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$ \theta $ is uniform [0,1]

E[ $ \theta^2 $ ] = $ \int\limits_{0}^{1} \theta^2 d\theta $ = 1/3

supp. $ \theta^{hat} $ = 1/3

MSE = E[ ($ \theta $ - 1/3)^2 ] = E[ $ \theta^2 $ - 2*(1/3)*$ \theta $ + (1/3)^2 ] = 1/3 - 2(1/3)(1/2) + (1/3)^2 = (1/3)^2 = 1/9

supp. instead $ \theta^{hat} $ = 1/2

MSE = E[ ($ \theta $ - 1/2)^2 ] = E[ $ \theta^2 $ - 2*(1/3)*$ \theta $ + (1/4) ] = 1/3 - 1/2 + 1/4 = 1/12


Back to ECE302 Fall 2008 Prof. Sanghavi

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva