t분포
t=\dfrac{Z}{\sqrt{U/v}} \sim t_{v}
사용처
- 분산을 다른 요소로 대체한 것.
- t-검정에서 사용한다.
공식 정리
\begin{align*}
\text{CDF} &\kern{10pt} \rightarrow
\kern{10pt}
F_T(t) = \int_{0}^{\infty}
f_{U}(u)~
\Phi \Big(
\dfrac{\sqrt{\mu}t}
{\sqrt{n-1}}
\Big)~
d\mu
\\[20pt]
\text{PDF} &\kern{10pt} \rightarrow
\kern{10pt} f_T(t)
= \int_{0}^{\infty}
f_{U}(u)~
\Phi \Big(
\dfrac{\sqrt{\mu}t}
{\sqrt{n-1}}
\Big)~
d\mu
\\[20pt]
\text{Average} &\kern{10pt} \rightarrow \kern{10pt}
E(T) = 0
\\[20pt]
\text{variance} &\kern{10pt} \rightarrow \kern{10pt} E(T^2)
= \dfrac{n-1}{n-3}
\end{align*}
CDF
\begin{align*}
F_T(t) &= P(T \le t) =
P\biggr(
\dfrac{\overline{X} - \mu}{s/\sqrt{n}}\biggr)
\\[15pt]
&= P\biggr(
\dfrac{\overline{X} - \mu}{\sigma/\sqrt{n}}
\le
\dfrac{s}{\sigma}~t
\biggr)
\\[15pt]
&= P\biggr(
Z \le
\dfrac{s}{\sigma}~t
\biggr)
\\[15pt]
&= P\biggr(
Z \le
\dfrac{\sqrt{\mu}}
{\sqrt{n-1}}~t
\biggr)
\\[15pt]
&=\int_{0}^{\infty}
\int_{0}^
{\frac{\sqrt{\mu}}
{\sqrt{n-1}}}
f_{U,~Z}(u, z)dzd\mu
\end{align*}
u = \dfrac{(n-1)s^2}{\sigma^2} \sim~\chi^2(n-1)
\\[10pt]
\dfrac{\mu}{n-1} =
\dfrac{s^2}{\sigma^2}_, \kern{10pt}
\dfrac{\sqrt{\mu}}{\sqrt{n-1}}
= \dfrac{s}{\sigma}
\begin{align*}
F_T(t) &= \int_{0}^{\infty}
\int_{0}^{\frac{\sqrt{\mu}}
{\sqrt{n-1}}}
f_{U}(u)~f_{Z}(z)~dzd\mu
\\[15pt]
&= \int_{0}^{\infty}
f_{U}(u)~
\Bigg\{
\int_{0}^{\frac{\sqrt{\mu}}
{\sqrt{n-1}}} f_{Z}(z)~dz
\Bigg\}
d\mu
\\[15pt]
&= \int_{0}^{\infty}
f_{U}(u)~
\Phi \Big(
\dfrac{\sqrt{\mu}t}
{\sqrt{n-1}}
\Big)~
d\mu
\end{align*}
즉, t분포는 카이제곱분포(감마분포)와 표준정규분포의 곱으로 만들어졌다.
\begin{align*}
f_T(t) &= \dfrac{d}{dt}~F_T(t)
\\[10pt]
&= \dfrac{d}{dt}~
\int_{0}^{\infty}
f_{U}(u)~
\Phi \Big(
\dfrac{\sqrt{\mu}t}
{\sqrt{n-1}}
\Big)~d\mu
\\[10pt]
&= \int_{0}^{\infty}
f_{U}(u)~
\phi \Big(
\dfrac{\sqrt{\mu}t}{\sqrt{n-1}}
\Big)
\dfrac{\sqrt{\mu}t}{\sqrt{n-1}}
~d\mu
\\[10pt]
\end{align*}
여기서 적분 기호 안의 함수들의 정체를 보면 다음과 같다.
- mu는 아까 보았듯이 카이제곱분포를 따르고, 이는 감마분포의 일종!
- phi 기호 = Z = 표준정규분포!
\begin{align*}
f_U(u) \kern{8pt} &=\kern{8pt}
\dfrac{1}{\Gamma(\frac{n-1}{2})~~
2 \raisebox{0.7em}
{$ \textstyle \frac{n-1}{2}$}~
}
~~\mu \raisebox{0.9em}
{$\textstyle \frac{n-1}{2}
- \scriptstyle 1$}
~~e \raisebox{3mm}
{$\textstyle -\frac{\mu}{2}$}
\\[40pt]
\phi \Big(
\dfrac{\sqrt{\mu}t}{\sqrt{n-1}}
\Big)
\dfrac{\sqrt{\mu}t}{\sqrt{n-1}}
\kern{8pt} &=\kern{8pt}
\dfrac{1}{\sqrt{2\pi}}
\cdot
\exp \Big(-\dfrac{1}{2}\cdot
\dfrac{\mu t^2}{n-1}
\Big) \cdot
\dfrac{\sqrt{\mu}}{\sqrt{n-1}}
\\[20pt]
\end{align*}
그래서 정리를 해보면…
우선 분모에 해당하는 부분을 가지고 온다.
\dfrac{1}{\Gamma(\frac{n-1}{2})~~
2 \raisebox{0.7em}
{$ \textstyle \frac{n-1}{2}$}~
}
\times
\dfrac{1}{\sqrt{2\pi}}
\times
\dfrac{1}{\sqrt{n-1}}
=
\dfrac{1}
{\Gamma(\frac{n-1}{2})~~
2 \raisebox{0.7em}
{$ \textstyle \frac{n-1}{2}$}~
\sqrt{2\pi}~\sqrt{n-1}}
\mu끼리 정리한다.
