conversion mahalanobis distance (multivariate distributions) to percentage

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• Jan 5th 2011, 11:52 PM
aznimah
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
• Jan 6th 2011, 02:04 PM
pickslides
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.
• Jan 6th 2011, 03:57 PM
aznimah
Quote:

Originally Posted by pickslides
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
• Jan 6th 2011, 04:13 PM
pickslides
Quote:

Originally Posted by aznimah
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
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
• Jan 6th 2011, 04:33 PM
aznimah
Quote:

Originally Posted by pickslides
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
• Jan 6th 2011, 04:36 PM
aznimah
• Jan 6th 2011, 05:09 PM
pickslides
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?
• Jan 6th 2011, 07:16 PM
aznimah
Quote:

Originally Posted by pickslides
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
• Jan 6th 2011, 07:26 PM
pickslides
If you sent me your data, what would you want done?

check normality?
• Jan 6th 2011, 07:29 PM
aznimah
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
• Jan 6th 2011, 07:36 PM
aznimah
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
• Jan 6th 2011, 08:07 PM
pickslides
Thanks for posting that, how did you find these distances?

You should have

$\displaystyle \begin{array}{|c|c|c|c|}
\text{Group}_n & x_{in} & y_{in} & d_n^2 \\ \hline
1 & x_{i1} & y_{i1} & d_i^2 \\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ 2 & x_{i2} & y_{i2} & d_i^2\\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ \hline \end{array}$
• Jan 6th 2011, 08:12 PM
pickslides
Quote:

Originally Posted by aznimah
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.
• Jan 6th 2011, 08:21 PM
aznimah
Quote:

Originally Posted by pickslides
Thanks for posting that, how did you find these distances?

You should have

$\displaystyle \begin{array}{|c|c|c|c|}
\text{Group}& x_n & y_n & d_n^2 \\ \hline
1 & x_{i1} & y_{i1} & d_i^2 \\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ 2 & x_{i2} & y_{i2} & d_i^2\\ \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots \\ \hline \end{array}$

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
• Jan 6th 2011, 08:35 PM
aznimah
Quote:

Originally Posted by pickslides
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
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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