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1

I would suggest looking at "On the Identification of Variances and Adaptive Kalman Filtering" by R. Mehra as a starting point. This paper is highly cited and the methods are not difficult to implement.


-2

$$\operatorname{Var}\left(\hat\theta\right)=\frac{3(1-θ)}{n}$$ $$\operatorname{Bias}\left(\hat\theta\right)=E(\hat\theta\right-theta)=0 $$\operatorname{Bias}^2\left(\hat\theta\right)=\frac{27(1-\theta)+4n\theta^2}{9n}$$


1

$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$ \begin{align} \E(\bar X) & = \E\left( \frac {X_1+\cdots+X_n} n \right) \\[8pt] & = \frac 1 n \E(X_1+\cdots+X_n) = \frac 1 n (\E(X_1)+\cdots+\E(X_n)) \\[8pt] & = \frac 1 n \Big( n\E(X_1) \Big) = \E(X_1). \\[25pt] \var( \bar X ) & = \var \left( \frac {X_1+\cdots+X_n} n ...


1

From what I understand, expectation is given as (assuming $X$ is continuous): $$ E[x] = \int_{-\infty}^{\infty}xf(x)dx$$ So your pdf should be something like this: $$F(x,\mu,\sigma)=\frac{1}{x\sigma\sqrt{2\pi}}e^{\left(-\frac{(\ln(x)-\mu)^2}{2\sigma^2}\right)}$$ You can then find expectation by using the above equation Can you solve this?



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