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\input{18440-preamble}

\begin{document}
\pset{7}{Dan Ports}{2003/10/27}{drkp@mit.edu}

\begin{ppart}{Problem 1}
\paragraph{Part a}
\begin{eqnarray*}
\E{R} &=& \int\int_S R(x,y) \,dA = \int_0^{2\pi}\int_0^1 \frac{r^2}{\pi} \,dr\,d\theta\\
      &=& \int_0^{2\pi} \frac{1}{3\pi} \,d\theta = \frac{2}{3}
\end{eqnarray*}
\paragraph{Part b}
The cumulative distribution function $F_R(r)$ is the area the subcircle with radius $r$ divided by $\pi$:
\[F_R(r) = \frac{\pi r^2}{\pi} = r^2\]
and so
\[f_R(r) = 2r\]
\paragraph{Part c}
\[\E{R^2} = \int_0^1r^2f_R(r)\,dr = \int_0^1 2r^3\,dr = \frac{2}{4}1^4 = \frac{1}{2}\]
\paragraph{Part d}
We expect the expected value of $R$ to be more than $\frac{1}{2}$. The area of the subcircle of radius $\frac{1}{2}$ is $\frac{1}{4}$th the total area of the circle. So there are more points in the sample space whose distance from the center is greater than $\frac{1}{2}$ than there are whose distance from the center is less than $\frac{1}{2}$.
\end{ppart}

\begin{ppart}{Problem 2}
\paragraph{Part a}
Since $f_X(x)$ is a probability density function, $\int_a^\infty f_X(x)\,dx = \lim_{b\to\infty}\int_a^bf_X(x)\,dx = 1$. So 
\[\lim_{b\to\infty}b(1-F_X(b)) = \lim_{b\to\infty} b-b\int_a^bf_X(x)\,dx = 0\]
\paragraph{Part b}
Applying integration by parts, for any $b \ge a$,
\begin{eqnarray*}
\int_a^bxf_X(x)\,dx &=&  bF_X(b) - aF_X(a)-\int_a^bF_X(x)\,dx\\
&=& -\int_a^bF_X(x)\,dx + b - a -b(1-F_X(b)) +a(1-F_X(a))\\
&=& \int_a^b1-F_X(x)\,dx - b(1-F_X(b)) +a(1-F_X(a))\\
\end{eqnarray*}
\paragraph{Part c}
\begin{eqnarray*}
\E{X} &=& \int_a^\infty xf_X(x)\,dx \\
&=& \lim_{b\to\infty}\int_a^b xf_X(x)\,dx \\
&=& \lim_{b\to\infty}\int_a^b1-F_X(x)\,dx - b(1-F_X(b)) + a(1-F_X(a))\\
&=& \lim_{b\to\infty}\int_a^b1-F_X(x)\,dx + a \\
\end{eqnarray*}
The last step is true because $\lim_{b\to\infty} b(1-F_X(b)) = 0$ from part a, and $F_X(a) = 0$ because it is the cumulative distribution function of a random variable that always takes values greater than or equal to $a$.
So we can write
\[\E{X} = \int_a^\infty1-F_X(x)\,dx + a\]
The improper integral converges because it is the limit $\lim_{b\to\infty}\int_a^b1-F_X(x)\,dx$ above, which exists because it is equal to $\E{X} -a$, and $\E{X}$ exists.
\paragraph{Part d}
Taking $a' < a$, we obtain
\begin{eqnarray*}
\E{X} &=& \int_{a'}^\infty 1-F_X(x)\,dx + a'\\
&=& \int_{a'}^\infty 1-F_X(x)\,dx + \int_{a}^\infty1-F_X(x)\,dx + a'\\
&=& \int_{a'}^\infty 1-\int_{a'}^\infty F_X(x)\,dx + \int_{a}^\infty1-F_X(x)\,dx + a'\\
&=& a-a'+\int_{a}^\infty1-F_X(x)\,dx + a' \\
&=&\int_a^\infty1-F_X(x)\,dx + a
\end{eqnarray*}
since $F_X(x) = 0$ for $x < a$ because X takes only values greater or equal to $a$
\end{ppart}
\begin{ppart}{Problem 3}
\paragraph{}
When tossing a fair coin 900 times, the expected number of heads is $900\frac{1}{2} = 450$ and the variance is $900\frac{1}{2}(1-\frac{1}{2}) = 225$. We approximate the experiment with a normal random variable with the same mean and variance. Since the standard deviation is $15$, the probability that the number of heads will be within $10 = \frac{2}{3}\sigma$ of the mean is \[\Phi(\frac{2}{3})-(1-\Phi(\frac{2}{3})) \approx .4972\]
\end{ppart}

\begin{ppart}{Problem 4}
\paragraph{}
Let $Y$ be the random variable with probability density function $f_Y(x)=\frac{1}{3}(e^{-x}+e^{-\frac{1}{2}x})$ for $x > 0$ and 0 otherwise. Note that $\int_0^\infty f_Y(x)\,dx = \frac{1}{3}\int_0^\infty e^{-x}+e^{-\frac{1}{2}x}\,dx = \frac{1}{3}3 = 1$, so this is a valid probability density function.

The cumulative distribution function is $F_Y(x) = \int_0^x f_Y(t)\,dt = 1-\frac{2}{3}e^{-\frac{x}{2}}-\frac{1}{3}e^{-x}$, so the hazard rate function is \[\lambda_Y(x) = \frac{f_Y(x)}{1-F_Y(x)} = \frac{\frac{1}{3}\left(e^{-x}+e^{-\frac{x}{2}}\right)}{\frac{2}{3}e^{-\frac{x}{2}}+\frac{1}{3}e^{-x}} = \frac{e^{\frac{x}{2}}+1}{2e^{\frac{x}{2}}+1}\]
The derivative of the hazard rate is
\[\lambda_Y'(x) = \frac{-e^{\frac{x}{2}}}{2\left(2e^{\frac{x}{2}}+1\right)^2}\]
which is clearly negative for all $x > 0$. So the hazard rate function is everywhere decreasing and the random variable $Y$ thus has the desired property that $\forall s,t > 0 \, \, \Pr{Y\ge s+t|Y\ge t} > \Pr{X>s}$.
\end{ppart}


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