4 Mar
2008
4 Mar
'08
3:13 p.m.
I was unsure in b and c how to treat N observations with T trials.
Are we just calculating the probability of k successes in N*T trials,
or what? I did it this way, but it seemed that they would have just
told us N trials since we would have done essentially the same thing.
I hope I made my question clear. Basically, I am wanting to know if
this information changes the problem from the standard way of
thinking about a binomial distribution (a number of trials, k, p) to
something else (for example, where a successful observation is one
with all successful trials, etc.)
Thanks,
Keith
5 Mar
5 Mar
3:03 p.m.
I have the same question. Reply-all, please.
On Tue, Mar 4, 2008 at 3:13 PM, Keith Schnakenberg <
keith.schnakenberg at gmail.com> wrote:
I was unsure in b and c how to treat N observations
with T trials.
Are we just calculating the probability of k successes in N*T trials,
or what? I did it this way, but it seemed that they would have just
told us N trials since we would have done essentially the same thing.
I hope I made my question clear. Basically, I am wanting to know if
this information changes the problem from the standard way of
thinking about a binomial distribution (a number of trials, k, p) to
something else (for example, where a successful observation is one
with all successful trials, etc.)
Thanks,
Keith
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
4:28 p.m.
I think we are trying to find the most likely value of pi, given that you
have N observations of the binomial distribution with T trials. That there
are T trials in each observation means that the binomial distribution has T
trials. It's not formally the same as calculating the probability of k
successes in N*T trials because you only get to see the sum of successful
trials in each observation. Hope this provides some clarification.
--Laurence
On Wed, Mar 5, 2008 at 3:03 PM, Sean Li <seanli at fas.harvard.edu> wrote:
I have the same question. Reply-all, please.
On Tue, Mar 4, 2008 at 3:13 PM, Keith Schnakenberg <
keith.schnakenberg at gmail.com> wrote:
I was unsure in b and c how to treat N
observations with T trials.
Are we just calculating the probability of k successes in N*T trials,
or what? I did it this way, but it seemed that they would have just
told us N trials since we would have done essentially the same thing.
I hope I made my question clear. Basically, I am wanting to know if
this information changes the problem from the standard way of
thinking about a binomial distribution (a number of trials, k, p) to
something else (for example, where a successful observation is one
with all successful trials, etc.)
Thanks,
Keith
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
4:36 p.m.
The best way to think about the full probability distribution is to first
think about the probability of a single observation.
One of the N observations, say y_j, is distributed according to a binomial
distribution. That y_j records the total number of successes out of T
trials. If our example were a flip of ten coins (where each flip lands
heads or tails independently of the others), then T would be 10 and our y_j
records the total number of successes (say, coins that land heads) out of
10. If we have 2 observations, then we have 2 instances of ten coins
flipped, and both observations tell the number that landed heads out of 10.
To find the probability of observing both observations, you would take the
product of both probability distributions.
This isn't exactly the same as modeling the total number of successes (the
sum of the ys) out of the total number of observations*trials (NT).
To convince yourself of the difference, you can consult pbinom(), which
returns the probability of observing a particular number of success out of a
trial of a particular size and the probability of success in each of the
trials. If it were true that the probability of observing a 3 out of 10
twice were the same as observing a 6 out of 20, then pbinom(q = 3, size =
10, prob= .2) * pbinom(q = 3, size = 10, prob = .2) should equal pbinom(q =
6, size = 20, prob = .2), which isn't the case.
Hope this helps,
Jenn
On Wed, Mar 5, 2008 at 4:28 PM, Laurence Tai <ltai at post.harvard.edu> wrote:
I think we are trying to find the most likely value of
pi, given that you
have N observations of the binomial distribution with T trials. That there
are T trials in each observation means that the binomial distribution has T
trials. It's not formally the same as calculating the probability of k
successes in N*T trials because you only get to see the sum of successful
trials in each observation. Hope this provides some clarification.
--Laurence
On Wed, Mar 5, 2008 at 3:03 PM, Sean Li <seanli at fas.harvard.edu> wrote:
I have the same question. Reply-all, please.
On Tue, Mar 4, 2008 at 3:13 PM, Keith Schnakenberg <
keith.schnakenberg at gmail.com> wrote:
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
I was unsure in b and c how to treat N
observations with T trials.
Are we just calculating the probability of k successes in N*T trials,
or what? I did it this way, but it seemed that they would have just
told us N trials since we would have done essentially the same thing.
I hope I made my question clear. Basically, I am wanting to know if
this information changes the problem from the standard way of
thinking about a binomial distribution (a number of trials, k, p) to
something else (for example, where a successful observation is one
with all successful trials, etc.)
Thanks,
Keith
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
5 p.m.
and of course all you careful readers caught that I referred to pbinom() as
returning the probability of a particular number of successes, when really
it's the probability of a particular number of successes or fewer.
You could perform the same test I suggested in the last email less
confusingly using the density value, dbinom(). Since the binomial
distribution is discrete, the density value is a probability.
