# Thread: distribution function with uniformity

1. ## distribution function with uniformity

I'm having trouble solving the following problem correctly:

The waiting time $Y$ until delivery of a new component for an industrial operation is uniformly distributed over the interval 1 to 5 days. The cost of this delay is given by $U = 2Y^2+3$. Find the probability density function for $U$.

so far I have:

$Y= \pm \sqrt{\frac{u-3}{2}}$

and based on an example from the book I have:

$f_U(u) = \left\{ \begin{array}{rcl}
\frac{1}{2\sqrt{u}} \left[f_Y\left(\sqrt{u}\right) + f_Y(y)(-\sqrt{u})\right] & \mbox{for} & u>0 \\
0 & \mbox{if} & \mbox{other}
\end{array}\right.$

then combining for the top part I would get:

$=\frac{1}{2\sqrt{\frac{u-3}{2}}} \left[\frac{1}{4}\left(\sqrt{\frac{u-3}{2}}\right) + \frac{1}{4}\left(-\sqrt{\frac{u-3}{2}}\right)\right]$

$=\frac{1}{8\sqrt{\frac{u-3}{2}}}$

and based on the solution in the back of the book it's:

$=\frac{1}{{\color{red}16}\sqrt{\frac{u-3}{2}}}$

if anyone can help me it it would be greatly appreciated

2. Originally Posted by lllll
I'm having trouble solving the following problem correctly:

The waiting time $Y$ until delivery of a new component for an industrial operation is uniformly distributed over the interval 1 to 5 days. The cost of this delay is given by $U = 2Y^2+3$. Find the probability density function for $U$.

so far I have:

$Y= \pm \sqrt{\frac{u-3}{2}}$

and based on an example from the book I have:

$f_U(u) = \left\{ \begin{array}{rcl}
\frac{1}{2\sqrt{u}} \left[f_Y\left(\sqrt{u}\right) + f_Y(y)(-\sqrt{u})\right] & \mbox{for} & u>0 \\
0 & \mbox{if} & \mbox{other}
\end{array}\right.$

then combining for the top part I would get:

$=\frac{1}{2\sqrt{\frac{u-3}{2}}} \left[\frac{1}{4}\left(\sqrt{\frac{u-3}{2}}\right) + \frac{1}{4}\left(-\sqrt{\frac{u-3}{2}}\right)\right]$

$=\frac{1}{8\sqrt{\frac{u-3}{2}}}$

and based on the solution in the back of the book it's:

$=\frac{1}{{\color{red}16}\sqrt{\frac{u-3}{2}}}$

if anyone can help me it it would be greatly appreciated
$F(u) = \Pr(U \leq u)$

$= \Pr(2Y^2 + 3 \leq u) = \Pr \left( Y^2 \leq \frac{u-3}{2} \right)$ $= \Pr \left( - \sqrt{\frac{u-3}{2}} \leq Y \leq \sqrt{\frac{u-3}{2}} \right)$

$= \int_{1}^{\sqrt{\frac{u-3}{2}}} \frac{1}{4} \, dy$

$= \frac{1}{4} \sqrt{\frac{u-3}{2}} - \frac{1}{4}$.

Therefore $f(u) = \frac{dF}{du} = \frac{1}{16\sqrt{\frac{u-3}{2}}}$ for $5 \leq u \leq 53$ and zero otherwise.

Note: You can see your answer is wrong because if you integrate it between u = 5 and u = 53 you get 2 rather than 1.