# coefficient of variation for probability functions

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?

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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

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.