Question

I have made a mistake in my calculations below. Can you please find my error? I cannot verify b = log(â) with the answer l've derived for b below: The exercise asked me to: Write the log-likelihood for a as a function of b, by substituting in b=log(a) (a)= (yiy2)log(a)- a(xX2) So I did the following, + £e)- (y1 +y2)b- e°(x +X) Then, log(e)) (y,+y2) log (b)-log(e)(x + x) Which gave me (b) (y1+y2)log(b)- b(x +x) Next I had to differentiatel(b) to find the MLE of b (b) (b) (y1y2)log(b)- b(x X) db Equating to zero: +Y2 (x + X2) b So, Yı+Y2which is the MLE b= Yı+Y2 b (xx2) (x X) Now I am supposed to verify that b = log(â), but I can't get this part to work which means that I messed up somewhere above in my work. The exercise states that â- Yi+Y2 X1+X2

Answer 1

The first step of your answer is absolutely correct. i.e, ereb) = (9 2 + Y2) b - eb (x2+x2) The problem is in the next step, taking logarithm in both sides. You wrote, e Jogleb)) = (ya+y2) Jog(b) - log (eb)(x1+x2) which is wpong. It should be, e (Jog (eb)) = log[(42+)6 - eb(x3 +22)], -é.e you can not simpiły this log[(4+4226-eb (24q+x22] to (yytyy) log (b) – log(es) It is not possible! Therefore, if we go forward from the correct step? eclog(eb) = log[(42+) b-eb(x2+x2)] 7 l (b) = log[(97+92)b-eb (x1+x2)]

d e(b) = Therefore, if we go forward from the correct step, eclog(eb) = log[(91+y) b- eb(x4+x2)] 7 l (b) = log[(94+ye) b-eb (x2+x2)] Now differentiating with respect to 'b', we get, - [Cyntyz7b-eb(x1 + x2)] (44+ y2)b-eb(Qq+x2) ..[(97+42) - eb(2g+x2)] (Yq+y)b-eb(221 +22) (41+Y2) - eb(Qcg+x2) (9, +42) b-eb (x1+x2) Now, equating with 'O' we get, (91 +92) - eb (x1+22) . (91+Y2) b-eb (x2 + x) (Yg + Y2) - eb (Ag +8₂)=0 -eb (x2+x2) = – (Y2+72) O

(Ys +Y2 x1 + x2 logarithm on both → log (et) = log () [Taking logarithm on both b = log (*) 8 = log (2 tk) = log (a [:a = 11 +7 2 1 21 +22 Y+Y2 21+22