Zmienna losowa \(\displaystyle{ \zeta}\) ma rozkład jednostajny na odcinku [a; b] o gęstości
\(\displaystyle{ f (x) = \frac{1}{b-a}, x \in [a; b]}\). Za pomocą metody momentów znaleźć estymatory
parametrów a i b.
estymatory parametrów
-
robin5hood
- Użytkownik
- Posty: 1676
- Rejestracja: 2 kwie 2007, o 14:43
- Płeć: Mężczyzna
- Lokalizacja: warszawa
- Podziękował: 178 razy
- Pomógł: 17 razy
-
bstq
- Użytkownik
- Posty: 319
- Rejestracja: 7 lut 2008, o 12:45
- Płeć: Mężczyzna
- Lokalizacja: Warszawa
- Pomógł: 67 razy
estymatory parametrów
\(\displaystyle{ \mathbb{E}\zeta=\frac{a+b}{2}}\)
\(\displaystyle{ Var\zeta=\frac{\left(b-a\right)^{2}}{12}=\mathbb{E}\zeta^{2}-\left(\mathbb{E}\zeta\right)^{2}\Rightarrow\mathbb{E}\zeta^{2}=\frac{\left(b-a\right)^{2}}{12}+\left(\mathbb{E}\zeta\right)^{2}}\)
Mamy układ równań:
\(\displaystyle{ \begin{cases}
\mathbb{E}\zeta=\frac{a+b}{2}\\
\mathbb{E}\zeta^{2}-\left(\mathbb{E}\zeta\right)^{2}=\frac{\left(b-a\right)^{2}}{12}\end{cases}\overset{\text{estymujemy}}{\rightsquigarrow}\begin{cases}
\widehat{\mathbb{E}\zeta}=\frac{\widehat{a}+\hat{b}}{2}\\
\widehat{\mathbb{E}\zeta^{2}}-\left(\widehat{\mathbb{E}\zeta}\right)^{2}=\frac{\left(\hat{b}-\hat{a}\right)^{2}}{12}\end{cases}\overset{\text{wstawiamy momenty empiryczne}}{=}\begin{cases}
M_{1}=\frac{\widehat{a}+\hat{b}}{2}\\
M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-\hat{a}\right)^{2}}{12}\end{cases}=\begin{cases}
\hat{a}=2M_{1}-\hat{b}\\
M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-2M_{1}+\hat{b}\right)^{2}}{12}\end{cases}}\)
\(\displaystyle{ M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-2M_{1}+\hat{b}\right)^{2}}{12}\Leftrightarrow M_{2}-M_{1}^{2}=\frac{\left(2\hat{b}-2M_{1}\right)^{2}}{12}\Leftrightarrow M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-M_{1}\right)^{2}}{3}\Leftrightarrow M_{2}-M_{1}^{2}=\frac{\left(\hat{b}\right)^{2}-2\cdot M_{1}\cdot\hat{b}+M_{1}^{2}}{3}\Leftrightarrow3M_{2}-3M_{1}^{2}=\left(\hat{b}\right)^{2}-2\cdot M_{1}\cdot\hat{b}+M_{1}^{2}}\)
\(\displaystyle{ \Leftrightarrow\left(\hat{b}\right)^{2}+\left(-2\cdot M_{1}\cdot\right)\hat{b}+\left(4M_{1}^{2}-3M_{2}\right)=0}\)
\(\displaystyle{ \Delta=4M_{1}^{2}-4\cdot\left(4M_{1}^{2}-3M_{2}\right)=4M_{1}^{2}-16M_{1}^{2}+12M_{2}=-12M_{1}^{2}+12M_{2}}\)
\(\displaystyle{ \hat{b}_{1}=\frac{2M_{1}+\sqrt{-12M_{1}^{2}+12M_{2}}}{2}\Rightarrow\hat{a}_{1}=2\cdot M_{1}-\frac{2M_{1}+\sqrt{-12M_{1}^{2}+12M_{2}}}{2}}\)
\(\displaystyle{ \hat{b}_{2}=\frac{2M_{1}-\sqrt{-12M_{1}^{2}+12M_{2}}}{2}\Rightarrow\hat{a}_{2}=2\cdot M_{1}-\frac{2M_{1}-\sqrt{-12M_{1}^{2}+12M_{2}}}{2}}\)
gdzie:
\(\displaystyle{ M_{1}=\frac{1}{n}\sum_{i=1}^{n}X_{i}}\)
\(\displaystyle{ M_{2}=\frac{1}{n}\sum_{i=1}^{n}X_{i}^{2}}\)-- 4 czerwca 2009, 23:33 --jesliby te estymatory nie byly okreslone, to mamy klopot - inaczej tego nie da sie wyznaczyc...
