\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
For a random variable \(X\) with \(\Expt (X) = \mu \) and \(\Var (X) = \sigma ^2\), we have \[ \kappa (X) = \frac {\Expt \left [(X - \mu )^4\right ]}{\sigma ^4} - 3 \]
We have \(Y = X - a\). Therefore, \(\Expt (Y) = \mu - a\) and \(\Var (Y) = \sigma ^2\).
\begin {align*} \kappa (Y) & = \frac {\Expt \left [(Y - (\mu - a))^4\right ]}{\sigma ^4} - 3 \\ & = \frac {\Expt \left [((X - a) - (\mu - a))^4\right ]}{\sigma ^4} - 3 \\ & = \frac {\Expt \left [(X - \mu )^4\right ]}{\sigma ^4} - 3 \\ & = \kappa (X), \end {align*}
as desired.
Let \(X \sim \Normal (0, \sigma ^2)\), \(\mu = 0\). Notice that \begin {align*} \kappa (X) = \frac {\Expt (X^4)}{\sigma ^4} - 3. \end {align*}
\(X\) has p.d.f. \[ f_X(x) = \frac {1}{\sigma \sqrt {2\pi }}\exp \left (-\frac {x^2}{2\sigma ^2}\right ). \]
Therefore, \[ \Expt (X^4) = \frac {1}{\sigma \sqrt {2\pi }} \int _{-\infty }^{\infty } x^4 \exp \left (-\frac {x^2}{2\sigma ^2}\right ) \Diff x. \]
Now, consider using integration by parts. Notice that \[ \Diff \exp \left (-\frac {x^2}{2\sigma ^2}\right ) = - \frac {x}{\sigma ^2} \exp \left (-\frac {x^2}{2\sigma ^2}\right ) \Diff x, \] and therefore, using integration by parts, we have \begin {align*} & \phantom {=} \int x^4 \exp \left (-\frac {x^2}{2\sigma ^2}\right ) \Diff x \\ & = - \sigma ^2 \int x^3 \Diff \exp \left (-\frac {x^2}{2\sigma ^2}\right ) \\ & = - \sigma ^2 \left [ x^3 \exp \left (-\frac {x^2}{2\sigma ^2}\right ) - \int \exp \left (-\frac {x^2}{2\sigma ^2}\right ) \Diff (x^3) \right ] \\ & = 3 \sigma ^2 \int x^2 \exp \left (-\frac {x^2}{2\sigma ^2}\right ) \Diff x - \sigma ^2 x^3 \exp \left (-\frac {x^2}{2\sigma ^2}\right ). \end {align*}
Therefore, considering the definite integral, we have \begin {align*} \Expt (X^4) & = \frac {1}{\sigma \sqrt {2\pi }} \int _{-\infty }^{\infty } x^4 \exp \left (-\frac {x^2}{2\sigma ^2}\right ) \Diff x \\ & = \frac {\sigma }{\sqrt {2\pi }} \left [3 \int _{-\infty }^{\infty } x^2 \exp \left (- \frac {x^2}{2\sigma ^2}\right )\Diff x - \left [x^3 \exp \left (- \frac {x^2}{2 \sigma ^2}\right )\right ]^{\infty }_{-\infty }\right ] \\ & = \frac {\sigma }{\sqrt {2\pi }} \left [3 \cdot \sigma \sqrt {2\pi } \cdot \sigma ^2 - 0\right ] \\ & = 3 \sigma ^4. \end {align*}
Therefore, \[ \kappa (X) = \frac {\Expt (X^4)}{\sigma ^4} - 3 = \frac {3 \sigma ^4}{\sigma ^4} - 3 = 0, \] as desired.
An alternative solution exists using generating functions.
Recall that a general normal distribution \(\Normal (\mu , \sigma ^2)\) has MGF \[ M(t) = \exp (\mu t + \frac {\sigma ^2}{2}t^2), \] and hence \begin {align*} M_X(t) & = \exp \left (\frac {\sigma ^2}{2}t^2\right ) \\ & = 1 + \left (\frac {\sigma ^2}{2}t^2\right ) + \frac {\left (\frac {\sigma ^2}{2}t^2\right )}{2!} + \ldots . \end {align*}
Therefore, \[ \Expt (X^4) = M^{(4)}_X(0) = \left (\frac {\sigma ^2}{2}\right )^4 \cdot 4! = 3 \sigma ^4, \] and the result follows.
