EE 278 Midterm: EE278 Midterm Exam Fall 2015 Solutions

• School: Stanford University
• Department: Electrical Engineering
• Course: EE 278
• Professor: All

Document Summary

E[x1, x2|x3] = cov(x1, x2|x3) + e[x1|x3] e[x2|x3], where the covariance term can be either positive (e. g. , by letting x3 = x1 x2), or negative (e. g. , by letting x3 = x1 + x2). By the law of conditional variances (and conditioning both sides on y ), it follows that. E[var(x|y )] = e[var(x|y, z)] + e[var(e(x|y, z)|y )]. This makes sense because with more obser- vations (y, z), the mse of the best estimate of the signal x should be less than or equal to that observing only y . But g(y ) is completely determined by y , thus e[var(x|y, g(y ))] = e[var(x|y )]. This result makes sense because in general y provides better information about the signal x than any function of it. First note that e(x 2|z) var(x|z) and e(y 2|z) var(y |z). E [var(x|z)var(y |z)] e(cid:2)(cov(x, y |z))2(cid:3) . But e [(cov(x, y |z))2] [e(cov(x, y |z))]2.