Versie van 28 nov 2008 21:46 bewerken Pjetter (overleg \| bijdragen) 10.553 bewerkingen →‎math ← Vorige wijziging		Versie van 28 nov 2008 22:00 bewerken ongedaan maken Pjetter (overleg \| bijdragen) 10.553 bewerkingen →‎math Volgende wijziging →
Regel 180: :<math>\operatorname{MSE}(\hat{\theta})=\operatorname{E}[(\hat{\theta}-\theta)^2]</math> Now we replace theta-hat with X-bar and theta with mu and you get: :<math>\operatorname{MSE}(\bar{X})=\operatorname{E}([(\bar{X}-\mu)^2)]</math> We will replace X-bar in this equation with the following: :<math>\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i = \frac{1}{n} (X_1+\cdots+X_n).</math> So you get: :<math>\operatorname{MSE}(\bar{X})=\operatorname{E}([\frac{1}{n} (X_1+\cdots+X_n)-\mu)^2)]</math> This you can write differently according to: :<math>\operatorname{MSE}(\bar{X})=\operatorname{E}([\frac{X_1+\cdots+X_n}{n}-\mu)^2)]</math> and :<math>\operatorname{MSE}(\bar{X})=\operatorname{E}([\frac{X_1+\cdots+X_n}{n}-\frac{n}{n}\mu)^2)]</math> which will lead to: :<math>\operatorname{MSE}(\bar{X})=\operatorname{E}([\frac{X_1+\cdots+X_n-{n}\mu)}{n}^2)]</math> We take the n in the denominator, ~~which~~out and this becomes squarred: :<math>\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\operatorname{E}([X_1+\cdots+X_n-{n}\mu)^2)]</math> which turns into: :<math>\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\operatorname{E}([(X_1-\mu) +\cdots+(X_n-\mu))^2)]</math> Now the expectation is redistributed in the equation: :<math>\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\operatorname{E}[(X_1-\mu)^2] +\cdots + \frac{1}{n^2}\operatorname{E}[(X_n-\mu)^2]</math> We know that the definition for sigma is :<math>\operatorname{E}[(X_1-\mu)^2] = \sigma_1^2</math> so you get: :<math>\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\sigma_1^2 +\cdots + \frac{1}{n^2}\sigma_n^2=\frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}</math> This is the variance of the unbiassed sample distribution of the mean ==Nav==

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