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Re: st: 2 dimensional graph for joint distribution
From
Nick Cox <[email protected]>
To
"[email protected]" <[email protected]>
Subject
Re: st: 2 dimensional graph for joint distribution
Date
Fri, 1 Nov 2013 15:33:15 +0000
Note that David's recommendation is essentially that recommended
earlier in the thread in terms of -tabplot- (SSC).
Nick
[email protected]
On 1 November 2013 12:01, David Hoaglin <[email protected]> wrote:
> Dear Lulu,
>
> I have not had contact with the literature of choice models, but I
> looked at the paper by Train. (Thank you for including the link.)
>
> I do not recall seeing an explanation of why odo and bill are independent draws.
>
> In your initial message you said that you had estimated those
> coefficients "from a nonparametric choice model using fixed mass point
> method following Kenneth Train's approach in his 2008 paper."
> According to the text of that paper, Figure 5 and Figure 6 show the
> joint distribution of two pairs of coefficients (out of the seven
> coefficients in his model). That is, the height of each bar is the
> sum of the values of share for the combinations of values of the two
> coefficients that fall in that particular bin. The pattern of heights
> of the bars in the two figures does not seem compatible with taking
> the product of independent marginal distributions. Indeed, I would
> not expect the distributions of a pair of coefficients to be
> independent.
>
> I don't know whether your data will produce a reasonably smooth
> surface, but Train's Figure 5 and Figure 6 are definitely based on
> (estimated) joint probabilities. In your last paragraph, your data
> will determine whether you have the joint probability for odo = 0.26
> and bill = 0.30 and also the joint probability for odo = 0.26 and bill
> = 0.56. You have displayed share as a column vector. For a figure
> like those in Train's paper, however, you should think of it as the
> height of the bar for the particular combination of values of odo and
> bill. The combinations of values of odo and bill in your data will
> determine the "grid" (which may be reduced to the bins of a
> histogram).
>
> As I recall, you have 1,000 combinations of odo and bill (with the
> share for each). It may be useful to look at the marginal
> distribution of odo and the marginal distribution of bill, to choose a
> reasonable set of bins for each (i.e., not leave the choice to a
> histogram command), and then produce (as panels of the same display) a
> separate histogram of bill for each category of odo. This approach is
> not elegant, but it avoids the shortcomings of 3D in Train's Figure 5
> and Figure 6.
>
> David Hoaglin
>
> On Fri, Nov 1, 2013 at 12:04 AM, Lulu Zeng <[email protected]> wrote:
>> Dear David, Alfonso and others,
>>
>> Thank you for your comments.
>>
>> My understanding of the fixed mass point choice model (using EM
>> algorithm) is that -- share here is the individual probability of odo
>> and bill (share is the same for odo and bill), not the joint
>> probability.
>>
>> odo and bill are independent draws (using the mdraws command),
>> therefore the joint probability is the product of their individual
>> probability.
>>
>> Please let me know if I have misunderstood the model. But if share is
>> the joint probability, then it would be just a colum vector, not a
>> grid (e.g., I can only have the joint probability for odo of 0.26 and
>> bill of 0.30, can't have the joint probability for odo of 0.26 and
>> bill of 0.56?) how can I produce a surface graph version of Train's
>> graph on page 65 of http://elsa.berkeley.edu/~train/EMtrain.pdf?
>>
>> It would be really appreciated if I could have your advice on this.
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