\mu \raisebox{0.9em}
{$\textstyle \frac{n-1}{2}
- \scriptstyle 1$} ~\times~ \sqrt{\mu}
~~=~ \mu \raisebox{0.9em}
{$\textstyle \frac{n-1}{2}
- \scriptstyle 1
+ \textstyle \frac{1}{2}$}
~~~=~~ \mu \raisebox{0.9em}
{$\textstyle \frac{n}{2}
- \scriptstyle 1 $}
지수함수(exp)끼리 정리한다.
\begin{align*}
&\exp \Big(\frac{\mu}{2}\Big)
~\times~
\exp \Big(-\dfrac{1}{2}\cdot
\dfrac{\mu t^2}{n-1}
\Big)
\\[20pt]
&= \exp \Big(
-\mu \Big\{~\dfrac{1}{2} +
\dfrac{t^2}{2(n-1)}~\Big\}
\Big)
\\[20pt]
&=\exp \Big(
-\mu~\Big\{
\dfrac{~n-1+t^2~}{2(n-1)}
\Big\}
\Big)
\\[20pt]
&=\exp \Big(
-\mu \div \Big\{
\dfrac{2(n-1)}{~n-1+t^2~}
\Big\}
\Big)
\\[20pt]
\end{align*}
f_T(t) = \dfrac{
\displaystyle
\int_{0}^{\infty}~
\mu \raisebox{0.9em}
{$\textstyle \frac{n}{2}
- \scriptstyle 1 $} ~~
\exp \Big(
-\mu \div \Big\{
\dfrac{2(n-1)}{~n-1+t^2~}
\Big\}
\Big)}
{\Gamma(\frac{n-1}{2})~~
2 \raisebox{0.7em}
{$ \textstyle \frac{n-1}{2}$}~
\sqrt{2\pi}~\sqrt{n-1}}
\\[30pt]
여기서 또 한번 정리에 들어간다.
\displaystyle
\int_{0}^{\infty}~
\mu \raisebox{0.9em}
{$\textstyle \frac{n}{2}
- \scriptstyle 1 $} ~~
\exp \Big(
-\mu \div \Big\{
\dfrac{2(n-1)}{~n-1+t^2~}
\Big\}
\Big)
\\[30pt]
=\displaystyle
\int_{0}^{\infty}~
\mu \raisebox{0.9em}
{$\textstyle \frac{n}{2}
- \scriptstyle 1 $} ~~
\exp \Bigg(
-\dfrac{\mu} {
\frac{2(n-1)}{~n-1+t^2~}
}\Bigg)
\\[30pt]
=\Gamma\Big(\frac{n}{2}\Big)~
\bigg\{\frac{2(n-1)}{~n-1+t^2~}\bigg\}
^{n/2}
f_T(t) = \frac{\Gamma(\frac{n}{2})}{
\Gamma(\frac{n-1}{2})
\Gamma(\frac{1}{2})}
\cdot (n-1)~^{(n-1)/2}
\cdot (n-1+t^2)~^{-\textstyle\frac{n}{2}}
n-1 = d로 치환하면 다음과 같다.
f_T(t) = \frac{\Gamma(\frac{d+1}{2})}{
\Gamma(\frac{d}{2})
\Gamma(\frac{1}{2})}
\cdot d~^{d/2}
\cdot (d+t^2)~^{-\textstyle\frac{(d+1)}{2}}
1차 적률 = 평균
E(T) = E\bigg( \dfrac
{\overline{X} -\mu}{s/\sqrt{n}} \bigg)
= E \bigg( \dfrac
{\overline{X} -\mu}{\sigma/\sqrt{n}}
\cdot \dfrac
{\sigma}{s} \bigg)
\\[10pt]
= E \bigg( \dfrac
{\overline{X} -\mu}{\sigma/\sqrt{n}}\bigg)
\cdot E \bigg( \dfrac{\sigma}{s} \bigg)
= 0
평균이 0이기 때문에, 2차적률이 곧 분산이다.
2차 적률 = 분산
\begin{align*}
E(T^2) &= E~\Bigg\{
\bigg( \dfrac{\overline{X} -\mu}{s/\sqrt{n}} \bigg)^2 \Bigg\}
\cdot
E~\bigg( \dfrac{\sigma^2}{s^2} \bigg)
=
E~\bigg( \dfrac{\sigma^2}{s^2} \bigg)
\\[25pt]
&=E\Bigg[ \dfrac{(n-1)s^2}{\sigma^2} \cdot
\dfrac{1}{n-1} \Bigg]
= E\Bigg[ (n-1)\cdot
\dfrac{s^2}{\sigma^2(n-1)}
\Bigg]
\\[25pt]
&=(n-1)~E~
\bigg( \dfrac{1}{\mu} \bigg)
\\[25pt]
&=(n-1) \cdot
\dfrac{1}{
\Big(\frac{n-1}{2}-1\Big)\cdot2}
= \dfrac{n-1}{n-3} \kern{10pt}
\{T \sim E(n-1)\}
\end{align*}
MGF
M_V(t) = E(e^{tV}) = E \Bigg(
\exp \bigg[
\dfrac{t(\overline{X} - \mu)}{s/\sqrt{n}}
\bigg] \Bigg)
- 너무 계산이 많아 생략