Alpha Geek
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9 Suppose that log(y) follows a linear model with a linear form of heteroskedasticity. We write this as
log(y) 5 b0 1 xb 1 u u|x ~ Normal(0,h(x)),
so that, conditional on x, u has a normal distribution with mean (and median) zero but with variance h(x) that depends on x. Because Med(u|x) 5 0, equation (9.48) holds: Med(y|x) 5 exp(b0 1 xb). Further, using an extension of the result from Chapter 6, it can be shown that
E(y|x) 5 exp[b0 1 xb 1 h(x)/2].
(i) Given that h(x) can be any positive function, is it possible to conclude ∂E(y|x)/∂xj is the same sign as bj?
(ii) Suppose h(x) 5 0 1 x (and ignore the problem that linear functions are not necessarily always positive). Show that a particular variable, say x1, can have a negative effect on Med(y|x) but a positive effect on E(y|x).
(iii) Consider the case covered in Section 6.4, where h(x) 5 s2. How would you predict y using an estimate of E(y|x)? How would you predict y using an estimate of Med(y|x)? Which prediction is always larger?