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2 votes
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Expectations of estimators

Jensen's Inequality for random variables is involved here. If, say, $X$ is a random variable and a function $f$ is convex then $$\mathbb E[f(X)] \geq f\left(\mathbb E[X]\right).$$ The reverse ...
Alecos Papadopoulos's user avatar
2 votes

Expectations of estimators

Specifically, why is the expected value of a ratio not equal to the ratio of expected values? Suppose $$ (X,Y) = \begin{cases} (1,1) \\ (1,2) \\ (2,1) \\ (2,2) \end{cases} \text{all with equal ...
Michael Hardy's user avatar
0 votes

Proof of $\mathbb{E} (|X-Y|) = 0 \implies X^2 = Y^2$

If you are allowed to use the standard fact that a measurable function vanishes almost everywhere iff the Lebesgue integral of its absolute value vanishes, then you can write $0 = \mathbb E(|X-Y|) = \...
statmerkur's user avatar
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1 vote

Bayes Predictor for linear regression with square loss and expected value properties

For the squared error loss $l(\theta, a)=(\theta-a)^2$, the Bayes estimate of $\theta$ after $X=x$ is observed is given by the value of $a$ which minimises the expected squared error loss $\mathbb{E}[(...
geoant's user avatar
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2 votes

(More complete) proof the Fisher information is additive

\begin{align} \mathbb E\left[\frac{\partial}{\partial\theta}\log f(X_1) \frac{\partial}{\partial\theta}\log f(X_2)\right] &= \mathbb E\left[\frac{\partial}{\partial\theta}\log f(X_1)\right] \...
Xi'an's user avatar
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4 votes
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(More complete) proof the Fisher information is additive

One has to utilize the regularity conditions which ensure that the family is stable meaning the gradient and hessian of the likelihood function are uniformly bounded in a nbd of $\theta$ by integrable ...
User1865345's user avatar
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