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1-Newbie

## Use of maximize function to iterate data

In the attached worksheet,I would like to find the parameters of the GEV that are generated by a booststrapping of the original data.

The Maximize function accomplishes this rather well for a data vector but as you can see I am stuggling to do this for an array.

I would appreciate any help in accomplishing this.

Dick.....

1 ACCEPTED SOLUTION

Accepted Solutions
1-Newbie
(To:DickCun)

Tthanks to both Werner and Richard. Your answers to my question really was a bug help to me and allows me to bootstrap or jackknife percentiles values of interest using several methods of establishing different pdf's.

Your quick and concise responses are also greatfully appreciated.

Dick....

6 REPLIES 6
24-Ruby V
(To:DickCun)

You are changing the value of ORIGIN in the midst of you sheet which is a very bad habit and confusing.

The solve block you have should not be closed with Maximize but with find (without LnL of course) and you get the same values as with the standalone Maximize (see below),

Its unclear to me what kind of *iteration* you have in mind.

You are calculating my, sigma and lambda for the first column of datab.

Are you trying to do so for all 1000 columns? Thats sure not an iteration.

19-Tanzanite
(To:Werner_E)

I think perhaps the requirement that the derivatives be equal to zero is redundant with the maximize? By that I mean Richard assumes that at the maximum the derivatives will be zero, and therefore a solve block with the derivatives set to zero is a way to find the maximum. However, niether is true. The data has a limited range, so the maximum may not be at a critical point, it may just be at the end of the data. Further, even if the function has a critical point for a given data column, it may not be a maximum. Certainly, there is at least one column where a solve block using the three derivatives and Find does not find a solution.

24-Ruby V
(To:RichardJ)
 Richard Jackson wrote:I think perhaps the requirement that the derivatives be equal to zero is redundant with the maximize? By that I mean Richard assumes that at the maximum the derivatives will be zero, and therefore a solve block with the derivatives set to zero is a way to find the maximum. However, niether is true. The data has a limited range, so the maximum may not be at a critical point, it may just be at the end of the data. Further, even if the function has a critical point for a given data column, it may not be a maximum. Certainly, there is at least one column where a solve block using the three derivatives and Find does not find a solution.

Certainly we see two different attempts to find the maximum in Richard's sheet, but the Solve block never worked anyway because of the typo in Given (3 instead of e).

The first column where we get different results is column 4 and of course Maximize() is the way to go.

19-Tanzanite
(To:DickCun)

See the attached

24-Ruby V
(To:RichardJ)

So here is my attempt.

I changed the range variable so only the first 5 columns are calculated (it took too long for me to wait for all 1000 to finish.

1-Newbie
(To:DickCun)

Tthanks to both Werner and Richard. Your answers to my question really was a bug help to me and allows me to bootstrap or jackknife percentiles values of interest using several methods of establishing different pdf's.

Your quick and concise responses are also greatfully appreciated.

Dick....

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