Parameter estimation

**Anonymous1** · March 29th 2010, 06:39 PM

Suppose $X$ has density $f_X (x) = \frac{ba^b}{x^{b+1}}$ if $x>a>0,$ otherwise $f_X (x)=0$

$(a)$ Describe how to sample from this distribution using the method of inversion.
$(b)$ Find the MLE estimates for $a$ and $b$ from $n$ $i.i.d.$ samples.
$(c)$ Determine the asymptotic variance of the MLE estimates.

**CaptainBlack** · March 30th 2010, 05:07 AM

Originally Posted by Anonymous1

Suppose $X$ has density $f_X (x) = \frac{ba^b}{x^{b+1}}$ if $x>a>0,$ otherwise $f_X (x)=0$

$(a)$ Describe how to sample from this distribution using the method of inversion.

First find the cumulative distriution function $F_X(x)$ , then find the inverse function of this:

$<br /> G_X(F_X(x))=x<br />$

where the domain of $G_X$ is [0,1) and range is $[a,\infty)$

Now generate a $U(0,1)$ random number $r$ and $x=G(r)$ has the required distribution, and I am absolutly certain that this is in your notes and or text book.

CB

**Anonymous1** · March 30th 2010, 11:21 AM

My problem is mostly with $(b).$ I cannot simply set $l'(\theta)=0$ to maximize. So what do I do?

**matheagle** · March 30th 2010, 08:39 PM

Does that mean me?
The likelihood function is...

$L(x_1,\cdots, x_n)=b^na^{bn}\left(\prod_{i=1}^nx_i\right)^{-(b+1)}I(x_{(1)}>a)$

so if you want to maximize this wrt a, you want a as big as possible, since bn>0.
The largest a can be is the smallest order stat, i.e., the min.
As for the max wrt b it looks like you can take the log and differentiate.

**Anonymous1** · March 30th 2010, 10:04 PM

Originally Posted by matheagle

Does that mean me?
The likelihood function is...

$L(x_1,\cdots, x_n)=b^na^{bn}\left(\prod_{i=1}^nx_i\right)^{-(b+1)}I(x_{(1)}>a)$

so if you want to maximize this wrt a, you want a as big as possible, since bn>0.
The largest a can be is the smallest order stat, i.e., the min.
As for the max wrt b it looks like you can take the log and differentiate.

Thanks!

**Anonymous1** · March 31st 2010, 07:35 PM

Sorry if this is a stupid question, but what exactly is this about?

$I(x_{(1)}>a)$

**matheagle** · March 31st 2010, 08:40 PM

Originally Posted by Anonymous1

Sorry if this is a stupid question, but what exactly is this about?

$I(x_{(1)}>a)$

It's an indicator function, some call it a characteristic function.
But I use that terminology for the Fourier transform.
And I combine what some people use in a subscript with the argument.

$I(x_{(1)}>a)=1$ if $x_{(1)}>a$

while $I(x_{(1)}>a)=0$ if $x_{(1)}\le a$

Most people write $I(x_{(1)}>a)$ as $I_{(a,\infty)}(x_{(1)})$
I combine the subscript and the argument.

**matheagle** · March 31st 2010, 08:53 PM

If $X\sim U(a,b)$ then instead of saying that f(x) is 0 otherwise, use...

$f_X(x)={1\over b-a}I(a<x<b)$

and this notation makes it easier to see how the max and min are suff stats in many problems.

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