From 2837c0e282832ef34c697b03501b1a2f50689c3e Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Tue, 6 Oct 2020 19:39:50 -0400
Subject: minor changes to q3 on hw5

---
 houdre/hw5.tex | 8 +++++---
 1 file changed, 5 insertions(+), 3 deletions(-)

(limited to 'houdre/hw5.tex')

diff --git a/houdre/hw5.tex b/houdre/hw5.tex
index a952e76..54b8ed4 100644
--- a/houdre/hw5.tex
+++ b/houdre/hw5.tex
@@ -76,9 +76,11 @@ QED
 \q3
 
 The density function $f(x)$ is proportional, so there is some constant
-$c$ such that $$1 = f(x) = c\infint g(x)dx = 2c\int_1^\infty x^{-n}dx =
-2c\left(-{x^{1-n}\over n-1}\right)\big|^\infty_1 = 2c\left({1\over
-n-1}\right) \to c = {n-1\over2}.$$
+$c$ such that $$1 = \infint f(x)dx = c\infint g(x)dx = 2c\int_1^\infty
+x^{-n}dx = 2c\left(-{x^{1-n}\over n-1}\right)\big|^\infty_1 =
+2c\left({1\over n-1}\right) \to c = {n-1\over2}.$$
+$$f(x) = {n-1\over2}g(x),$$
+and this will look like a vertically stretched $1/x^2$ graph.
 
 The mean and variance of $X$ exist when $\E(X)$ and $\E(X^2)$ exist,
 respectively. The first exists when $n>2$ because for $n=2,$
-- 
cgit 

