Gebruiker:Pjetter/klad: verschil tussen versies
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Regel 153:
==math==
:<math>\operatorname{MSE}(\hat{\theta})=\operatorname{E}((\hat{\theta}-\theta)^2)</math>
Normal multiplication turns this into:
:<math>\operatorname{MSE}(\hat{\theta})=\operatorname{E}(\hat{\theta}^2 - 2\hat{\theta}{\theta} +\theta^2)</math>
The expectation of theta is exactly theta, so you can write it like that:
:<math>\operatorname{MSE}(\hat{\theta})=\operatorname{E}(\hat{\theta}^2) - 2{\theta}\operatorname{E}(\hat{\theta}) +\theta^2</math>
Add two parameters: one with a negative sign and the other with a positive sign (the sum of both is 0, but you can use that in additional calculations)
:<math>\operatorname{MSE}(\hat{\theta})=\operatorname{E}(\hat{\theta}^2) - (\operatorname{E}(\hat{\theta}))^2 + (\operatorname{E}(\hat{\theta}))^2 - 2{\theta}\operatorname{E}(\hat{\theta}) +\theta^2</math>
Now the second part is the usual (a-b)^2, which can be written as a^2 - 2ab + b^2. You just work the other way around.
:<math>\operatorname{MSE}(\hat{\theta})=\operatorname{E}(\hat{\theta}^2) - (\operatorname{E}(\hat{\theta}))^2 + (\operatorname{E}(\hat{\theta}) -\theta)^2</math>
And the first part is the variance of theta-hat and the second part is the bias. This can be written as:
:<math>\operatorname{MSE}(\hat{\theta})=\operatorname{V}(\hat{\theta}) + (bias(\hat{\theta}))^2</math>
Now when theta-hat is unbiassed this means that the expectation of theta-hat is equal to theta and therefore the bias is zero.<br />
And that means that the MSE of theta-hat is exactly the variance of theta-hat.
If we take for theta the mu and for theta-hat the X-bar, you will use the normal variance for a n-sample from a population (X) with mean μ and variance σ2 you will get:
<math>\operatorname{MSE}=\left(\frac{\sigma}{\sqrt{n}}\right)^2</math>
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