\(\displaystyle{ Var\zeta=\frac{\left(b-a\right)^{2}}{12}=\mathbb{E}\zeta^{2}-\left(\mathbb{E}\zeta\right)^{2}\Rightarrow\mathbb{E}\zeta^{2}=\frac{\left(b-a\right)^{2}}{12}+\left(\mathbb{E}\zeta\right)^{2}}\)
Mamy układ równań:
\(\displaystyle{ \begin{cases}
\mathbb{E}\zeta=\frac{a+b}{2}\\
\mathbb{E}\zeta^{2}-\left(\mathbb{E}\zeta\right)^{2}=\frac{\left(b-a\right)^{2}}{12}\end{cases}\overset{\text{estymujemy}}{\rightsquigarrow}\begin{cases}
\widehat{\mathbb{E}\zeta}=\frac{\widehat{a}+\hat{b}}{2}\\
\widehat{\mathbb{E}\zeta^{2}}-\left(\widehat{\mathbb{E}\zeta}\right)^{2}=\frac{\left(\hat{b}-\hat{a}\right)^{2}}{12}\end{cases}\overset{\text{wstawiamy momenty empiryczne}}{=}\begin{cases}
M_{1}=\frac{\widehat{a}+\hat{b}}{2}\\
M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-\hat{a}\right)^{2}}{12}\end{cases}=\begin{cases}
\hat{a}=2M_{1}-\hat{b}\\
M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-2M_{1}+\hat{b}\right)^{2}}{12}\end{cases}}\)
\(\displaystyle{ M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-2M_{1}+\hat{b}\right)^{2}}{12}\Leftrightarrow M_{2}-M_{1}^{2}=\frac{\left(2\hat{b}-2M_{1}\right)^{2}}{12}\Leftrightarrow M_{2}-M_{1}^{2}=\frac{\left(\hat{b}-M_{1}\right)^{2}}{3}\Leftrightarrow M_{2}-M_{1}^{2}=\frac{\left(\hat{b}\right)^{2}-2\cdot M_{1}\cdot\hat{b}+M_{1}^{2}}{3}\Leftrightarrow3M_{2}-3M_{1}^{2}=\left(\hat{b}\right)^{2}-2\cdot M_{1}\cdot\hat{b}+M_{1}^{2}}\)
\(\displaystyle{ \Leftrightarrow\left(\hat{b}\right)^{2}+\left(-2\cdot M_{1}\cdot\right)\hat{b}+\left(4M_{1}^{2}-3M_{2}\right)=0}\)
\(\displaystyle{ \Delta=4M_{1}^{2}-4\cdot\left(4M_{1}^{2}-3M_{2}\right)=4M_{1}^{2}-16M_{1}^{2}+12M_{2}=-12M_{1}^{2}+12M_{2}}\)
\(\displaystyle{ \hat{b}_{1}=\frac{2M_{1}+\sqrt{-12M_{1}^{2}+12M_{2}}}{2}\Rightarrow\hat{a}_{1}=2\cdot M_{1}-\frac{2M_{1}+\sqrt{-12M_{1}^{2}+12M_{2}}}{2}}\)
\(\displaystyle{ \hat{b}_{2}=\frac{2M_{1}-\sqrt{-12M_{1}^{2}+12M_{2}}}{2}\Rightarrow\hat{a}_{2}=2\cdot M_{1}-\frac{2M_{1}-\sqrt{-12M_{1}^{2}+12M_{2}}}{2}}\)
gdzie:
\(\displaystyle{ M_{1}=\frac{1}{n}\sum_{i=1}^{n}X_{i}}\)
\(\displaystyle{ M_{2}=\frac{1}{n}\sum_{i=1}^{n}X_{i}^{2}}\)-- 4 czerwca 2009, 23:33 --jesliby te estymatory nie byly okreslone, to mamy klopot - inaczej tego nie da sie wyznaczyc...