{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe argue by induction on the number of events in the intersection.\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]] and let $A_1, \\dots, A_n \\in \\mathcal{F}$ with $n \\ge 2$ and $P(A_1 \\cap \\dots \\cap A_{n-1}) \u003e 0$.\n2. For $1 \\le k \\le n$ let $C_k = A_1 \\cap \\dots \\cap A_k$. Each $C_k$ is in $\\mathcal{F}$, because $\\mathcal{F}$ is a [[σ-algebra|04WSW8T3SY]].\n3. Let $1 \\le k \\le n - 1$. Then $C_{n-1} \\subseteq C_k$. By [[monotonicity of probability|0464Z0JP2E]], $P(C_k) \\ge P(C_{n-1}) \u003e 0$.\n4. So for $1 \\le k \\le n - 1$ the [[conditional probability|0FJB06AD2P]] $P(A_{k+1} \\mid C_k) = P(A_{k+1} \\cap C_k) / P(C_k)$ is defined.\n5. Since $A_{k+1} \\cap C_k = C_{k+1}$, step 4 gives $P(C_{k+1}) = P(C_k) \\, P(A_{k+1} \\mid C_k)$ for $1 \\le k \\le n - 1$.\n6. We claim that $P(C_m) = P(A_1) \\prod_{k=1}^{m-1} P(A_{k+1} \\mid C_k)$ for $1 \\le m \\le n$. For $m = 1$ the product is empty and the claim reads $P(C_1) = P(A_1)$, which holds since $C_1 = A_1$.\n7. Suppose the claim holds for some $m$ with $1 \\le m \\le n - 1$. Step 5 with $k = m$ multiplies $P(C_m)$ by $P(A_{m+1} \\mid C_m)$, so the claim holds for $m + 1$.\n8. The case $m = n$ of the claim reads $P(A_1 \\cap \\dots \\cap A_n) = P(A_1) \\, P(A_2 \\mid A_1) \\cdots P(A_n \\mid A_1 \\cap \\dots \\cap A_{n-1})$.\n\n## Idea\n\nEach factor $P(A_{k+1} \\mid C_k)$ is the ratio $P(C_{k+1}) / P(C_k)$. Multiplying all these ratios and $P(A_1)$ gives a telescoping product that leaves $P(C_n)$. The hypothesis on the last intersection makes all the denominators positive.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "60A05"
    ],
    "name": "Proof of the multiplication rule",
    "proves": "0BS3RMX8HM",
    "sources": [
      {
        "locator": "Chain rule for events",
        "url": "https://en.wikipedia.org/wiki/Chain_rule_(probability)"
      }
    ],
    "tag": "SSKG51ZSEY",
    "uses": [
      "0ZYBZ34RT2",
      "0FJB06AD2P",
      "0464Z0JP2E",
      "04WSW8T3SY"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/SSKG51ZSEY/",
  "links": [
    {
      "field": "proves",
      "target": "0BS3RMX8HM"
    },
    {
      "field": "uses",
      "target": "0ZYBZ34RT2"
    },
    {
      "field": "uses",
      "target": "0FJB06AD2P"
    },
    {
      "field": "uses",
      "target": "0464Z0JP2E"
    },
    {
      "field": "uses",
      "target": "04WSW8T3SY"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/SSKG51ZSEY.md",
  "msc": [
    "60A05"
  ],
  "name": "Proof of the multiplication rule",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/SSK/SSKG51ZSEY.md",
  "status": null,
  "tag": "SSKG51ZSEY",
  "uri": "m2://SSKG51ZSEY",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-multiplication-rule/",
  "writing": {
    "tag": "SSKG51ZSEY",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}