From c8e67614bbf6945d457a808ecf9e8c4a7c8ce607 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Mon, 25 Feb 2019 16:16:01 -0500
Subject: [PATCH] Add first part of 4.6

---
 _psets/3.md | 29 +++++++++++++++++++++++++++++
 1 file changed, 29 insertions(+)

diff --git a/_psets/3.md b/_psets/3.md
index 9e0472f..076e4f6 100644
--- a/_psets/3.md
+++ b/_psets/3.md
@@ -423,3 +423,32 @@ $$\num{1.5e8} \si{Hz}$$.
 Let $$(x_1, x_2, \ldots, x_n)$$ be drawn from a Gaussian distribution with variance $$\sigma^2$$ and
 unknown mean value $$x_0$$. Show that $$f(x_1, \ldots, x_n) = n^{-1} \sum_{i = 1}^n x_i$$ is an
 estimator for $$x_0$$ that is unbiased and achieves the Cram&eacute;r–Rao lower bound.
+
+The probability of seeing a particular sequence $$x_1, \ldots, x_n$$ of samples is
+
+$$
+p(x_1, \ldots, x_n)
+= \prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}} \mathrm{d} x
+$$
+
+So the expected value of this estimator is
+
+$$
+\begin{align*}
+\left \langle \frac{1}{n} \sum_{i = 1}^n x_i \right \rangle
+&= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty
+    \left( \prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}}
+    \right) \left( \frac{1}{n} \sum_{i = 1}^n x_i \right) \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
+&= \frac{1}{n} \sum_{i = 1}^n \prod_{j = 1}^n
+    \int_{-\infty}^\infty
+    \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_j - x_0)^2}{2 \sigma^2}}
+    \left( 1 - \delta_{ij} (1 + x_i) \right)
+    \mathrm{d} x_j \\
+&= \frac{1}{n} \sum_{i = 1}^n x_0 \\
+&= x_0
+\end{align*}
+$$
+
+The [Kronecker delta](https://en.wikipedia.org/wiki/Kronecker_delta) is used to indicate that only
+one term in the product contains $$x_i$$. This term integrates to $$x_0$$ (since it is the expected
+value of a single Gaussian), and the others integrate to one.
-- 
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