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MathGuy
03-18-2002, 03:38 PM
[I am working from Weishaus's ASM notes, so this problem is in Lesson 7, page 45]:

7.6 Given claim sizes: 1500, 3500, 1800, 4800, 3900, 6000, 3800, 5500, 4200, 3000

The underlying distribution is assumed to be Gamma with α = 12. What is the variance of the MLE of θ?

Anyway, I used the formula Var = -(d<sup>2</sup>l/dθ<sup>2</sup>)<sup>-1</sup>, where l is the log-likelihood function. I don't get the right answer, which is θ<sup>2</sup>/120.

My guess is that the formula is wrong, but I got it from the same page (the method worked for problem 7.9).

L(x) = Π [(x/θ)<sup>12</sup>e<sup>-x/θ</sup>/(xΓ(12))]

l(x) = log (L(x)) = 12Σx - 120log (θ) - Σx/θ - Σ(log x) -10Γ(12)

l'(x) = -120/θ + Σxθ<sup>-2</sup>

l''(x) = 120θ<sup>-2</sup> -2Σxθ<sup>-3</sup>

[i]Corrected according to AW's note below.

<font size=-1>[ This Message was edited by: MathGuy on 2002-03-18 16:10 ]</font>

<font size=-1>[ This Message was edited by: MathGuy on 2002-03-19 08:26 ]</font>

Abraham Weishaus
03-18-2002, 05:26 PM
You got the sign wrong (120/theta^2-2sum x_i/theta^3), but otherwise you're fine. Now take the expected value of this (remember E[X]=12 theta), negate it, and invert it, and you've got the right answer!

Strictly speaking, as I indicate in the manual, this is only the asymptotic variance, not THE variance. You have to use my method to get the actual variance.

New at pd
03-19-2002, 07:54 AM
To build on what Dr. Weishaus said:

l'(x) = -120/theta + sum(x)/(theta^2)
Set to zero to get theta = sum(x)/120, the MLE.

The question asks for the variance of the MLE, so

var(theta) = var(sum(x)/120)
= var(sum(x)) / 120^2
= 10*var(x) / 120^2

We know that for a gamma dist,
var(x) = alpha*theta^2 = 12 * theta^2

so var(theta) = 10*12*theta^2/120^2 = theta^2/120

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<font size=-1>[ This Message was edited by: New at c6 on 2002-03-19 07:57 ]</font>

MathGuy
03-19-2002, 08:28 AM
I see. Thanks for the assistance.