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June 26th, 2017, 08:49 AM   #1
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what choice of b minimizes E(|X −b|)?

The Pilsdorff Beer Company runs a fleet of trucks along the $100$ mile road
from Hangtown to Dry Gulch. The trucks are old, and are apt to break
down at any point along the road with equal probability. Where should the
company locate a garage so as to minimize the expected distance from a
typical breakdown to the garage? In other words, if $X$ is a random variable
giving the location of the breakdown, measured, say, from Hangtown, and $b$
gives the location of the garage, what choice of $b$ minimizes $E(|X − b|)$?
Now, suppose $X$ is not distributed uniformly over $[0, 100]$, but instead has density function $f_X(x) = \frac{2x}{10000}$. Then what choice of $b$ minimizes $E(|X − b|)$?

**Attempt:** For this, I thought about taking the definite integral $$\int_{0}^{100}((2x^2)/10,000)dx,$$
and got $200/3$, but it did not seem to be leading to the right answer. Later, I was really stuck.
, but it did not seem to be leading to the right answer. Later, I was really stuck.
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June 26th, 2017, 10:25 AM   #2
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Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
$\displaystyle
\begin{align}
E(|X - b|) &= E(X - b|X \geq b) P(X \geq b) + E(b - X|X < b) P(X < b)\\
&= [E(X|X \geq b) - b] P(X \geq b) + [b - E(X|X < b)] P(X < b)\\
&= \left[\int_{b}^{100} x \frac{2x/10000}{P(X \geq b)} \mathop{} \mathrm{d}x - b\right]P(X \geq b) + \left[b-\int_{0}^{b} x \frac{2x/10000}{P(X < b)} \mathop{} \mathrm{d}x \right]P(X <b)
\end{align}$

Noting that

$\displaystyle P(X < b) = \int_0^b \frac{2x}{10000} \mathop{}\mathrm{d}x= \frac{b^2}{10000}$

we thus have

$\displaystyle
\begin{align}
E(|X - b|) &= \left[\int_{b}^{100} \frac{2x^2}{10000}\mathop{} \mathrm{d}x - b \left(1-\frac{b^2}{1000}\right)\right] + \left[ \frac{b^3}{10000}-\int_{0}^{b} \frac{2x^2}{10000} \mathop{} \mathrm{d}x \right]\\
& = \left[\frac{200}{3} - \frac{b^3}{15000} -b \left(1-\frac{b^2}{1000}\right) \right] + \left[ \frac{b^3}{10000}-\frac{b^3}{15000}\right]\\
&= \frac{200}{3} + \frac{b^3}{3000} - b
\end{align}$

and thus the expectation is minimised when $\displaystyle \frac{b^2}{1000} - 1 = 0$, i.e. $\displaystyle b = \sqrt{1000}$.
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Last edited by 123qwerty; June 26th, 2017 at 10:37 AM.
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June 26th, 2017, 02:36 PM   #3
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Hey, I was curious, what is the formula for P(X is greater than or = to b)?
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June 26th, 2017, 07:59 PM   #4
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Quote:
Originally Posted by poopeyey2 View Post
Hey, I was curious, what is the formula for P(X is greater than or = to b)?
$\displaystyle P(X \geq b) = \int_b^{100} \frac{2x}{10000} \mathop{}\mathrm{d}x$
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June 27th, 2017, 04:04 PM   #5
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If *X is a random variable
giving the location of the breakdown, measured, say, from Hangtown, and *b
gives the location of the garage, what choice of *b* minimizes *E(|X−b|)?
Also, for this part of the question, is a different formula needed?

Last edited by skipjack; June 28th, 2017 at 08:49 AM.
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June 27th, 2017, 06:52 PM   #6
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Quote:
Originally Posted by poopeyey2 View Post
If *X is a random variable
giving the location of the breakdown, measured, say, from Hangtown, and *b
gives the location of the garage, what choice of *b* minimizes *E(|X−b|)?
Also, for this part of the question, is a different formula needed?
You mean the first part of the question with equal probability? Just replace $\displaystyle \frac{2x}{10000}$ in the formulas with $\displaystyle \frac{1}{100}$.
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Last edited by skipjack; June 28th, 2017 at 08:50 AM.
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June 28th, 2017, 09:59 AM   #7
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Also is the equation to find the minimum P(X < b)-1=o? And we have to find b?
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June 28th, 2017, 07:21 PM   #8
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Quote:
Originally Posted by poopeyey2 View Post
Also is the equation to find the minimum P(X < b)-1=o? And we have to find b?
I'm not sure I understood what you meant here. Which equation are you referring to?
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