Notice that \begin {align*} & \phantom {=} T^4 \\ & = \sum _{a} \binom {4}{4} Y_a^4 + \sum _{a < b} \left [\binom {4}{1, 3}Y_a Y_b^3 + \binom {4}{2, 2}Y_a^2 Y_b^2\right ] \\ & \phantom {=} + \sum _{a < b < c} \binom {4}{1, 1, 2}Y_a Y_b Y_c^2 + \sum _{a < b < c < d} \binom {4}{1, 1, 1, 1} Y_a Y_b Y_c Y_d \\ & = \sum _{a}Y_a^4 + \sum _{a < b} (4Y_a Y_b^3 + 6Y_a^2 Y_b^2) + \sum _{a < b < c} 12Y_a Y_b Y_c^2 + \sum _{a < b < c < d} 24 Y_a Y_b Y_c Y_d, \end {align*}
where \[ \binom {n}{a_1, a_2, \ldots , a_k} = \frac {n!}{a_1!a_2!\ldots a_k!}, \sum _{i = 1}^{k} a_i = n \] stands for the multinomial coefficient.
Note that \(\Expt (Y_r) = 0\) for any \(r = 1, 2, \ldots , n\). Therefore, \begin {align*} \Expt (Y_a Y_b^3) & = \Expt (Y_a) \Expt (Y_b^3) = 0, \\ \Expt (Y_a Y_b Y_c^2) & = \Expt (Y_a) \Expt (Y_b) \Expt (Y_c^2) = 0, \\ \Expt (Y_a Y_b Y_c Y_d) & = \Expt (Y_a) \Expt (Y_b) \Expt (Y_c) \Expt (Y_d) = 0. \end {align*}
Therefore, \begin {align*} \Expt (T^4) & = \sum _a \Expt (Y_a^4) + \sum _{a < b} 6 \Expt (Y_a^2 Y_b^2) \\ & = \sum _{r = 1}^{n} \Expt (Y_r^4) + 6 \sum _{r = 1}^{n - 1} \sum _{s = r + 1}^{n} \Expt (Y_a^2) \Expt (Y_b^2), \end {align*}
as desired.
Let \(Y_i = X_i - \mu \) for \(i = 1, 2, \ldots , n\), and \(\mu = \Expt (X), \sigma ^2 = \Var (X) = \Var (Y)\) with \(\Expt (Y) = 0\)
Therefore, let \(T = \sum _{i}^{n} Y_i = \sum _{i}^{n} X_i - n\mu \), we must have \(\Expt (T) = 0\) and \(\Var (T) = n \sigma ^2\).
But since the kurtosis remains constant with shifts, we must have that \(\kappa (Y_i) = \kappa \), and \[ \kappa (T) = \kappa \left [\sum _{i}^{n} X_i \right ]. \]
Hence, we have \begin {align*} \kappa \left [\sum _{i}^{n} X_i \right ] & = \kappa (T) \\ & = \frac {\Expt (T^4)}{(n \sigma ^2)^2} - 3 \\ & = \frac {\sum _{r = 1}^{n} \Expt (Y_r^4) + 6 \sum _{r = 1}^{n - 1} \sum _{s = r + 1}^{n} \Expt (Y_a^2) \Expt (Y_b^2)}{n^2 \sigma ^4} - 3 \\ & = \frac {1}{n^2} \sum _{r = 1}^{n} \frac {\Expt (Y_r^4)}{\sigma ^4} + \frac {6}{n^2} \sum _{r = 1}^{n - 1} \sum _{s = r + 1}^{n} \frac {\sigma ^4}{\sigma ^4} - 3 \\ & = \frac {1}{n^2} n \cdot (\kappa + 3) + \frac {6}{n^2} \binom {n}{2} - 3 \\ & = \frac {\kappa }{n} + \frac {3n + 3n(n - 1) - 3n^2}{n^2} \\ & = \frac {\kappa }{n} + 0 \\ & = \frac {\kappa }{n}, \end {align*}
as desired.