Gebruiker:Pjetter/klad: verschil tussen versies

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Pjetter (overleg | bijdragen)
Pjetter (overleg | bijdragen)
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>
Regel 174:
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>
 
Another calculation will be as follows:
We start again with:
:<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 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>
 
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