probability would be the best one
– Gokuruto
Nov 28 '18 at 10:15