I saw the post that describe the evaluation metric function (https://www.kaggle.com/c/g-research-crypto-forecasting/discussion/291845) but I noticed that the mean is calculated in a strange way:
\nmean_a = np.sum(a * w) / sum_w
\n mean_b = np.sum(b * w) / sum_w
Is not the mean of x, for example, is : sum(x)/n , where n = len(x)
\nand then if values in x are weighted by another vector w, then it becomes:
\nsum(x*w) / sum_w / n
\nwhere sum_w = sum(w), and sum(w/sum_w) = 1
\nIf so, then the calculation of mean_a & mean_b should be like
\nmean_a = np.sum(a * w) / sum_w /n
\nmean_b = np.sum(b * w) / sum_w / n
where n = len(a) = len(b) = len(w)
\nWhy there is no n?
", "messages": [ { "id": "1668228", "postDate": "01/29/2022 13:02:40", "content": "I saw the post that describe the evaluation metric function (https://www.kaggle.com/c/g-research-crypto-forecasting/discussion/291845) but I noticed that the mean is calculated in a strange way:
\nmean_a = np.sum(a * w) / sum_w
\n mean_b = np.sum(b * w) / sum_w
Is not the mean of x, for example, is : sum(x)/n , where n = len(x)
\nand then if values in x are weighted by another vector w, then it becomes:
\nsum(x*w) / sum_w / n
\nwhere sum_w = sum(w), and sum(w/sum_w) = 1
\nIf so, then the calculation of mean_a & mean_b should be like
\nmean_a = np.sum(a * w) / sum_w /n
\nmean_b = np.sum(b * w) / sum_w / n
where n = len(a) = len(b) = len(w)
\nWhy there is no n?
", "rawMarkdown": "I saw the post that describe the evaluation metric function (https://www.kaggle.com/c/g-research-crypto-forecasting/discussion/291845) but I noticed that the mean is calculated in a strange way:\n\n\n mean_a = np.sum(a * w) / sum_w\n mean_b = np.sum(b * w) / sum_w\n\n\nIs not the mean of x, for example, is : sum(x)/n , where n = len(x)\n\nand then if values in x are weighted by another vector w, then it becomes:\n\nsum(x*w) / sum_w / n\n\nwhere sum_w = sum(w), and sum(w/sum_w) = 1\n\nIf so, then the calculation of mean_a & mean_b should be like\n\nmean_a = np.sum(a * w) / sum_w /**n**\nmean_b = np.sum(b * w) / sum_w / **n**\n\nwhere n = len(a) = len(b) = len(w)\n\nWhy there is no **n**?", "votes": null }, { "id": "1668303", "postDate": "01/29/2022 14:12:45", "content": "I think you might be muddling things a bit. Take the case where all of the weights are 1, i.e. that w is a vector with all (n) of its entries = 1. Then using the competition's definition of weighted mean:
\nnp.sum(a * w) = np.sum(a),
\nand
\nsum(w) = n,
\nso this reduces to the case:
\nmean_a = np.sum(a) / n
\nas desired.
If we were to use your proposed normalisation and also divide by n then that would mean mean_a = np.sum(a), which is inconsistent with the standard definition of the mean.
", "rawMarkdown": "I think you might be muddling things a bit. Take the case where all of the weights are 1, i.e. that w is a vector with all (n) of its entries = 1. Then using the competition's definition of weighted mean:\n\nnp.sum(a * w) = np.sum(a),\nand\nsum(w) = n,\nso this reduces to the case:\nmean_a = np.sum(a) / n\nas desired.\n\nIf we were to use your proposed normalisation and also divide by n then that would mean mean_a = np.sum(a), which is inconsistent with the standard definition of the mean.", "votes": null }, { "id": "1668413", "postDate": "01/29/2022 15:44:31", "content": "Oh yes you are right! Thanks for replying.
", "rawMarkdown": "Oh yes you are right! Thanks for replying.", "votes": null } ], "comments": [ { "id": 1668303, "author_name": "jagofc", "author_url": "", "post_date": "01/29/2022 14:12:45", "content": "I think you might be muddling things a bit. Take the case where all of the weights are 1, i.e. that w is a vector with all (n) of its entries = 1. Then using the competition's definition of weighted mean:
\nnp.sum(a * w) = np.sum(a),
\nand
\nsum(w) = n,
\nso this reduces to the case:
\nmean_a = np.sum(a) / n
\nas desired.
If we were to use your proposed normalisation and also divide by n then that would mean mean_a = np.sum(a), which is inconsistent with the standard definition of the mean.
", "votes": null, "replies": [ { "id": 1668413, "author_name": "mohammadabdelwahab", "author_url": "", "post_date": "01/29/2022 15:44:31", "content": "Oh yes you are right! Thanks for replying.
", "votes": null, "replies": [] } ] } ], "raw_markdown_by_id": { "1668228": "I saw the post that describe the evaluation metric function (https://www.kaggle.com/c/g-research-crypto-forecasting/discussion/291845) but I noticed that the mean is calculated in a strange way:\n\n\n mean_a = np.sum(a * w) / sum_w\n mean_b = np.sum(b * w) / sum_w\n\n\nIs not the mean of x, for example, is : sum(x)/n , where n = len(x)\n\nand then if values in x are weighted by another vector w, then it becomes:\n\nsum(x*w) / sum_w / n\n\nwhere sum_w = sum(w), and sum(w/sum_w) = 1\n\nIf so, then the calculation of mean_a & mean_b should be like\n\nmean_a = np.sum(a * w) / sum_w /**n**\nmean_b = np.sum(b * w) / sum_w / **n**\n\nwhere n = len(a) = len(b) = len(w)\n\nWhy there is no **n**?", "1668303": "I think you might be muddling things a bit. Take the case where all of the weights are 1, i.e. that w is a vector with all (n) of its entries = 1. Then using the competition's definition of weighted mean:\n\nnp.sum(a * w) = np.sum(a),\nand\nsum(w) = n,\nso this reduces to the case:\nmean_a = np.sum(a) / n\nas desired.\n\nIf we were to use your proposed normalisation and also divide by n then that would mean mean_a = np.sum(a), which is inconsistent with the standard definition of the mean.", "1668413": "Oh yes you are right! Thanks for replying." }, "source": "meta" }