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Math Help - conversion mahalanobis distance (multivariate distributions) to percentage

  1. #1
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    conversion mahalanobis distance (multivariate distributions) to percentage

    hi,

    i have data that would return vertex points of shape (coordinate x and y) example A x1,y1), (x2,y2), (x3,y3), (x4,y4),(x1,y1), (x2,y2) and group B: (x5,y5),(x6,y6). my data group would have two group, A group (39 coordinates) and group B (4 coordinates).

    i applied mahalanobis distance to calculate the dissimilarity computation and mahalanobis distance return standard deviation results from expected data.
    how im going to get the percentage of the mahalanobis distance and what method is better use for my data? now i already trying statistical normalization (z-score):
     z=(X-u)/standard deviation

    but my problem is, im getting confuse to place the input data for X, u and standard deviation based on my data. really appreciate for any ideas since im really urgent to know this. Anyway thank you for the great site

    thanks
    aznimah
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  2. #2
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    In multivariate Statistics you can use Mahalanobis distance to test for multivariate normality. From your post, this is not what you are trying to achieve. Is it?

    Given you are talking about 2 groups of data are you looking for a classification process? i.e. use Mahalanobis distance from the centre of each group to potentially re-classify each data point? You can form a classification rule, either linear or otherwise as a classification rule. To proceed you first need to test if Cov(Group A) = Cov(Group B). After this you can decide with rule fits best.
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    Quote Originally Posted by pickslides View Post
    i.e. use Mahalanobis distance from the centre of each group to potentially re-classify each data point? You can form a classification rule, either linear or otherwise as a classification rule. To proceed you first need to test if Cov(Group A) = Cov(Group B). After this you can decide with rule fits best.
    hi, thanks for the reply, can you give more explanation with example and what is the classification rule,

    im not clearer

    thanks
    Aznimah
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  4. #4
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    Quote Originally Posted by aznimah View Post
    i applied mahalanobis distance to calculate the dissimilarity computation and mahalanobis distance return standard deviation results from expected data.
    Are you saying that Mahalanobis distance has given you the standard deviation? As the sample covariance matrix is required to compute these distances you should be able to find the stanard deviations before this.

    Quote Originally Posted by aznimah View Post
    how im going to get the percentage of the mahalanobis distance and what method is better use for my data? now i already trying statistical normalization (z-score):
    Of the distances \displaystyle d_i^2 = (X_i-\bar{X})^TS^{-1}(X_i-\bar{X}) if approximately 50% are \displaystyle \leq \chi_{df,\alpha}^2= \chi_{2,0.5}^2 = 1.39 then assume multivariate normality
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    Quote Originally Posted by pickslides View Post
    Of the distances \displaystyle d_i^2 = (X_i-\bar{X})^TS^{-1}(X_i-\bar{X}) if approximately 50% are \displaystyle \leq \chi_{df,\alpha}^2= \chi_{2,0.5}^2 = 1.39 then assume multivariate normality
    thanks, the results 1.39 is refering to? im so sorry because im not in mathematical fields and it really quiet bit hard for me... is that 1.39 is the standard deviation and i need to refer to z-score table index or else? really can't get it.
    i took the mahalanobis distance from this site : Mahalanobis Distance
    and someone told me that mahalanobis distance is essentially give standard deviation from expected data and ask me to refer to z-score for percentage conversion.

    so sorry for the lot of questions but i really appreciate and need help on this urgently.

    thank you so much

    aznimah
    Last edited by aznimah; January 6th 2011 at 04:34 PM. Reason: link: http://people.revoledu.com/kardi/tutorial/Similarity/MahalanobisDistance.html
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  7. #7
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    1.39 is the chi-squared value with 2 degrees of freedom and signifcance of 0.5. i.e. \displaystyle  \chi_{2,0.5}^2 = 1.39

    Are these terms firmilar?
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  8. #8
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    Quote Originally Posted by pickslides View Post
    1.39 is the chi-squared value with 2 degrees of freedom and signifcance of 0.5. i.e. \displaystyle  \chi_{2,0.5}^2 = 1.39

