\documentclass{article}
\input{18440-preamble}

\begin{document}
\pset{9}{Dan Ports}{2003/11/19}{drkp@mit.edu}

\begin{ppart}{Problem 1}
Since $X$ is distributed uniformly over $\left[-1,1\right]$, $f_X(x) = \frac{1}{2}$ for all $x$ in this range and 0 otherwise. Therefore, the $k$th moment of $X$ is
\begin{eqnarray*}
m_k(X) &=& \E{X^k} = \int_{-1}^1 \frac{1}{2}x^k\;dx \\
&=& \left[\frac{1}{2}\frac{x^{k+1}}{k+1}\right]^1_{-1} \\
&=&
\begin{cases}
0 \; &k \mbox{ odd} \\
\frac{1}{k+1} \; &k \mbox{ even}
\end{cases}
\end{eqnarray*}
\end{ppart}

\begin{ppart}{Problem 2}
Let $X$ and $Y$ be random variables with moments $m_p(X)$ and $m_q(Y)$ defined. Then
\begin{eqnarray*}
m_k(X+Y) &=& \E{\left(X+Y\right)^k} \\
&=& \E{\sum_{i=0}^k \binom{k}{i} X^i Y^{k-i}} \mbox{ by the binomial theorem} \\
&=& \sum_{i=0}^k \binom{k}{i} \E{X^i Y^{k-i}} \mbox{ by linearity of expectation}\\
&=& \sum_{i=0}^k \binom{k}{i} \E{X^i} \E{Y^{k-i}} \mbox{ by independence of $X$ and $Y$}\\
&=& \sum_{i=0}^k \binom{k}{i} m_i(X) m_{k-i}(Y) \mbox{ by definition}\\
\end{eqnarray*}
\end{ppart}

\begin{ppart}{Problem 3}
\paragraph{Part a}
\[\E{S_n} = \E{X_1 + X_2 + \cdots + X_n} = \E{X_1} + \E{X_2} + \cdots + \E{X_n} = 0\]
\paragraph{Part b}
First, note that since $\E{X_i}$ = 0, $\Var X_i = \E{X_i^2} = m_2(X_i) = \frac{1}{3}$. Since the variables are independent,
\[\Var{S_n} = \Var X_1 + X_2 + \cdots + X_n  = \Var{X_1} + \Var{X_2} + \cdots + \Var{X_n} = \frac{n}{3}\]
\paragraph{Part c}
Consider first the case of even $k$. Then for odd $i$, $m_i(X) = m_{k-i} = 0$. Hence,
\begin{align*}
m_k(S_2) &= \sum_{i=0}^{k} \binom{k}{i} m_i(X_1) m_{k-i}(X_2)\\
&= \sum_{i=0 \mbox{ even}}^k \binom{k}{i} m_i(X_1) m_{k-i}(X_2)\\
&= \sum_{i=0}^\frac{k}{2} \binom{k}{i} \frac{1}{i+1} \frac{1}{k-i+1}\\
&\mbox{Performing a change of variables, with $p=i$ and $q=\frac{k}{2}$,}\\
&= \sum_{p=0}^q \binom{2q}{2p} \frac{1}{2p+1} \frac{1}{2q-2p+1} \\
&= \sum_{p=0}^q \frac{(2q)!}{(2p)! (2q-2p)!} \frac{1}{2p+1} \frac{1}{2q-2p+1}\\
&= \sum_{p=0}^q \frac{(2q)!}{(2p+1)! (2q-2p+1)!}\\
&= \sum_{p=0}^q \binom{2q}{2p+1} = 2^{2q-1} = 2^{k-1}\\
\end{align*}
Then, consider the case of $k$ odd. Then if $i$ is odd, $m_i(X) = 0$ and if $i$ is even, $k-i$ is odd and $m_{k-i}(X) = 0$. Hence, each term in the sum is zero
\begin{align*}
m_k(S_2) &= \sum_{i=0}^{k} \binom{k}{i} m_i(X_1) m_{k-i}(X_2)\\
&= \sum_{i=0}^k 0 = 0\\
\end{align*}
So
\[m_k(X) = \begin{cases} 2^{k-1} \; & \mbox{$k$ even} \\ 0 & \mbox{$k$ odd}\\ \end{cases}\]
\end{ppart}

\begin{ppart}{Problem 4}
From above, 
\[f_{S_1} = f_X = \begin{cases} \frac{1}{2} \; & -1 \le x \le 1 \\ 0 & \mbox{ otherwise} \end{cases}\]

