{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\nLet $n \\ge 1$, let $(X_k)_{0 \\le k \\le n}$ be a [[submartingale|EQ055QMACW]] with respect to a [[filtration|HNK1Y62G7V]] $(\\mathcal{F}_k)_{0 \\le k \\le n}$, and let $a \u003c b$ be real numbers.\nLet $U_n[a, b]$ be the number of upcrossings of $[a, b]$ by $X_0, \\dots, X_n$.\nThis is the largest $m \\ge 0$ with times $0 \\le s_1 \u003c t_1 \u003c \\dots \u003c s_m \u003c t_m \\le n$ such that $X_{s_i} \\le a$ and $X_{t_i} \\ge b$ for each $i$.\nThen\n$$(b - a)\\, E\\bigl[U_n[a, b]\\bigr] \\le E\\bigl[(X_n - a)^+\\bigr] - E\\bigl[(X_0 - a)^+\\bigr],$$\nwhere $y^+ = \\max(y, 0)$.\n\n## Remarks\n\nThe proof uses a betting strategy: buy one unit when the process is at most $a$, and sell when it reaches at least $b$.\nWith $Y_k = (X_k - a)^+$, which is again a submartingale, each completed upcrossing earns at least $b - a$, so the gain is at least $(b - a) U_n[a, b]$.\nThe opposite strategy holds exactly when the first one does not.\nIt is predictable and bounded, so its expected gain on the submartingale $Y$ is non-negative.\nThe two gains add up to $Y_n - Y_0$.\nAs a consequence, if $\\sup_n E[X_n^+] \u003c \\infty$, then the total number of upcrossings of every interval $[a, b]$ is finite almost surely.\nThis is the key step of [[Doob's martingale convergence theorems|MRXBC4ZNRR]]: a sequence that crosses no rational interval infinitely often has a limit in $[-\\infty, \\infty]$.\n\n## Explanations\n\n### L1 human\n\nAn upcrossing is a trip of the process from below $a$ to above $b$.\nThe inequality says that a process with an upward drift cannot make many such trips, on average, unless it ends up high.\nIn other words, a submartingale with controlled size cannot oscillate forever.\n\n### L2 human\n\nThe inequality turns a question about convergence into a counting question.\nA real sequence fails to converge in $[-\\infty, \\infty]$ exactly when it crosses some interval with rational endpoints infinitely often.\nThe upcrossing bound and monotone convergence show this happens with probability zero for $L^1$-bounded submartingales.\nFor a [[supermartingale|KYW1PV0SX5]] the analogous bound is $(b - a)\\, E[U_n[a, b]] \\le E[(X_n - a)^-]$, where $y^- = \\max(-y, 0)$.\n",
  "commit": "efe85cee8dbc57d85bedbacfb97b99d1b7bee843",
  "front": {
    "about": [
      "6M852B4RA8"
    ],
    "kind": "statement",
    "msc": [
      "60G42"
    ],
    "name": "Doob's upcrossing inequality",
    "role": "theorem",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Doob%27s_martingale_convergence_theorems"
      }
    ],
    "tag": "10HFW1GRV2",
    "uses": [
      "EQ055QMACW",
      "HNK1Y62G7V",
      "0DCT0QZVQ1"
    ]
  },
  "kind": "statement",
  "link": "https://math2.tamnd.com/node/10HFW1GRV2/",
  "links": [
    {
      "field": "about",
      "target": "6M852B4RA8"
    },
    {
      "field": "uses",
      "target": "EQ055QMACW"
    },
    {
      "field": "uses",
      "target": "HNK1Y62G7V"
    },
    {
      "field": "uses",
      "target": "0DCT0QZVQ1"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/10HFW1GRV2.md",
  "msc": [
    "60G42"
  ],
  "name": "Doob's upcrossing inequality",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/10H/10HFW1GRV2.md",
  "status": null,
  "tag": "10HFW1GRV2",
  "uri": "m2://10HFW1GRV2",
  "url": "https://math2.tamnd.com/wiki/doobs-upcrossing-inequality/",
  "writing": {
    "tag": "10HFW1GRV2",
    "kind": "statement",
    "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
      }
    ]
  }
}