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Drew75
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It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong. There are two definitions for the pdf of a geometric distribution. The one you use, where $E(X)=\frac{1}{p}$ is defined from 1 to infinity. At zero it is not defined. So, the generating function needs to take this into account, as well.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$

If you use the alternative definition, where $P(Y=y)=q^ip$, then the pdf is defined at zero. In this case the generating function converges to $\frac{p}{1-qs}$.

As for what $s$ represents, as far as I know it represents nothing. Generating functions are derived functions that hold information in their coefficients. They are sometimes left as an infinite sum, sometimes they have a closed form expression. Take a look at the wikipedia article, which give some examples of how they can be used. Here and here.wiki article probability generating functions and wiki article generating functions

It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong. There are two definitions for the pdf of a geometric distribution. The one you use, where $E(X)=\frac{1}{p}$ is defined from 1 to infinity. At zero it is not defined. So, the generating function needs to take this into account, as well.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$

If you use the alternative definition, where $P(Y=y)=q^ip$, then the pdf is defined at zero. In this case the generating function converges to $\frac{p}{1-qs}$.

It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong. There are two definitions for the pdf of a geometric distribution. The one you use, where $E(X)=\frac{1}{p}$ is defined from 1 to infinity. At zero it is not defined. So, the generating function needs to take this into account, as well.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$

If you use the alternative definition, where $P(Y=y)=q^ip$, then the pdf is defined at zero. In this case the generating function converges to $\frac{p}{1-qs}$.

As for what $s$ represents, as far as I know it represents nothing. Generating functions are derived functions that hold information in their coefficients. They are sometimes left as an infinite sum, sometimes they have a closed form expression. Take a look at the wikipedia article, which give some examples of how they can be used. Here and here.wiki article probability generating functions and wiki article generating functions

added 400 characters in body
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Drew75
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It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong. There are two definitions for the pdf of a geometric distribution. The one you use, where $E(X)=\frac{1}{p}$ is defined from 1 to infinity. At zero it is not defined. So, the generating function needs to take this into account, as well.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$

If you use the alternative definition, where $P(Y=y)=q^ip$, then the pdf is defined at zero. In this case the generating function converges to $\frac{p}{1-qs}$.

It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$

It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong. There are two definitions for the pdf of a geometric distribution. The one you use, where $E(X)=\frac{1}{p}$ is defined from 1 to infinity. At zero it is not defined. So, the generating function needs to take this into account, as well.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$

If you use the alternative definition, where $P(Y=y)=q^ip$, then the pdf is defined at zero. In this case the generating function converges to $\frac{p}{1-qs}$.

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Drew75
  • 1.2k
  • 10
  • 12

It's normal you'd arrive at the wrong answer in this case. The problem is that your index is wrong.

$$\pi(s)=E(S^X)=\sum^\infty_{i=1}q^{i-1}ps^i$$ $$= ps\sum^\infty_{i=1}(qs)^{i-1}=ps\sum^\infty_{i=0}(qs)^i$$ $$=\frac{ps}{1-qs}$$