Cheers,
Jenn
On Wed, Mar 5, 2008 at 4:36 PM, Jenn Larson <jmlarson at fas.harvard.edu>
wrote:
The best way to think about the full probability
distribution is to first
think about the probability of a single observation.
One of the N observations, say y_j, is distributed according to a binomial
distribution. That y_j records the total number of successes out of T
trials. If our example were a flip of ten coins (where each flip lands
heads or tails independently of the others), then T would be 10 and our y_j
records the total number of successes (say, coins that land heads) out of
10. If we have 2 observations, then we have 2 instances of ten coins
flipped, and both observations tell the number that landed heads out of 10.
To find the probability of observing both observations, you would take the
product of both probability distributions.
This isn't exactly the same as modeling the total number of successes (the
sum of the ys) out of the total number of observations*trials (NT).
To convince yourself of the difference, you can consult pbinom(), which
returns the probability of observing a particular number of success out of a
trial of a particular size and the probability of success in each of the
trials. If it were true that the probability of observing a 3 out of 10
twice were the same as observing a 6 out of 20, then pbinom(q = 3, size =
10, prob= .2) * pbinom(q = 3, size = 10, prob = .2) should equal pbinom(q =
6, size = 20, prob = .2), which isn't the case.
Hope this helps,
Jenn
On Wed, Mar 5, 2008 at 4:28 PM, Laurence Tai <ltai at post.harvard.edu>
wrote:
I think we are trying to find the most likely
value of pi, given that
you have N observations of the binomial distribution with T trials. That
there are T trials in each observation means that the binomial distribution
has T trials. It's not formally the same as calculating the probability of k
successes in N*T trials because you only get to see the sum of successful
trials in each observation. Hope this provides some clarification.
--Laurence
On Wed, Mar 5, 2008 at 3:03 PM, Sean Li <seanli at fas.harvard.edu> wrote:
I have the same question. Reply-all, please.
On Tue, Mar 4, 2008 at 3:13 PM, Keith Schnakenberg <
keith.schnakenberg at gmail.com> wrote:
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
I was unsure in b and c how to treat N
observations with T trials.
Are we just calculating the probability of k successes in N*T
trials,
or what? I did it this way, but it seemed that they would have just
told us N trials since we would have done essentially the same
thing.
I hope I made my question clear. Basically, I am wanting to know if
this information changes the problem from the standard way of
thinking about a binomial distribution (a number of trials, k, p) to
something else (for example, where a successful observation is one
with all successful trials, etc.)
Thanks,
Keith
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
6 Mar
6 Mar
4:46 p.m.
New subject: [gov2001-l] code for this week (with apologies)
Hi Jenn,
I'm still not officially registered for the class (I definitely will by next
week), so unfortunately, I have to email my code to you again.
Here it is attached.
Thanks!
Shahrzad
Quoting Jenn Larson <jmlarson at fas.harvard.edu>:
and of course all you careful readers caught that I
referred to pbinom() as
returning the probability of a particular number of successes, when really
it's the probability of a particular number of successes or fewer.
You could perform the same test I suggested in the last email less
confusingly using the density value, dbinom(). Since the binomial
distribution is discrete, the density value is a probability.
Cheers,
Jenn
On Wed, Mar 5, 2008 at 4:36 PM, Jenn Larson <jmlarson at fas.harvard.edu>
wrote:
The best way to think about the full probability
distribution is to first
think about the probability of a single observation.
One of the N observations, say y_j, is distributed according to a binomial
distribution. That y_j records the total number of successes out of T
trials. If our example were a flip of ten coins (where each flip lands
heads or tails independently of the others), then T would be 10 and our y_j
records the total number of successes (say, coins that land heads) out of
10. If we have 2 observations, then we have 2 instances of ten coins
flipped, and both observations tell the number that landed heads out of 10.
To find the probability of observing both observations, you would take the
product of both probability distributions.
This isn't exactly the same as modeling the total number of successes (the
sum of the ys) out of the total number of observations*trials (NT).
To convince yourself of the difference, you can consult pbinom(), which
returns the probability of observing a particular number of success out of
a
trial of a particular size and the probability of
success in each of the
trials. If it were true that the probability of observing a 3 out of 10
twice were the same as observing a 6 out of 20, then pbinom(q = 3, size =
10, prob= .2) * pbinom(q = 3, size = 10, prob = .2) should equal pbinom(q =
6, size = 20, prob = .2), which isn't the case.
Hope this helps,
Jenn
On Wed, Mar 5, 2008 at 4:28 PM, Laurence Tai <ltai at post.harvard.edu>
wrote:
> I think we are trying to find the most likely value of pi, given that
> you have N observations of the binomial distribution with T trials. That
> there are T trials in each observation means that the binomial
distribution
> has T trials. It's not formally the
same as calculating the probability
of k
successes
in N*T trials because you only get to see the sum of successful
trials in each observation. Hope this provides some clarification.
--Laurence
On Wed, Mar 5, 2008 at 3:03 PM, Sean Li <seanli at fas.harvard.edu> wrote:
I have the same question. Reply-all, please.
