{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe sandwich $F_n(x)$ between values of $F$ near $x$.\n\n1. Let $F, F_1, F_2, \\dots$ be the [[cumulative distribution functions|0MXMS3AE13]] of $X, X_1, X_2, \\dots$. Fix a point $x \\in \\R$ at which $F$ is continuous, and fix $\\varepsilon \u003e 0$.\n2. Claim: $\\set{X_n \\le x} \\subseteq \\set{X \\le x + \\varepsilon} \\cup \\set{\\abs{X_n - X} \u003e \\varepsilon}$. If $X_n \\le x$ and $X \u003e x + \\varepsilon$, then $X - X_n \u003e \\varepsilon$.\n3. By step 2 and [[Boole's inequality|0EGSEE9J9R]], $F_n(x) \\le F(x + \\varepsilon) + P(\\abs{X_n - X} \u003e \\varepsilon)$.\n4. Claim: $\\set{X \\le x - \\varepsilon} \\subseteq \\set{X_n \\le x} \\cup \\set{\\abs{X_n - X} \u003e \\varepsilon}$. If $X \\le x - \\varepsilon$ and $X_n \u003e x$, then $X_n - X \u003e \\varepsilon$.\n5. By step 4 and Boole's inequality, $F(x - \\varepsilon) \\le F_n(x) + P(\\abs{X_n - X} \u003e \\varepsilon)$.\n6. By [[convergence in probability|0EFDDXTVD6]], $P(\\abs{X_n - X} \u003e \\varepsilon) \\to 0$ as $n \\to \\infty$.\n7. Steps 3, 5 and 6 give $F(x - \\varepsilon) \\le \\liminf_{n} F_n(x) \\le \\limsup_{n} F_n(x) \\le F(x + \\varepsilon)$.\n8. Step 7 holds for every $\\varepsilon \u003e 0$. Since $F$ is continuous at $x$, $F(x - \\varepsilon) \\to F(x)$ and $F(x + \\varepsilon) \\to F(x)$ as $\\varepsilon \\to 0^+$.\n9. So $\\liminf_n F_n(x) = \\limsup_n F_n(x) = F(x)$, that is, $F_n(x) \\to F(x)$.\n10. This holds at every continuity point $x$ of $F$. So $X_n \\to X$ [[in distribution|0JDAQHWA9N]].\n\n## Idea\n\nIf $X_n$ is within $\\varepsilon$ of $X$ with high probability, then the event $X_n \\le x$ is squeezed between $X \\le x - \\varepsilon$ and $X \\le x + \\varepsilon$, up to a small error. At a continuity point of $F$, both bounds come together as $\\varepsilon$ shrinks.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "60F05"
    ],
    "name": "Proof that convergence in probability implies convergence in distribution",
    "proves": "0EPTDZPJ83",
    "sources": [
      {
        "locator": "Convergence in probability implies convergence in distribution",
        "url": "https://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables"
      }
    ],
    "tag": "QC1D2SRDFX",
    "uses": [
      "0EFDDXTVD6",
      "0JDAQHWA9N",
      "0MXMS3AE13",
      "0EGSEE9J9R"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/QC1D2SRDFX/",
  "links": [
    {
      "field": "proves",
      "target": "0EPTDZPJ83"
    },
    {
      "field": "uses",
      "target": "0EFDDXTVD6"
    },
    {
      "field": "uses",
      "target": "0JDAQHWA9N"
    },
    {
      "field": "uses",
      "target": "0MXMS3AE13"
    },
    {
      "field": "uses",
      "target": "0EGSEE9J9R"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/QC1D2SRDFX.md",
  "msc": [
    "60F05"
  ],
  "name": "Proof that convergence in probability implies convergence in distribution",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/QC1/QC1D2SRDFX.md",
  "status": null,
  "tag": "QC1D2SRDFX",
  "uri": "m2://QC1D2SRDFX",
  "url": "https://math2.tamnd.com/wiki/proof-that-convergence-in-probability-implies-convergence-in-distribution/",
  "writing": {
    "tag": "QC1D2SRDFX",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}