{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Counterexample\n\nLet $P$ be the uniform [[probability measure|059TJN0NZ1]] on the unit sphere $S^2 \\subset \\R^3$.\nDescribe a point by its longitude $\\lambda \\in (-\\pi, \\pi]$ and its latitude $\\varphi \\in [-\\pi/2, \\pi/2]$, as in [[spherical coordinates|01QFQAQ2PA]].\nThe area element of $S^2$ is $\\cos \\varphi \\, d\\lambda \\, d\\varphi$, so under $P$ the pair $(\\lambda, \\varphi)$ has the [[density|0ZY9DX8FBF]] $\\frac{1}{4\\pi} \\cos \\varphi$.\nHence $\\lambda$ and $\\varphi$ are independent, $\\lambda$ is uniform on $(-\\pi, \\pi]$, and $\\varphi$ has density $\\frac{1}{2} \\cos \\varphi$.\n\nCondition on the event $\\{\\varphi = 0\\}$, the equator.\nThe [[conditional distribution|0ZN9EZTEM6]] of $\\lambda$ given $\\varphi = 0$ is the law of $\\lambda$, which is uniform.\nSo the point is uniform on the equator.\n\nNow let $\\Theta = \\lambda \\bmod \\pi \\in [0, \\pi)$ and condition on the event $\\{\\Theta = 0\\}$, the great circle through the two poles.\nSince $\\Theta$ is independent of $\\varphi$, the conditional distribution of $\\varphi$ given $\\Theta = 0$ still has density $\\frac{1}{2} \\cos \\varphi$.\nSo the point is not uniform on this great circle: it is rarely near the poles.\n\nA rotation of the sphere maps the equator onto the great circle through the poles and preserves $P$.\nYet the two conditional laws on these two great circles are not images of each other under the rotation.\n\n## Remarks\n\nBoth events have probability $0$.\nThe conditional law given a null event depends on the random variable whose level set it is, not on the event alone.\nA [[regular conditional probability|PHHT5W3BEJ]] given $\\varphi$, or given $\\Theta$, is defined only up to null sets of values, so it fixes no answer for a single null event.\n\n## Explanations\n\n### L1 human\n\nPick a point at random on the surface of the Earth, and suppose you are told that it lies on a given great circle.\nOne may expect the point to be uniform on that circle.\nIf the circle is seen as a line of constant latitude (the equator), the answer is uniform.\nIf the same kind of circle is seen as a line of constant longitude (a meridian circle), points near the poles are less likely, because meridians crowd together there.\nThe lesson is that one cannot condition on an event of probability zero by itself.\nOne must say which family of events it belongs to, and this is what a [[regular conditional probability|PHHT5W3BEJ]] does.\n",
  "commit": "547b6eddb8f2feb8d579f2d633d22a27b7e132fc",
  "front": {
    "kind": "counterexample",
    "msc": [
      "60A05",
      "60A10"
    ],
    "name": "Borel–Kolmogorov paradox",
    "same_as": {
      "wikidata": "Q3363312",
      "wikipedia": "Borel–Kolmogorov paradox"
    },
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Borel%E2%80%93Kolmogorov_paradox"
      }
    ],
    "tag": "9RMS0ZJ618",
    "uses": [
      "059TJN0NZ1",
      "01QFQAQ2PA",
      "0ZY9DX8FBF",
      "0ZN9EZTEM6",
      "PHHT5W3BEJ"
    ]
  },
  "kind": "counterexample",
  "link": "https://math2.tamnd.com/node/9RMS0ZJ618/",
  "links": [
    {
      "field": "uses",
      "target": "059TJN0NZ1"
    },
    {
      "field": "uses",
      "target": "01QFQAQ2PA"
    },
    {
      "field": "uses",
      "target": "0ZY9DX8FBF"
    },
    {
      "field": "uses",
      "target": "0ZN9EZTEM6"
    },
    {
      "field": "uses",
      "target": "PHHT5W3BEJ"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/9RMS0ZJ618.md",
  "msc": [
    "60A05",
    "60A10"
  ],
  "name": "Borel–Kolmogorov paradox",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/9RM/9RMS0ZJ618.md",
  "status": null,
  "tag": "9RMS0ZJ618",
  "uri": "m2://9RMS0ZJ618",
  "url": "https://math2.tamnd.com/wiki/borel-kolmogorov-paradox/",
  "writing": {
    "tag": "9RMS0ZJ618",
    "kind": "counterexample",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      },
      {
        "what": "edge refutes or not_instance_of out",
        "have": 0,
        "want": 1
      }
    ]
  }
}