I have a scatter plot of states and I want to label each point as the
respective state. How do I do it?
--
Andrew Reeves
reeves(a)fas.harvard.edu
617.493.3485 tel.
301.639.8369 cell.
http://people.fas.harvard.edu/~reeves/
Hi all,
I want Latex to show multiplication of matrices, i.e. to put several matrices
one next to each other, and not one under the other. How do I do that?
Thanks,
Asif
on 2b, do we take the integral from 0 -> x?
If so, one seems to get:
y = 1-e^(-alpha*x^beta)
solving for x ultimately produces:
x = (ln(y-1)/alpha)^(1/beta)
But since y is [0,1], ln(y-1) is undefined. Is there some way to get out of this?
Thanks,
Phillip.
-------------------------------------------------
Phillip Y. Lipscy
Perkins Hall Room #129
35 Oxford Street
Cambridge, MA 02138
(617)493-4893 DORM
(617)851-8220 CELL
lipscy(a)fas.harvard.edu
http://www.people.fas.harvard.edu/~lipscy/
First Year Student, Ph.D. Program
Harvard University, FAS, Department of Government
-------------------------------------------------
yeah, that's a typo. A good find!
Kosuke
Quoting Traci Burch <tburch(a)fas.harvard.edu>:
> For 2a:
>
> In Gary's lecture notes on log-likelihood of poisson, he lists the equation
>
>
> ln L(lambda |y) = the sum of{y* lambda - lambda - ln(y!)}
>
>
>
> but how do you get that from taking ln of both sides? We get
>
>
>
> the sum of -lambda + y ln( lambda) - ln (y!)
>
>
>
> why is our center term (y ln (lambda) ) different from what gary gets, which
> is y*lambda?
>
>
>
>
>
> Traci and Vanessa
>
>
----- End forwarded message -----
This is a multi-part message in MIME format.
------=_NextPart_000_00AF_01C2E1D5.DB1DEB90
Content-Type: text/plain;
charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable
For 2a:
In Gary's lecture notes on log-likelihood of poisson, he lists the =
equation=20
ln L(lambda |y) =3D the sum of{y* lambda - lambda - ln(y!)}
but how do you get that from taking ln of both sides? We get
the sum of -lambda + y ln( lambda) - ln (y!)
why is our center term (y ln (lambda) ) different from what gary gets, =
which is y*lambda? =20
Traci and Vanessa
------=_NextPart_000_00AF_01C2E1D5.DB1DEB90
Content-Type: text/html;
charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
<META content=3D"MSHTML 6.00.2719.2200" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>For 2a:</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>In Gary's lecture notes on =
log-likelihood of=20
poisson, he lists the equation</FONT> </DIV>
<DIV><FONT face=3DArial></FONT> </DIV>
<DIV><FONT face=3DCMR10>
<P align=3Dleft>ln </FONT><FONT face=3DEURM10>L</FONT><FONT=20
face=3DEUFM10>(lambda</FONT><FONT face=3DEURM10> </FONT><FONT=20
face=3DEUSM10>|</FONT><FONT face=3DEURM10>y</FONT><FONT face=3DEUFM10>) =
=3D the sum=20
of</FONT><FONT face=3DEUSM10>{</FONT><FONT face=3DEURM10>y</FONT><FONT =
face=3DEURM10>*=20
lambda</FONT><FONT face=3DEURM10> </FONT><FONT face=3DEUFM10>- =
lambda</FONT><FONT=20
face=3DEURM10> </FONT><FONT face=3DEUFM10>- </FONT><FONT =
face=3DCMR10>ln</FONT><FONT=20
face=3DEUFM10>(</FONT><FONT face=3DEURM10>y</FONT><FONT =
face=3DEUFM10>!)</FONT><FONT=20
face=3DEUSM10>}</FONT></P>
<P align=3Dleft><FONT face=3DEUSM10></FONT> </P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial size=3D2>but how =
do you get that=20
from taking ln of both sides? We get</FONT></FONT></P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial =
size=3D2></FONT></FONT> </P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial size=3D2>the sum =
of -lambda + y=20
ln( lambda) - ln (y!)</FONT></FONT></P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial =
size=3D2></FONT></FONT> </P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial size=3D2>why is =
our center term=20
(y ln (lambda) ) different from what gary gets, which is y*lambda? =
</FONT></FONT></P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial =
size=3D2></FONT></FONT> </P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial =
size=3D2></FONT></FONT> </P>
<P align=3Dleft><FONT face=3DEUSM10><FONT face=3DArial size=3D2>Traci =
and=20
Vanessa</FONT></P></FONT></DIV></BODY></HTML>
------=_NextPart_000_00AF_01C2E1D5.DB1DEB90--
Hi, Kosuke,
Another basic math question:
On 1(a), I'm taking a draw of length n from the normal distribution to get
Z, then another draw from the normal distribution to get W. In the
context of the first sentence (which states that X is a k x 1 vector),
does this mean that X is (Z-W, Z+W), such that each element of X is a 1xn
vector and that X is a kxn vector?
Thanks,
Olivia.
p.s. -- Do you know of a basic math couse I could audit in conjunction
with Gov2001?
Kosuke,
I'm not clear on 2(a). When you say "plot the log liklihood function
using the Poisson distribution", are you asking us to plot
log (pr(lambda | Y))?
Where pr(lambda | Y) = k(Y) * pr(Y | lambda)
and pr(Y | lambda) is distributed according to poisson(Y | lambda),
with unknown lambda, which makes:
pr(lambda-hat | Y) [proportional to] pois(Y | lambda-given)
where lambda-give is the mean of Y?
Am I interpreting the likelihood function correctly?
Thanks,
Olivia.
Kosuke,
Isn't the density function for Weibull distirubtion an exponential
function when *beta* = 1? The problem set says alpha, but that doesn't
seem to make sense to me with the equation you have.
Thanks,
Olivia.