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MLE of Bivariate Vector Random Variable: Proof & Explanation
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[QUOTE="Ray Vickson, post: 5492251, member: 330118"] In one case you observe ##(X,Y)=(x,y)## and estimate ##\theta## from ##f_{XY}(x,y)##, which requires finding the maximum of a function of the form ##a \theta e^{-x \theta}##, where ##a = x e^{-xy}## is a number. In the other case you observe ##X=x## and estimate ##\theta## from ##f_X(x)##, which requires finding the maximum of a function of the form ##a \theta e^{-x \theta}##, where ##a = 1##. The "scale factor" ##a## does not affect the optimization. Put it another way: the MLS estimator of ##\theta##, based on ##f_{XY}(x,y)## is independent of ##y##. Think of ##f_X(x)## as a sum over ##y## of curves ##f_{XY}(x,y)##, where the variable is ##\theta## and ##x, y## are just an input parameters. Each of these constituent curves has a maximum at the same point ##\theta = 1/x##, so the sum over ##y## is maximized at the same point. [/QUOTE]
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MLE of Bivariate Vector Random Variable: Proof & Explanation
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