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=Proof that <math>I(θ) = E[(s(θ;X))^2]</math>= | =Proof that <math>I(θ) = E[(s(θ;X))^2]</math>= | ||
| − | + | Using the definition of Variance, as proved below: | |
<div style="margin-left: 3em;"> | <div style="margin-left: 3em;"> | ||
<math> | <math> | ||
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\bar Var(Y) &= E[(Y-E(Y))^2]\\ | \bar Var(Y) &= E[(Y-E(Y))^2]\\ | ||
&= E[Y^2-2YE[Y]+(E[Y])^2]\\ | &= E[Y^2-2YE[Y]+(E[Y])^2]\\ | ||
| − | &= | + | &= E[Y^2]-2(E[Y])^2+(E[Y])^2\\ |
| + | &= E[Y^2] - (E[Y])^2 | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
</div> | </div> | ||
Revision as of 21:57, 6 December 2020
Proof that $ I(θ) = E[(s(θ;X))^2] $
Using the definition of Variance, as proved below:
$ \begin{align} \bar Var(Y) &= E[(Y-E(Y))^2]\\ &= E[Y^2-2YE[Y]+(E[Y])^2]\\ &= E[Y^2]-2(E[Y])^2+(E[Y])^2\\ &= E[Y^2] - (E[Y])^2 \end{align} $
