2016.3.13 Question 13

For a random variable $X$ with $\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X)=\mu$ and $\text{Var}<!--\; FUNCTION\; APPLICATION\; -->(X)={\sigma }^2$, we have

We have $Y=X-a$. Therefore, $\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y)=\mu -a$ and $\text{Var}<!--\; FUNCTION\; APPLICATION\; -->(Y)={\sigma }^2$.

as desired.

1. Let $X\sim \text{N}<!--\; FUNCTION\; APPLICATION\; -->(0,{\sigma }^2)$, $\mu =0$. Notice that

   $$\begin{array}{lll}\hfill \kappa (X)=\frac{\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X^4)}{{\sigma }^4}-3.& & \hfill \end{array}$$

   $X$ has p.d.f.

   $$f_X(x)=\frac{1}{\sigma \sqrt{2\pi }}\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right).$$

   Therefore,

   $$\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X^4)=\frac{1}{\sigma \sqrt{2\pi }}{\int \; <!--\; nolimits\; -->}_{-\infty }^{+\infty }{x}^4\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)<!--\; FUNCTION\; APPLICATION\; -->dx.$$

   Now, consider using integration by parts. Notice that

   $$d\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)=-\frac{x}{{\sigma }^2}\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)<!--\; FUNCTION\; APPLICATION\; -->dx,$$

   and therefore, using integration by parts, we have

   $$\begin{array}{llll}\hfill & \int \; <!--\; nolimits\; -->x^4\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)<!--\; FUNCTION\; APPLICATION\; -->dx& \hfill & \\ \hfill & =-{\sigma }^2\int \; <!--\; nolimits\; -->x^{3}d\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)& \hfill & \\ \hfill & =-{\sigma }^2\left[x^3\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)-\int \; <!--\; nolimits\; -->\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)<!--\; FUNCTION\; APPLICATION\; -->d(x^3)\right]& \hfill & \\ \hfill & =3{\sigma }^2\int \; <!--\; nolimits\; -->x^2\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)<!--\; FUNCTION\; APPLICATION\; -->dx-{\sigma }^2{x}^3\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right).& \hfill & \end{array}$$

   Therefore, considering the definite integral, we have

   $$\begin{array}{llll}\hfill \text{E}<!--\; FUNCTION\; APPLICATION\; -->(X^4)& =\frac{1}{\sigma \sqrt{2\pi }}{\int \; <!--\; nolimits\; -->}_{-\infty }^{+\infty }{x}^4\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)<!--\; FUNCTION\; APPLICATION\; -->dx& \hfill & \\ \hfill & =\frac{\sigma }{\sqrt{2\pi }}\left[3{\int \; <!--\; nolimits\; -->}_{-\infty }^{+\infty }{x}^2\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)<!--\; FUNCTION\; APPLICATION\; -->dx-{\left[x^3\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(-\frac{x^2}{2{\sigma }^2}\right)\right]}_{-\infty }^{+\infty }\right]& \hfill & \\ \hfill & =\frac{\sigma }{\sqrt{2\pi }}\left[3\cdot \sigma \sqrt{2\pi }\cdot {\sigma }^2-0\right]& \hfill & \\ \hfill & =3{\sigma }^4.& \hfill & \end{array}$$

   Therefore,

   $$\kappa (X)=\frac{\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X^4)}{{\sigma }^4}-3=\frac{3{\sigma }^4}{{\sigma }^4}-3=0,$$

   as desired.

   An alternative solution exists using generating functions.

   Recall that a general normal distribution $\text{N}<!--\; FUNCTION\; APPLICATION\; -->(\mu ,{\sigma }^2)$ has MGF

   $$M(t)=\text{exp}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{\mu t}+\frac{{\sigma }^2}{2}{t}^2),$$

   and hence

   $$\begin{array}{llll}\hfill M_X(t)& =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{{\sigma }^2}{2}{t}^2\right)& \hfill & \\ \hfill & =1+\left(\frac{{\sigma }^2}{2}{t}^2\right)+\frac{\left(\frac{{\sigma }^2}{2}{t}^2\right)}{2!}+\dots .& \hfill & \end{array}$$

   Therefore,

   $$\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X^4)=M_X^{(4)}(0)={\left(\frac{{\sigma }^2}{2}\right)}^4\cdot 4!=3{\sigma }^4,$$

   and the result follows.

