0
$\begingroup$

I have a question which says: "Show that the probability generating function of Poisson varied with parameter $\lambda$ is $e^{-\lambda(1-t)}$.

I am not quite sure about the question because the paper I got has the last term of the question blurred.

$\endgroup$
  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Dec 11 '15 at 23:26
  • $\begingroup$ Are you sure that's "varied" rather than "variable", say? $\endgroup$ – Glen_b -Reinstate Monica Dec 19 '15 at 5:19
  • $\begingroup$ I have corrected the pgf (there was a sign missing) and formatted the mathematics. $\endgroup$ – Glen_b -Reinstate Monica Dec 19 '15 at 6:21
  • $\begingroup$ Please add the self-study tag, read its tag-wiki and modify your question to follow the guidelines on asking such questions. In particular, you'll need to clearly identify what you've done to solve the problem yourself, and indicate the specific help you need at the point you struck difficulty. [Note that the pgf is defined as $\sum_{x=0}^\infty p_X(x)t^x$] $\endgroup$ – Glen_b -Reinstate Monica Dec 19 '15 at 6:24
2
$\begingroup$

An outline of the approach:

The pgf is defined as $G(t) = \sum_{x=0}^\infty p_X(x)\,t^x$.

So for the Poisson you get

$$G(t) = \sum_{x=0}^\infty \frac{e^{-\lambda} \lambda^x}{x!}\cdot t^x$$

To simplify things, pull the constant out the front, then combine the two terms than have powers of $x$.

You now play "spot the pmf", put the necessary normalizing constant in the sum and its inverse outside. After the obvious simplification you're left with the desired pgf.

$\endgroup$
1
$\begingroup$

Let me use the variable nomenclature in Wikipedia, which hopefully will be around forever if everybody donates money, as in here and here.

So we depart from the probability mass function ($\text{pmf}$) of the Poisson distribution as:

$\displaystyle p_X(k) = \frac{\lambda^k}{k!}\,e^{-\lambda}$

and the definition of probability generating function ($\text{pgf}$) as:

$\displaystyle G(z)\,= \,\mathbb{E}[z^k]\,=\,\sum_{k=0}^\infty p_X(k) \,\,z^k=$

$=p(0)z^0+p(1)z^1+p(2)z^2+p(3)z^3+\cdots=$

$=\displaystyle\frac{\lambda^0}{0!}\,e^{-\lambda}\,z^0+\frac{\lambda^1}{1!}\,e^{-\lambda}\,z^1+\frac{\lambda^2}{2!}\,e^{-\lambda}\,z^2+\frac{\lambda^3}{3!}\,e^{-\lambda}\,z^3+\cdots\tag1$

Knowing that the Maclaurin series of $e^z$ evaluated at zero is

$\displaystyle e^z= \frac{z^0}{0!}+\frac{z^1}{1!}+\frac{z^2}{2!}+\frac{z^3}{2!}+...$

And for $\displaystyle e^{z\lambda}$,

$\displaystyle e^{z\lambda}=\frac{\lambda^0 z^0}{0!}+\frac{\lambda^1 z^1}{1!}+\frac{\lambda^2 z^2}{2!}+\frac{\lambda^3 z^3}{2!}+...\tag2$

We can compare equation (2) to (1), where $z^{-\lambda}$ is a common factor, to obtain the more practical expression of $G(z)$:

$\displaystyle G(z)\,=\,e^{-\lambda}\,e^{z\lambda}\,=\,e^{\lambda(z-1)}\,=\,e^{-\lambda(1-z)}$. And this is the expression on the OP, ($e^{-\lambda(1-t)}$), using $z$ as opposed to $t$ as the variable in the power series.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.