I am trying to solve the following:

The nonnegative, integer-valued, random variable X has generating function $g_{X(t)}=log(\frac{1}{1-qt})$. Determine $P(X = k)$ for $k = 0, 1, 2, . . .$, $E[X]$, and $Var[X]$.

I know the probability generating function is of the form $g_{X(t)}=E[t^{x}]=\sum_{n=1}^{\infty}t^{n}*P(X=n)$. I thought I could do a Taylor Series expansion of the $g_{X(t)}$ and make maniuplate it into the form of $\sum_{n=1}^{\infty}t^{n}*P(X=n)$, but I'm not sure I can get it to factor. When I take the Taylor Series expansion I get $log(\frac{1}{1-qt}) = -\sum_{k=1}^{\infty} \frac{(-1)^{k} (\frac{qt}{1-qt})^k}{k}$ for $|{\frac{x}{1-x}}|<1$. I'm not sure what to do about the alternating term, and this doesn't look like anyone of the discrete distributions I am familiar with.

To get the variance and expected value I should be able to use $E[X]=g'_x(1)$ and $Var[X]=g''_x(1)+g'_x(1)-(g'(1))^2$, where I substitute $qt=1$ into the probability generating function.

  • $\begingroup$ The $(t)$ should not be in the subscript. Should be $g_X(t)$ $\endgroup$ – Thomas Andrews Jun 22 '15 at 0:11

You've got the wrong Taylor series.

$$\log\left(\frac{1}{1-z}\right)=\sum_{n=1}^\infty \frac{z^n}{n}$$

for $|z|<1$.

Setting $z=qt$ gives you $P(X=n)=\frac{q^n}{n}$.

Note that this is not a probability function for all $q$, because you'd need $g_X(1)=1$, that is $\log\left(\frac{1}{1-q}\right) = 1$, which means you can only have one specific $q=1-e^{-1}$, or you need to scale $\log\left(\frac{1}{1-qt}\right)$.

  • $\begingroup$ Why is it a requirement that $log(\frac{1}{1-q})=1$? Further more when I go to solve for the expected value using the derivatives of the generating function, am I substituting t=1 and then the specific value of q in? $\endgroup$ – user75514 Jun 22 '15 at 0:36
  • $\begingroup$ Because $g_X(1)=\sum p(X=n)=1$, or $p$ is not a probability on the natural numbers. For your second question, the expected value of what? @user75514 $\endgroup$ – Thomas Andrews Jun 22 '15 at 0:39
  • $\begingroup$ If you substitute $t=1$, then $g_X(1)$ is computing the expected value of $1^X$, which is obviously going to be $1$. $\endgroup$ – Thomas Andrews Jun 22 '15 at 0:41
  • $\begingroup$ The questions asks to compute $E[X]$ and $Var[X]$. I have the formulas $E[X]=g'(1)$ and $Var[X]=g''(1)+g'(1)-(g'(1))^2$. To use these do I let $t=1$ and $q=1-e^{-1}$? $\endgroup$ – user75514 Jun 22 '15 at 0:45
  • $\begingroup$ Yes. $g'(t)=\frac{q}{1-qt}$. You get $g'(1)=e-1$ assuming that the question did not want you to scale the given generating function to make it a valid probability generating function - it is possible that is what the problem wanted you to do. $\endgroup$ – Thomas Andrews Jun 22 '15 at 0:48

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