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The exponential distribution is given by: $$PDF: \lambda e^{\lambda x}$$ And the formula for probability generating function is given by: $$G(z) = \sum_{x=0}^\infty p(x)z^x$$ where $p(x)$ is a probability mass function. I know that $p(x)$ is only defined for discrete random variables, so $G(z)$ for exp. distr. doesn't exist. Can I prove it some other way though? Can I substitute $p(x)$ for something else that does exist, and mathematically prove that this sum is divergent?

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  • $\begingroup$ There is a typo in your PDF. It should be $\lambda e^{-\lambda x}$. $\endgroup$ – drhab Apr 19 at 15:11
  • $\begingroup$ A sum over an uncountable set is ill-defined, you surely mean $\int_0^\infty z^x f(x)\mathrm{d}x$ where $f$ is the PDF. $\endgroup$ – Nap D. Lover Apr 19 at 20:41
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The probability generating function of a random variable $X$ that has a PDF $f_X$ is :$$G(z)=\mathbb Ez^X=\int z^x f_X(x)dx$$

So applying that here we must find:$$G(z)=\mathbb Ez^X=\lambda \int_0^\infty z^x e^{-\lambda x}dx$$ Give it a try.

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If the distribution of $X$ has a probability generating function $G_X(z)=\mathbb E\left(z^X\right)$ and a moment generating function $M_X(t)=\mathbb E\left(e^{tX}\right)$ then you can say $$G_X\left(e^t\right)=M_X\left(t\right)$$ and $$G_X\left(z\right)=M_X\left(\log_e(z)\right)$$ though the latter has issues when $z \le 0$

So for example with a binomial distribution with parameters $n,p$ you get $G(z)=(1-p+pz)^n$ and $M(t)=(1-p+pe^{t})^n$

Now an exponential distribution with rate parameter $\lambda$ has $M(t)=\dfrac{\lambda}{\lambda -t}$ for $t \lt \lambda$ so you might guess it might have a probability generating function $G(z) =\dfrac{\lambda}{\lambda - \log_e(z)}$ at least for $0 \lt z \lt e^\lambda$

But this does not work, as there is no series expansion of that $G(z)$ at $z=0$

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The mean of $z^X$ can be computed, for suitable $z$, even when the distribution isn't discrete. The "suitable" $z$ always include $1$, but what else qualifies depends on the distribution considered. Sometimes, nothing else works; sometimes, any $z>0$ works. For the exponential distribution, something intermediate happens; if $0<z<\exp\lambda$, $$\int_0^\infty\lambda z^x\exp -\lambda xdx=\int_0^\infty\lambda\exp -(\lambda -\ln z)xdx=\frac{\lambda}{\lambda -\ln z}.$$When people say the probability generating function "only exists" in the discrete case, they're not denying that the mean of $z^X$ exists; they're just pointing out it lacks the usual probability interpretation, wherein the result is a power series with $x^k$ having coefficient $P(X=k)$. Meanwhile, outside the discrete case $\sum_{x\ge 0}p(x)z^x$, with $p$ the PDF, isn't the mean of $z^X$ at all, so two "obviously equivalent" definitions of the PGF are not, in fact, equivalent. (That's at least part of why we don't bother thinking of PGFs in this context.) Instead your sum converges (for the same range of $z$ as in the previous calculation) to $$\lambda\sum_{x\ge 0}(z\exp -\lambda)^x=\frac{\lambda}{1-z\exp -\lambda}.$$This calculation doesn't require the theory of PGFs at all, just an understanding of geometric series.

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