From 9ff4a07c9b3b644e4485f827cd9b54b672a08c10 Mon Sep 17 00:00:00 2001
From: Bruno BELANYI <bruno@belanyi.fr>
Date: Fri, 2 Aug 2024 21:04:49 +0100
Subject: [PATCH] posts: reservoir-sampling: add one-element sample

---
 .../2024-08-02-reservoir-sampling/index.md    | 56 +++++++++++++++++++
 1 file changed, 56 insertions(+)

diff --git a/content/posts/2024-08-02-reservoir-sampling/index.md b/content/posts/2024-08-02-reservoir-sampling/index.md
index 386a840..eabeab6 100644
--- a/content/posts/2024-08-02-reservoir-sampling/index.md
+++ b/content/posts/2024-08-02-reservoir-sampling/index.md
@@ -25,3 +25,59 @@ amount of space and a single pass through the stream.
 [online]: https://en.wikipedia.org/wiki/Online_algorithm
 
 <!--more-->
+
+## Sampling one element
+
+As an introduction, we'll first focus on fairly sampling one element from the
+stream.
+
+```python
+def sample_one[T](stream: Iterable[T]) -> T:
+    stream_iter = iter(stream)
+    # Sample the first element
+    res = next(stream_iter)
+    for i, val in enumerate(stream_iter, start=1):
+        j = random.randint(0, i)
+        # Replace the sampled element with probability 1/(i + 1)
+        if j == 0:
+            res = val
+    # Return the randomly sampled element
+    return res
+```
+
+### Proof
+
+Let's now prove that this algorithm leads to a fair sampling of the stream.
+
+We'll be doing proof by induction.
+
+#### Hypothesis $H_N$
+
+After iterating through the first $N$ items in the stream,
+each of them has had an equal $\frac{1}{N}$ probability of being selected as
+`res`.
+
+#### Base Case $H_1$
+
+We can trivially observe that the first element is always assigned to `res`,
+$\frac{1}{1} = 1$, the hypothesis has been verified.
+
+#### Inductive Case
+
+For a given $N$, let us assume that $H_N$ holds. Let us now look at the events
+of loop iteration where `i = N` (i.e: observation of the $N + 1$-th item in the
+stream).
+
+`j = random.randint(0, i)` uniformly selects a value in the range $[0, i]$,
+a.k.a $[0, N]$. We then have two cases:
+
+* `j == 0`, with probability $\frac{1}{N + 1}$: we select `val` as the new
+reservoir element `res`.
+
+* `j != 0`, with probability $\frac{N}{N + 1}$: we keep the previous value of
+`res`. By $H_N$, any of the first $N$ elements had a $\frac{1}{N}$ probability
+of being `res` before at the start of the loop, each element now has a
+probability $\frac{1}{N} \cdot \frac{N}{N + 1} = \frac{1}{N + 1}$ of being the
+element.
+
+And thus, we have proven $H_{N + 1}$ at the end of the loop.