I made up the following problem. Is my answer correct? Any comments about my style?
Thanks,
Bob
Problem:
Find the family of densities functions which are parabolas of the form
$y = Ax^2 + Bx + C$, have their vertex on the y axis, intersect the x-axis at the points $(\pm c, 0)$ where $c$ is a positive real number and $A < 0$. In addition, the density function should be $0$ for $x < -c$ and $x > c$. Also find $E(x^2)$,
$E(x^3)$ and the standard deviation of $x$.
Answer:
The first step in the process is to find the density function. Since we have
$A < 0$, the vertex is going to be a maximum, not a minimum.
\begin{eqnarray*}
Ac^2 + Bc + C &=& 0 \\
A(-c)^2 + B(-c) + C &=& 0 \\
Ac^2 - Bc + C &=& 0 \\
C &=& -Ac^2 \\
\end{eqnarray*}
Now since we are looking for a density function, the area under the curve must
be $1$.
\begin{eqnarray*}
\int_{-c}^{c} Ax^2 + Bx + C \, dx &=& 1 \\
\frac{Ax^3}{3} + \frac{Bx^2}{2} + Cx \Big|_{-c}^{c} &=& 1 \\
\frac{Ac^3}{3} + \frac{Bc^2}{2} + Cc - ( \frac{-Ac^3}{3} + \frac{Bc^2}{2} - Cc )
&=& 1 \\
\frac{2Ac^3}{3} + 2Cc &=& 1 \\
\frac{2Ac^3}{3} + 2(-Ac^2)c &=& 1 \\
\end{eqnarray*}
\begin{eqnarray*}
\frac{Ac^3}{3} - Ac^3 &=& 1 \\
-\frac{2Ac^3}{3} &=& 1 \\
A &=& \frac{-3}{2c^3} \\
C &=& -Ac^2 = \Big( \frac{3}{2c^2} \Big) (c^2 ) \\
C = \frac{3}{2c}
\end{eqnarray*}
Now we need to solve for $B$.
\begin{eqnarray*}
Ac^2 + Bc + C &=& 0 \\
\Big( \frac{-3}{2c^3}\Big)c^2 + Bc + \frac{3}{2c} &=& 0 \\
\frac{-3}{2c} + Bc + \frac{3}{2c} &=& 0 \\
Bc &=& 0 \\
B &=& 0 \\
\end{eqnarray*}
Hence our density function is:
\begin{eqnarray*}
y &=& \frac{-3}{2c^3}x^2 + \frac{3}{2c} \,\,\, \text{ for } -c <= x <= c \\
E(x^2) &=& \int_{-c}^{c} x^2 \Big( \frac{-3}{2c^3}x^2 + \frac{3}{2c} \Big) \,\, dx \\
E(x^2) &=& \int_{-c}^{c} \frac{-3x^4}{2c^3} + \frac{3x^2}{2c} \,\, dx =
\frac{-3x^5}{10c^3} + \frac{3x^3}{6c} \Big|_{-c}^{c} \\
E(x^2) &=&
\frac{-3c^5}{10c^3} + \frac{3c^3}{6c} - ( \frac{3c^5}{10c^3} + \frac{-3c^3}{6c} ) \\
E(x^2) &=&
\frac{-3c^5}{10c^3} + \frac{c^3}{2c} - \frac{3c^5}{10c^3} + \frac{c^3}{2c} \\
E(x^2) &=&
\frac{-3c^2}{10} + \frac{c^3}{2c} - \frac{3c^2}{10} + \frac{c^3}{2c} \\
E(x^2) &=& \frac{c^3}{c} - \frac{3c^2}{10} - \frac{3c^2}{10} \\
E(x^2) &=& c^2 - \frac{6c^2}{10} \\
E(x^2) &=& \frac{2c^2}{5} \\
\end{eqnarray*}
\begin{eqnarray*}
E(x^3) &=& \int_{-c}^{c} x^3 \Big( \frac{-3}{2c^3}x^2 + \frac{3}{2c} \Big) \,\, dx \\
E(x^3) &=& \int_{-c}^{c} \frac{-3}{2c^3}x^5 + \frac{3x^3}{2c} \,\, dx \\
E(x^3) &=& \frac{-3x^6}{12c^3} + \frac{3x^4}{8c} \Big|_{-c}^{c} \\
E(x^3) &=&
\frac{-3c^6}{12c^3} + \frac{3c^4}{8c} -
\Big( \frac{-3c^6}{12c^3} + \frac{3c^4}{8c} \Big) \\
E(x^3) &=& 0 \\
\sigma^2 &=& E(x^2) - u^2 = E(x^2 ) \\
\sigma^2 &=& \frac{2c^2}{5} \\
\sigma &=& \frac{\sqrt{2}c}{\sqrt{5}} \\
\end{eqnarray*}