added proof "ind-self"

Author: JoramSoch

P/ind-self.md

@@ -0,0 +1,58 @@
+---
+layout: proof
+mathjax: true
+
+author: "Joram Soch"
+affiliation: "BCCN Berlin"
+e_mail: "joram.soch@bccn-berlin.de"
+date: 2024-09-20 14:04:20
+
+title: "Self-independence of random event"
+chapter: "General Theorems"
+section: "Probability theory"
+topic: "Probability"
+theorem: "Self-independence"
+
+sources:
+  - authors: "Wikipedia"
+    year: 2024
+    title: "Independence (probability theory)"
+    in: "Wikipedia, the free encyclopedia"
+    pages: "retrieved on 2024-09-20"
+    url: "https://en.wikipedia.org/wiki/Independence_(probability_theory)#Self-independence"
+  - authors: "JoramSoch"
+    year: 2023
+    title: "Suppose A is an event. Can A be independent of itself?"
+    in: "X"
+    pages: "Aug 7, 2023, 03:59 PM"
+    url: "https://x.com/JoramSoch/status/1688550557034651648"
+
+proof_id: "P470"
+shortcut: "ind-self"
+username: "JoramSoch"
+---
+
+
+**Theorem:** Let $E$ be a [random event](/D/reve). Then, $E$ is [independent of itself](/D/ind), if and only if its [probability](/D/prob) is zero or one:
+
+$$ \label{eq:ind-self}
+E \text{ self-independent} \quad \Leftrightarrow \quad P(E) = 0 \quad \text{or} \quad P(E) = 1 \; .
+$$
+
+
+**Proof:** According to the definition of [statistical independence](/D/ind), it must hold that:
+
+$$ \label{eq:ind}
+\begin{split}
+P(E,E) &= P(E) \cdot P(E) \\
+P(E)   &= \left( P(E) \right)^2 \; .
+\end{split}
+$$
+
+This is only fulfilled, if
+
+$$ \label{eq:ind-self-qed}
+P(E) = 0 \quad \text{or} \quad P(E) = 1 \; .
+$$
+
+Both is possible, since the [lower bound of probability is zero](/D/prob-ax) and the [upper bound of probability is one](/P/prob-range).