{
  "id": 1313,
  "title": "The Expected Value of the Capped Binomial Deviance",
  "url": "/competitions/WhatDoYouKnow/discussion/1313",
  "author_name": "Mike L.",
  "post_date": "2012-01-29T03:23:05.563000",
  "votes": 0,
  "comment_count": 0,
  "views": 1499,
  "content": "<p>The Expected Value of the Capped Binomial Deviance (CBD) when the data fit the model</p>\r\n<p><em>Does this make sense to you?</em></p>\r\n<p>Let x = CBD = -mean(log10-likelihood)<br>\r\nx / loge(10) = -mean(loge-likelihood)<br>\r\n2*x * loge(10) = mean(log-likelihood chi-square) = chi-square / degrees of freedom = mean-square statistic</p>\r\n<p>The expected value of a mean-square statistic when the data fit the model with large degrees of freedom = 1.0.\r\n</p>\r\n<p>So expected value of a CBD when the data fit the model = 1.0 / (2*loge(10)) = 0.2171</p>\r\n<p>When the CBD is less than 0.2171, then the model could be diagnosed to be over-parameterized.</p>\r\n<p>Current leaderboard value = 0.24706 corresponding to mean-square = 0.24706/0.2171 = 1.1380, i.e., there is 14% excess variance in the data from the perspective of the model. This model could be diagnosed to be under-parameterized for these data.</p>",
  "messages": [
    {
      "id": 8328,
      "postDate": "2012-01-29T03:23:05.563Z",
      "content": "<p>The Expected Value of the Capped Binomial Deviance (CBD) when the data fit the model</p>\r\n<p><em>Does this make sense to you?</em></p>\r\n<p>Let x = CBD = -mean(log10-likelihood)<br>\r\nx / loge(10) = -mean(loge-likelihood)<br>\r\n2*x * loge(10) = mean(log-likelihood chi-square) = chi-square / degrees of freedom = mean-square statistic</p>\r\n<p>The expected value of a mean-square statistic when the data fit the model with large degrees of freedom = 1.0.\r\n</p>\r\n<p>So expected value of a CBD when the data fit the model = 1.0 / (2*loge(10)) = 0.2171</p>\r\n<p>When the CBD is less than 0.2171, then the model could be diagnosed to be over-parameterized.</p>\r\n<p>Current leaderboard value = 0.24706 corresponding to mean-square = 0.24706/0.2171 = 1.1380, i.e., there is 14% excess variance in the data from the perspective of the model. This model could be diagnosed to be under-parameterized for these data.</p>"
    }
  ],
  "comments": [],
  "raw_markdown_by_id": {
    "8328": ""
  }
}