    Are these terms firmilar?
    hi, thanks again, i already check those terms, i understand from the given example but having a problem to relate with my data. can i send my data to you since im totally need help on this. really urgent and really appreciate if you could help me. thanks

    aznimah
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  9. #9
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    If you sent me your data, what would you want done?

    check normality?
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  10. #10
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    here is my data:
    average coordinate (x,y)


    x1 y1 x2 y2 Mahalanobis Distance
    70.8 58.4 81.3 65.3 0.33
    70.8 58.4 74 69.3 0.37
    70.8 58.4 72 57.3 0.05
    70.8 58.4 0 0 0
    70.8 58.4 76.8 64 0.23
    70.8 58.4 82.8 60 0.28
    70.8 58.4 0 0 0
    70.8 58.4 79.8 68.5 0.4
    70.8 58.4 80.8 58 0.23
    70.8 58.4 68.3 64.8 0.22
    70.8 58.4 71.8 62 0.12
    70.8 58.4 72 57.3 0.05
    70.8 58.4 71 63.3 0.16
    70.8 58.4 82.5 56.5 0.27
    70.8 58.4 82.8 57.8 0.27
    70.8 58.4 76.5 56.3 0.15
    70.8 58.4 75 64 0.21
    70.8 58.4 67.8 56.3 0.1
    70.8 58.4 68.5 56.3 0.06
    70.8 58.4 71.8 62.3 0.13
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  11. #11
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    i need to get the accuray (percentage) of mahalanobis distance, im totally clueless, i already get the result of mahalanobis distance but can't manage to compute the accuray (%)

    thanks
    aznimah
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    Thanks for posting that, how did you find these distances?

    You should have

    \displaystyle \begin{array}{|c|c|c|c|}<br />
\text{Group}_n & x_{in} & y_{in} & d_n^2 \\ \hline<br />
1 & x_{i1} &  y_{i1} & d_i^2 \\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ 2 & x_{i2} &  y_{i2} & d_i^2\\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ \hline \end{array}
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  13. #13
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    Quote Originally Posted by aznimah View Post
    i need to get the accuray (percentage) of mahalanobis distance, im totally clueless, i already get the result of mahalanobis distance but can't manage to compute the accuray (%)

    thanks
    aznimah
    I'm not sure what you mean by the "accuracy of Mahalanobis distance". The distance itself is just a calculation, you either have it right or have it wrong.

    Given the distances you have calculated are correct then you find the percentage of these that are \leq 1.39 as discussed. If approximately 50% are \leq 1.39 then normality is a fair assumption.
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  14. #14
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    Quote Originally Posted by pickslides View Post
    Thanks for posting that, how did you find these distances?

    You should have

    \displaystyle \begin{array}{|c|c|c|c|}<br />
\text{Group}& x_n & y_n & d_n^2 \\ \hline<br />
1 & x_{i1} &  y_{i1} & d_i^2 \\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ 2 & x_{i2} &  y_{i2} & d_i^2\\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ \hline \end{array}
    hi, so sorry, i attach image of my data. i refering this to this link ://people.revoledu.com/kardi/tutorial/Similarity/MahalanobisDistance.html

    and i have coordinate data of vertex points. here is my data link image : FreeImageHosting.net Hosting Service

    thanks

    aznimah
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  15. #15
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    Quote Originally Posted by pickslides View Post
    I'm not sure what you mean by the "accuracy of Mahalanobis distance". The distance itself is just a calculation, you either have it right or have it wrong.

    Given the distances you have calculated are correct then you find the percentage of these that are \leq 1.39 as discussed. If approximately 50% are \leq 1.39 then normality is a fair assumption.
    can you help me clear my understanding, so sorry again for cause you a lot of questions since i need to finish it by today. so sorry:

    as you are mentioned earlier:

    Quote Originally Posted by pickslides View Post
    Given the distances you have calculated are correct then you find the percentage of these that are \leq 1.39 as discussed. If approximately 50% are \leq 1.39 then normality is a fair assumption.
    Steps required:
    1) get the distance value (mahalanobis distance)
    2) find the percentage by looking at chi-square?
    what should i do on step 2?
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