On Tue, Mar 4, 2008 at 3:13 PM, Keith Schnakenberg <
keith.schnakenberg at gmail.com> wrote:
> I was unsure in b and c how to treat N observations with T trials.
> Are we just calculating the probability of k successes in N*T
> trials,
> or what? I did it this way, but it seemed that they would have just
> told us N trials since we would have done essentially the same
> thing.
>
> I hope I made my question clear. Basically, I am wanting to know if
> this information changes the problem from the standard way of
> thinking about a binomial distribution (a number of trials, k, p) to
> something else (for example, where a successful observation is one
> with all successful trials, etc.)
>
> Thanks,
> Keith
> _______________________________________________
> gov2001-l mailing list
> gov2001-l at lists.fas.harvard.edu
> http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
>
>
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
6:02 p.m.
New subject: [gov2001-l] i know, i know
It took six months, but now I've done it too.
Sorry for sending you all an email to Jenn...with my code unattached.
S
Quoting Shahrzad Sabet <ssabet at fas.harvard.edu>:
Hi Jenn,
I'm still not officially registered for the class (I definitely will by next
week), so unfortunately, I have to email my code to you again.
Here it is attached.
Thanks!
Shahrzad
Quoting Jenn Larson <jmlarson at fas.harvard.edu>:
successful
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
and of course all you careful readers caught that
I referred to pbinom() as
returning the probability of a particular number of successes, when really
it's the probability of a particular number of successes or fewer.
You could perform the same test I suggested in the last email less
confusingly using the density value, dbinom(). Since the binomial
distribution is discrete, the density value is a probability.
Cheers,
Jenn
On Wed, Mar 5, 2008 at 4:36 PM, Jenn Larson <jmlarson at fas.harvard.edu>
wrote:
> The best way to think about the full probability distribution is to first
> think about the probability of a single observation.
>
> One of the N observations, say y_j, is distributed according to a
binomial
> distribution. That y_j records the total
number of successes out of T
> trials. If our example were a flip of ten coins (where each flip lands
> heads or tails independently of the others), then T would be 10 and our
y_j
> records the total number of successes (say,
coins that land heads) out of
> 10. If we have 2 observations, then we have 2 instances of ten coins
> flipped, and both observations tell the number that landed heads out of
10.
> To find the probability of observing both
observations, you would take
the
> product of both probability distributions.
>
> This isn't exactly the same as modeling the total number of successes
(the
> sum of the ys) out of the total number of
observations*trials (NT).
>
> To convince yourself of the difference, you can consult pbinom(), which
> returns the probability of observing a particular number of success out
of
a
> trial of a particular size and the probability of success in each of the
> trials. If it were true that the probability of observing a 3 out of 10
> twice were the same as observing a 6 out of 20, then pbinom(q = 3, size =
> 10, prob= .2) * pbinom(q = 3, size = 10, prob = .2) should equal pbinom(q
=
> 6, size = 20, prob = .2), which isn't
the case.
>
> Hope this helps,
>
> Jenn
>
>
> On Wed, Mar 5, 2008 at 4:28 PM, Laurence Tai <ltai at post.harvard.edu>
> wrote:
>
> > I think we are trying to find the most likely value of pi, given that
> > you have N observations of the binomial distribution with T trials.
That
>
there are T trials in each observation means that the binomial
distribution
> has T trials. It's not formally the
same as calculating the probability
of k
> > successes in N*T trials because you only get to see the sum of > > trials in each observation. Hope this
provides some clarification.
> >
> > --Laurence
> >
> >
> > On Wed, Mar 5, 2008 at 3:03 PM, Sean Li <seanli at fas.harvard.edu>
wrote:
> >
> > > I have the same question. Reply-all, please.
> > >
> > >
> > > On Tue, Mar 4, 2008 at 3:13 PM, Keith Schnakenberg <
> > > keith.schnakenberg at gmail.com> wrote:
> > >
> > > > I was unsure in b and c how to treat N observations with T trials.
> > > > Are we just calculating the probability of k successes in N*T
> > > > trials,
> > > > or what? I did it this way, but it seemed that they would have just
> > > > told us N trials since we would have done essentially the same
> > > > thing.
> > > >
> > > > I hope I made my question clear. Basically, I am wanting to know if
> > > > this information changes the problem from the standard way of
> > > > thinking about a binomial distribution (a number of trials, k, p)
to
> > something else (for example, where a
successful observation is one
> > with all successful trials, etc.)
> >
> > Thanks,
> > Keith
> > _______________________________________________
> > gov2001-l mailing list
> > gov2001-l at lists.fas.harvard.edu
> > http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
> >
> >
>
> _______________________________________________
> gov2001-l mailing list
> gov2001-l at lists.fas.harvard.edu
> http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
>
>
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
5913
days inactive
5915
days old
gov2001@lists.gking.harvard.edu
6 comments
5 participants
participants (5)
-
jmlarson＠fas.harvard.edu -
keith.schnakenberg＠gmail.com -
ltai＠post.harvard.edu -
seanli＠fas.harvard.edu -
ssabet＠fas.harvard.edu