2. Notice that

   $$\begin{array}{llll}\hfill & T^4& \hfill & \\ \hfill & =\sum _a\left(\genfrac{}{}{0.0pt}{}{4}{4}\right)Y_a^4+\sum _{a<b}\left[\left(\genfrac{}{}{0.0pt}{}{4}{1,3}\right)Y_a{Y}_b^3+\left(\genfrac{}{}{0.0pt}{}{4}{2,2}\right)Y_a^2{Y}_b^2\right]& \hfill & \\ \hfill & +\sum _{a<b<c}\left(\genfrac{}{}{0.0pt}{}{4}{1,1,2}\right)Y_a{Y}_b{Y}_c^2+\sum _{a<b<c<d}\left(\genfrac{}{}{0.0pt}{}{4}{1,1,1,1}\right)Y_a{Y}_b{Y}_c{Y}_d& \hfill & \\ \hfill & =\sum _a{Y}_a^4+\sum _{a<b}(4Y_a{Y}_b^3+6Y_a^2{Y}_b^2)+\sum _{a<b<c}12Y_a{Y}_b{Y}_c^2+\sum _{a<b<c<d}24Y_a{Y}_b{Y}_c{Y}_d,& \hfill & \end{array}$$

   where

   $$\left(\genfrac{}{}{0.0pt}{}{n}{a_1,a_2,\dots ,a_k}\right)=\frac{n!}{a_1!a_2!\dots a_k!},\sum _{i=1}^k{a}_i=n$$

   stands for the multinomial coefficient.

   Note that $\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_r)=0$ for any $r=1,2,\dots ,n$. Therefore,

   $$\begin{array}{llll}\hfill \text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a{Y}_b^3)& =\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_b^3)=0,& \hfill & \\ \hfill \text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a{Y}_b{Y}_c^2)& =\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_b)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_c^2)=0,& \hfill & \\ \hfill \text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a{Y}_b{Y}_c{Y}_d)& =\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_b)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_c)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_d)=0.& \hfill & \end{array}$$

   Therefore,

   $$\begin{array}{llll}\hfill \text{E}<!--\; FUNCTION\; APPLICATION\; -->(T^4)& =\sum _a\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a^4)+\sum _{a<b}6\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a^2{Y}_b^2)& \hfill & \\ \hfill & =\sum _{r=1}^n\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_r^4)+6\sum _{r=1}^{n-1}\sum _{s=r+1}^n\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a^2)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_b^2),& \hfill & \end{array}$$

   as desired.

3. Let $Y_i=X_i-\mu$ for $i=1,2,\dots ,n$, and $\mu =\text{E}<!--\; FUNCTION\; APPLICATION\; -->(X),{\sigma }^2=\text{Var}<!--\; FUNCTION\; APPLICATION\; -->(X)=\text{Var}<!--\; FUNCTION\; APPLICATION\; -->(Y)$ with $\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y)=0$

   Therefore, let $T={\sum <!--\; FUNCTION\; APPLICATION\; -->}_i^n{Y}_i={\sum }_i^n{X}_i-\mathit{n\mu }$, we must have $\text{E}<!--\; FUNCTION\; APPLICATION\; -->(T)=0$ and $\text{Var}<!--\; FUNCTION\; APPLICATION\; -->(T)=n{\sigma }^2$.

   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

   $$\begin{array}{llll}\hfill \kappa \left[\sum _i^n{X}_i\right]& =\kappa (T)& \hfill & \\ \hfill & =\frac{\text{E}<!--\; FUNCTION\; APPLICATION\; -->(T^4)}{{(n{\sigma }^2)}^2}-3& \hfill & \\ \hfill & =\frac{\sum _{r=1}^n\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_r^4)+6\sum _{r=1}^{n-1}\sum _{s=r+1}^n\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_a^2)\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_b^2)}{n^2{\sigma }^4}-3& \hfill & \\ \hfill & =\frac{1}{n^2}\sum _{r=1}^n\frac{\text{E}<!--\; FUNCTION\; APPLICATION\; -->(Y_r^4)}{{\sigma }^4}+\frac{6}{n^2}\sum _{r=1}^{n-1}\sum _{s=r+1}^n\frac{{\sigma }^4}{{\sigma }^4}-3& \hfill & \\ \hfill & =\frac{1}{n^2}n\cdot (\kappa +3)+\frac{6}{n^2}\left(\genfrac{}{}{0.0pt}{}{n}{2}\right)-3& \hfill & \\ \hfill & =\frac{\kappa }{n}+\frac{3n+3n(n-1)-3n^2}{n^2}& \hfill & \\ \hfill & =\frac{\kappa }{n}+0& \hfill & \\ \hfill & =\frac{\kappa }{n},& \hfill & \end{array}$$

   as desired.