Performing the convolutions graphically (see the attached figures),
\begin{align*}
f_{S_2}(x) &= f_{S_1}(x) * f_{S_1}(x) \\
&= \int_{-\infty}^\infty f_{S_1}(t) f_{S_1}(x-t)\;dt \\
&=  
\begin{cases}
\frac{x}{8} + \frac{1}{4} \; & -2 \le x \le 0\\
\frac{1}{4} - \frac{x}{8} \; & 0 < x \le 2\\
0 & \mbox{otherwise}
\end{cases}
\end{align*}

\begin{align*}
f_{S_3}(x) &= f_{S_1}(x) * f_{S_1}(x) * f_{S_1}(x) \\
&= f_{S_2}(x) * f_{S_1}(x) \\
&= \int_{-\infty}^\infty f_{S_2}(t) f_{S_1}(x-t)\;dt \\
&= 
\begin{cases}
\frac{1}{36}(x-3)(x+3)\;&-3 \le x \le 3\\
0 & \mbox{otherwise}
\end{cases}
\end{align*}
\end{ppart}

\begin{ppart}{Problem 5}
\paragraph{Part a}
Since $f(x) = \frac{1}{2}$ for $x$ between $-1$ and $1$ and 0 otherwise, $f(x-t) = \frac{1}{2}$ for $t$ between $x-1$ and $x+1$ and 0 otherwise. Hence,
\begin{align*}
(f*f_X)(x) &= \int_{-\infty}^\infty f(t)f_X(x-t)\,dt\\
&= \int_{x-1}^{x+1} f(t) \frac{1}{2} \,dt\\
&= \frac{1}{2} \int_{x-1}^{x+1} f(t)  \,dt\\
\end{align*}
\paragraph{Part b}
\begin{align*}
(f*f_X)(x+\Delta x) - (f*f_X)(x) &=
 \frac{1}{2} \int_{x+\Delta x-1}^{x+\Delta x+1} f(t)  \:dt - \frac{1}{2} \int_{x-1}^{x+1} f(t)  \:dt\\
&= \frac{1}{2} \int_{x+\Delta x-1}^{x+1} f(t)  \:dt + \frac{1}{2} \int_{x+1}^{x+\Delta x+1} f(t)  \:dt \\
&- \frac{1}{2} \int_{x-1}^{x+\Delta x-1} f(t)  \:dt - \frac{1}{2} \int_{x+\Delta x-1}^{x+1} f(t)  \:dt\\
&= \frac{1}{2} \int_{x+1}^{x+\Delta x+1} f(t)  \:dt - \frac{1}{2} \int_{x-1}^{x+\Delta x-1} f(t)  \:dt\\
\end{align*}
\paragraph{Part c}
\begin{align*}
\frac{d}{dx}\left(f*f_X\right)(x) &=
\lim_{\Delta x \to 0} \frac{\frac{1}{2} \int_{x+\Delta x-1}^{x+\Delta x+1} f(t)  \:dt - \frac{1}{2} \int_{x-1}^{x+1} f(t)  \:dt}{\Delta x}\\
&= \lim_{\Delta x \to 0}  \frac{\frac{1}{2} \int_{x+1}^{x+\Delta x+1} f(t)  \:dt - \frac{1}{2} \int_{x-1}^{x+\Delta x-1} f(t)  \:dt}{\Delta x}\\
&\mbox{Noting that as $\Delta x$ becomes small, the integrals can be approximated by their width}\\
&\mbox{multiplied by the value at one of the endpoints,}\\
&= \lim_{\Delta x \to 0}  \frac{\frac{1}{2} f(x+1)\Delta x - \frac{1}{2} f(x-1)\Delta x}{\Delta x}\\
&= \frac{1}{2} f(x+1) - \frac{1}{2} f(x-1)
\end{align*}
\paragraph{Part d}
Assuming $g$ is a function on $\R$ having a continuous derivative that vanishes outside some finite interval,
\begin{align*}
\frac{d}{dx}[f*x] &= \frac{d}{dx}\int_{-\infty}^\infty f(t) g(x-t)\:dt\\
&= \int_{-\infty}^\infty \frac{\partial}{\partial x} f(t) g(x-t)\:dt\\
&= \int_{-\infty}^\infty f(t) \frac{\partial}{\partial x} g(x-t)\:dt\\
&= \int_{-\infty}^\infty f(t) g'(x-t)\:dt\\
&= (f*g')(x)
\end{align*}
Noting that the function $f_X$ is a square pulse, its derivative $f_X'$ is a pair of impulses: a positive impulse of magnitude $\frac{1}{2}$ at $x=-1$ and a negative impulse of the same magnitude at $x=1$. Since the convolution of an impulse with a function is simply the function shifted and scaled, the convolution with the derivative of $f_X$ is the sum of two copies of the function scaled by $\frac{1}{2}$, one positive and shifted a unit to the left, and another negative and shifted to the right. The formula from part c follows:
\[\frac{d}{dx}\left(f*f_X\right)(x) = \frac{1}{2} f(x+1) - \frac{1}{2} f(x-1)\]
\end{ppart}

\end{document}