| Line 17: | Line 17: | ||
<math>I(θ) = var(s(θ;X)) = E[(s(θ;X))^2] - (E[s(θ;X)])^2</math><br /> | <math>I(θ) = var(s(θ;X)) = E[(s(θ;X))^2] - (E[s(θ;X)])^2</math><br /> | ||
| − | As you can see we already have our E[(s(θ;X))^2] term which is part of our Fischer's Information definition. From here, we can use integrals to make <math>(E[s(θ;X)])^2</math> 0, therefore satisfying the equation<br /> | + | As you can see we already have our E[(s(θ;X))^2] term which is part of our Fischer's Information definition. From here, we can use integrals to make <math>(E[s(θ;X)])^2</math> equal 0, therefore satisfying the equation<br /> |
Recall that the score function is equal to the gradient with respect to θ of the natural log of the likelihood function with parameters θ and X. Also denoted like this: | Recall that the score function is equal to the gradient with respect to θ of the natural log of the likelihood function with parameters θ and X. Also denoted like this: | ||
Revision as of 21:47, 6 December 2020
Proof of Fischer's Information
Variance can also be shown as differences between the expected value of the square of event Y and the square of the expected value of Y. This theorem is proven 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} $
note: This definition is used often in statistics and therefore I will not be explaining the derivation of this identity. If you would like to know more, check our "More Sources" tab.
Using the identity of variance above, we can show that:
$ I(θ) = var(s(θ;X)) = E[(s(θ;X))^2] - (E[s(θ;X)])^2 $
As you can see we already have our E[(s(θ;X))^2] term which is part of our Fischer's Information definition. From here, we can use integrals to make $ (E[s(θ;X)])^2 $ equal 0, therefore satisfying the equation
Recall that the score function is equal to the gradient with respect to θ of the natural log of the likelihood function with parameters θ and X. Also denoted like this:
$ s(θ;X) = \nabla [ln(L(θ,X))] $
From here,
$ \begin{align} \bar E[(s(θ;X))] &= E[\nabla [ln(L(θ,X))]]\\ &=\int \nabla [ln(L(θ,X))] f(X;θ)dx\\ &=\int \frac{1}{f(X;θ)}\nabla [f(X;θ)]f(X;θ)dx\\ &=\int \nabla [f(X;θ)]dx\\ &=\nabla [\int[f(X;θ)]dx]\\ &=\nabla[1]\\ &=0 \end{align} $
note: $ \int[f(X;θ)]dx $ simplifies out to one because recall that the integral of a probability function is always 1.
Coming back to our intial identity:
$ I(θ) = E[(s(θ;X))^2] - (E[s(θ;X)])^2 $, if we plug in 0 into the second term and we get:
