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@rossmeier rossmeier requested a review from thielpa January 22, 2020 20:14
\textbf{Lösung:}\\
$\alpha = \dfrac{E[\X\Y]-E[\X]E[\Y]}{E[\X^2]-E[\X]^2} = \dfrac{c_{\X,\Y}}{\sigma^2_{\X}} = \dfrac{c_{\X,\Y}}{\sigma^2_{\X}}\dfrac{\sigma_{\Y}}{\sigma_{\Y}} = \rho_{\X,\Y}\dfrac{\sigma_{\Y}}{\sigma_{\X}}$\\
$\beta = E[\Y] - \alpha E[\X] = E[\Y] - \rho_{\X,\Y}\dfrac{\sigma_{\Y}}{\sigma_{\X}}E[\X]$\\
$\Rightarrow \hat{\Y} = \rho_{\X,\Y}\dfrac{\sigma_{\Y}}{\sigma_{\X}}(\X-E[\X])+E[\Y]$
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\begin{itemize}
\item $\X_n\xrightarrow{\text{a.s.}}\X \Rightarrow \X_n\xrightarrow{\text{p.}}\X$
\item $\X_n\xrightarrow{\text{m.s.}}\X \Rightarrow \X_n\xrightarrow{\text{p.}}\X$
\item $P(\{|\X_n|\leq\Y\}) = 1 \forall n \wedge E[Y^2]<\infty\wedge \X_n\xrightarrow{\text{p.}}\X \Rightarrow \X_n\xrightarrow{\text{m.s.}}\X$
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@dariusptrs dariusptrs requested a review from stevensalloum March 5, 2025 11:07
@stevensalloum stevensalloum merged commit ab3f1f7 into master Mar 7, 2025
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5 participants