Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

the random variable that models the return on a block of business(in millions) is denoted by R. the generating function is $(.8+.2s)^3 $. what is the coefficient of variation? first off I get different answers if I expand and take the derivative or if I just use the chain rule. Im getting over 100% as a answer but does not match the 86.6% they get.

$E[X]=h^{'}_{X}(s)= 3(0.2)((0.8+0.2s)^2= 0.6(1)^2=0.6$

$V[X]=h^{''}_{X}(s)+E[X]-E[X]^2= 3(2)(0.2)^2(0.8+0.2(1))^1+0.6-0.6^2=.48$

$\dfrac{\sigma_X(100)}{\mu_X}=\dfrac{\sqrt{.48}100}{0.6}\approx 115$% So what am I doing wrong?

share|cite|improve this question
When I learned this, I learned that $E[X] = h'_X(0)$ and $E[X^2] = h''_X(0)$ where $s$ is evaluated at $0$, not $1$ (see here). And also, that $V[X] = E[X^2]-E[X]^2$, rather than what you have written. Do we have different definitions of generating function? – angryavian Dec 17 '13 at 3:16
In the book they define $V[X]=h^{''}_X(1)+h^{'}_X(1)-[h^{'}_X]^2$. The reason you add $E[X]$ is because you need to realize that $h^{''}_X(1)=E[X(X-1)] = E[X^2]-E[X]$ therefore we need to add $E[X]$ to account for the $E[X]$ subtracted which gives us our result – adam Dec 17 '13 at 3:41
Thanks for the explanation. I see we have different definitions of generating function. What is the definition of $h_X(s)$ for you? I mistook it to be $h_X(s) = E\left[e^{sX}\right]$ as found here. – angryavian Dec 17 '13 at 3:45
@blf: You are confusing moment generating functions with probability generating functions. The Wikipedia article you linked to establishes the identity $G(e^t)=M(t)$, which for $t=0$ shows that evaluating the generating function at $e^0=1$ is the correct approach. – baudolino Dec 17 '13 at 3:46
@baudolino Oh ok, thanks! That was very helpful. – angryavian Dec 17 '13 at 3:48
up vote 1 down vote accepted

Your calculation is correct. What's probably happening, given that $1/1.15 \approx 0.866$, is that "they" simply thought that $CV=\mu/\sigma$ instead of the other way around. "They" are wrong.

share|cite|improve this answer
I also had the same issue with another problem where I am given a table x: 0,50,100,200. and p(x) .74,.12 .09, .05 respectively. I do not get the answer in the back of the book as well for this problem so are they wrong here as well? – adam Dec 17 '13 at 3:06
Maybe. See if you take the reciprocal of your result leads to the published one, i.e., if they made the same conceptual mistake. – baudolino Dec 17 '13 at 3:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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