3rd Cumulant Equals to 3rd moment, 4th Cumulant Equalts to 4th moment${}- 3$ Under given standardized random variable $Z = (X-\mu)/\sigma$ Show that 
$$C_3(Z) = M_3(Z)\;\;\text{and}\;\; C_4(Z) = M_4(Z) -3$$
where $C_r(Z)$ refer to $r$-th cumulant, $M_r(Z)$ refer to $r$-th moment.

Question
I've been required to show above relationship by my textbook, but sadly, my textbook doesnot provide well-constructed mathematical definition of moment and cumulant. I've checked Hogg's book, but this book also, is insufficient rigourously prove above statement. Any advice or recommendation for above proof?
 A: The validity of an answer might depend on which characterizations of cumulants you're looking for.
The $r$th cumulant is a polynomial function of the first $r$ moments. The $r$th cumulant is shift-invariant if $r>1.$ "Shift invariant" means that for random variables $X$ and constants $c$ ("constant" = not random) the $r$th cumulant of the probability distribution of $X+c$ is the same as the $r$th cumulant of the probability distribution of $X,$ i.e. $C_r(X+c) = C_r(X).$ Because of shift invariance, you can take the $r$th cumulant to be a polynomial in the first $r$ central moments.
Thus the $4$th cumulant must be a polynomial function of the second central moment (i.e. the variance), the $3$rd central moment, and the $4$th central moment (the $4$th central moment of the probability distribution of $X$ is $\operatorname{E}((X-\mu)^4)$ where $\mu = \operatorname{E}(X)$).
The $4$th cumulant should also be $4$th-degree homogeneous, i.e. $C_4(cX) = c^4 C_4(X).$ That means the third central moment will not appear in the polynomial and the $2$nd central moment needs to get squared. So we have
$$
C_4(X) = \operatorname{E}((X-\mu)^4) + \left( \text{coefficient} \times \left( \operatorname{E}((X-\mu)^2) \right)^2 \right).
$$
(The coefficient of the $4$th central moment is $1$; that is how cumulants get scaled.)
The question now is, what must the “coefficient” be in order that the $4$th cumulant be “cumulative”, i.e. for independent random variables $X_1,\ldots,X_n$ we have
$$
C_4(X_1+\cdots+X_n) = C_4(X_1)+\cdots+C_4(X_n) \text{ ?}
$$
Some algebra using the binomial theorem will show that it must be $-3,$ so that we have
$$
C_4(X) = \operatorname{E}((X-\mu)^4) - 3\big( \operatorname{E}((X-\mu)^2) \big)^2
$$
To see that, first not that no generality is lost by assuming that $\mu=0.$ So we have $X_1,X_2$ independent and $\operatorname{E}(X_1) = \operatorname{E}(X_2) =0.$
Then$\require{cancel}$
\begin{align}
C_4(X_1+X_2) = {} & \operatorname{E}((X_1+X_2)^4) + \left( \text{coefficient} \times \left( \operatorname{E}((X_1+X_2)^2) \right)^2 \right) \\[10pt]
= {} & \operatorname{E}(X_1^4) + 4\operatorname{E}(X_1^3) \cancelto{0}{\operatorname{E}(X_2)} + 6\operatorname{E}(X_1^2)\operatorname{E}(X_2^2) + 4\cancelto{0}{\operatorname{E}(X_1)} \operatorname{E}(X_2^3) + \operatorname{E}(X_2^4) \\
& {} - \text{coefficient} \times (\operatorname{E}(X_1^2) + 2\cancelto{0}{\operatorname{E}(X_1) \operatorname{E}(X_2)} + \operatorname{E}(X_2^2))^2 \\[10pt]
= {} & \operatorname{E}(X_1^4) + 6\operatorname{E}(X_1^2) \operatorname{E}(X_2^2) + \operatorname{E}(X_2^4) \\
& {} - \left( (\text{coefficient}) \times \left( (\operatorname{E}X_1^2))^2 + 2\operatorname{E}(X_1^2) \operatorname{E}(X_2^2) + (\operatorname{E}(X_2^2))^2 \right) \right)
\\[12pt]
= {} & \overbrace{\operatorname{E}(X_1^4) + \text{coefficient}\times (\operatorname{E}(X_1^2))^2} {} + {} \overbrace{\operatorname{E}(X_2^4) + \text{coefficient}\times (\operatorname{E}(X_2^2))^2} \\
& {} + 6 \operatorname{E}(X_1^2) \operatorname{E}(X_2^2) + (\text{coefficient}) \times 2 \operatorname{E}(X_1^2) \operatorname{E}(X_2^2) \\[10pt]
= {} & C_4(X_1) + C_4(X_2) + \underbrace{6 \operatorname{E}(X_1^2) \operatorname{E}(X_2^2) + (\text{coefficient}) \times 2 \operatorname{E}(X_1^2) \operatorname{E}(X_2^2)} \\[10pt]
= {} & C_4(X_1) + C_4(X_2) + \underbrace{(6 + 2\times\text{coefficient}) \operatorname{E}(X_1^2) \operatorname{E}(X_2^2)} \\[10pt]
= {} & C_4(X_1) + C_4(X_2) \text{ if and only if coefficient} = -3.
\end{align}
$$
\text{Conclusion: } C_4(X) = \operatorname{E}\big((X-\mu)^4\big) - 3 \Big( \operatorname{E}\big((X-\mu)^2\big) \Big)^2.
$$
A: Note that $Z\sim N(0,1)$. The MGF of a standard normal is
$$
M(t)=\exp\left(\mu t+\frac{1}{2}\sigma^2t^2\right)=\exp\left(\frac{1}{2}t^2\right).
$$
In particular the cumulant generating function of a standard normal is 
$$
K(t)=\frac{1}{2}t^2.
$$
It follows that the nth cumulant $\kappa_n=K^{(n)}(0)=0$ for $n>2$. The first identity is clear since
$$
E(Z^3)=0=\kappa_3
$$
as the the density of a standard normal is even. The second identity will follow from showing that $E(Z^4)=3$ from our comments above. Note that
$$
E(Z^4)=V(Z^2)+E(Z^2)^2=2+1=3
$$
since $Z^2\sim \chi^2_{(1)}=\text{Gamma}(1/2,1/2)$ where we used the shape, rate parametrization of the gamma distribution. 
