From a7f4d5c2ce9ee09879b4183d18169d5ba45cf422 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Fri, 18 Sep 2015 15:52:14 -0400
Subject: [HW1] fixing typos in problem 1

---
 hw1/main.tex | 24 ++++++++++++------------
 1 file changed, 12 insertions(+), 12 deletions(-)

(limited to 'hw1')

diff --git a/hw1/main.tex b/hw1/main.tex
index cd5c1f9..e05594e 100644
--- a/hw1/main.tex
+++ b/hw1/main.tex
@@ -60,7 +60,7 @@ $X+Y$ respectively, we have:
     g(z) = \int_{\mathbb{R}} f_X(u)f_Y(z-u)du,\quad z\in\mathbb{R}
 \end{displaymath}
 
-(b) Using the formula for the cdf of a gaussain, we obtain:
+(b) Using the formula for the cdf of a Gaussian variable, we obtain:
 \begin{displaymath}
     g(z) = \frac{1}{2\pi\sigma_X\sigma_Y}\int_{\mathbb{R}}
     \exp\left(-\frac{u^2}{2\sigma_X^2} - \frac{(z-u)^2}{2\sigma_y^2}\right)du
@@ -78,8 +78,8 @@ Hence:
     g(z) = \frac{1}{2\pi\sigma_X\sigma_Y}
     \exp\left(- \frac{z^2}{2(\sigma_X^2+\sigma_Y^2)}\right)
     \int_{\mathbb{R}}
-    -\frac{\sigma_X^2+ \sigma_Y^2}{2\sigma_X^2\sigma_Y^2}\left(u
-    - \frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}z\right)^2du
+    \exp\left(-\frac{\sigma_X^2+ \sigma_Y^2}{2\sigma_X^2\sigma_Y^2}\left(u
+    - \frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}z\right)^2\right)du
 \end{displaymath}
 We now recognize the integral of the un-normalized cdf of a Gaussian variable.
 Hence the integral evaluates to:
@@ -91,12 +91,12 @@ after simplification we obtain the following formula for $g(z)$:
     g(z) = \frac{1}{\sqrt{2\pi}\sqrt{\sigma_X^2+\sigma_Y^2}}
     \exp\left(- \frac{z^2}{2(\sigma_X^2+\sigma_Y^2)}\right)
 \end{displaymath}
-That is, $X+Y$ is a Gaussian variable of mean $0$ and variance
+That is, $X+Y$ is a Gaussian variable with mean $0$ and variance
 $\sigma_X^2+\sigma_Y^2$.
 
-(c) If $X$ has mean $\mu_X$ and $Y$ has mean $\mu_Y$, $X-\mu_X$ and $Y-\mu_Y$
-are still independent variables with the same variance as before and zero mean.
-Applying the previous result. $X+Y-\mu_X-\mu_Y$ is a Gaussian variable with
+(c) If $X$ has mean $\mu_X$ and $Y$ has mean $\mu_Y$, $X-\mu_X$ and $Y-\mu_Y$
+are still independent variables with unchanged variance and zero mean.
+Applying the previous result, $X+Y-\mu_X-\mu_Y$ is a Gaussian variable with
 mean $0$ and variance $\sigma_X^2 + \sigma_Y^2$. Hence, $X+Y$ is normally
 distributed with mean $\mu_X+\mu_Y$ and variance $\sigma_X^2+\sigma_Y^2$.
 
@@ -106,16 +106,16 @@ for any interval $[a,b]$ with $a,b>0$:
 \begin{align*}
     \prob(Y\in[a,b]) = \prob(X\in [\log a, \log b])
     &=\frac{1}{\sqrt{2\pi}\sigma_X}\int_{\log a}^{\log b}
-    \exp\left(-\frac{1}{2\sigma_X}(x-\mu)^2\right)dx\\
-    &=\frac{1}{\sqrt{2\pi}\sigma_X}\int_{a}^{b}
-    \exp\left(-\frac{1}{2z\sigma_X}(\log z-\mu)^2\right)dz
+    \exp\left(-\frac{1}{2\sigma_X^2}(x-\mu)^2\right)dx\\
+    &=\frac{1}{\sqrt{2\pi}\sigma_X^2}\int_{a}^{b}
+    \exp\left(-\frac{1}{2z\sigma_X^2}(\log z-\mu)^2\right)dz
 \end{align*}
 where we used the change of variable $z = e^x$ in the last equality. Hence the
 pdf of $Y$:
 \begin{displaymath}
-    f_Y = \begin{cases}
+    f_Y(z) = \begin{cases}
     \frac{1}{\sqrt{2\pi}\sigma_X}
-    \exp\left(-\frac{1}{2z\sigma_X}(\log z-\mu)^2\right)&\text{if $x>0$}\\
+    \exp\left(-\frac{1}{2z\sigma_X^2}(\log z-\mu)^2\right)&\text{if $z>0$}\\
     0&\text{otherwise}
     \end{cases}
 \end{displaymath}
-- 
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