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Re: st: Graph for longitudinal data - calculation based on mortality approach


From   Nick Cox <[email protected]>
To   [email protected]
Subject   Re: st: Graph for longitudinal data - calculation based on mortality approach
Date   Tue, 22 Feb 2011 11:23:02 +0000

The problem is not the graphics, as once you have created these
results, the graph is standard. The problem is that you may need to
code these calculations yourself.

Some starter hints:

The Stata function for the cumulative sum is -sum()-

Cumulative products are often best calculated as -exp(sum(ln()))- so
long as all arguments are positive.

Nick

On Tue, Feb 22, 2011 at 11:14 AM, Timo Beck <[email protected]> wrote:

> I have a question regarding graphs for longitudinal data. I am working with a data set on payments on outstanding loans in long format containing one observation for each person and payment of that person, i.e.
>
> Person     Cum. Payment    Loan     Period
> 1                 10         50        3
> 1                 20         50        7
> 1                 30         50        8
> 2                100        500        2
> 2                200        500       17
>
> I am trying to create a graph with the cumulative average fraction of the loan paid back over time (cumulative average over all persons).
>
> The mortality approach I want to use was introduced for bond defaults in Altman, E., 1989. Measuring Corporate Bond Mortality and Performance. Journal of Finance, 44, 909-922, p. 912.) and the calculation works as follows:
>
> Marginal fraction paid at time t (MFT)= Sum(all payments)/sum(all loans outstanding) at time t
>
> Fraction unpaid at t (FUT) = 1 - MFT
>
> Cumulative Fraction paid = 1 - product(previous FUT)
>
>
> I am relatively new to Stata and not sure how to approach this problem. I tried several related ideas that I found in previous threats, e.g. transforming the data set in a person period form and go from there by using -xtgraph-. However, nothing worked so far. I use Stata 10 (Stata/SE 10.1 for Windows, Born 01 Oct 2009).
>
> I would be grateful for any hint on how to approach this. Thanks a lot in advance!

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