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 March 4th, 2014, 10:27 AM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Help understanding/putting together this type of problem Let X be a random variable with support {1,2,3,5,15,25,50}, each point of which has the same probabiliy, 1/7. Argue that c=5 is the value that minimizes $h(c)= E( |X-c|)$. Compare c with the value of b that minimizes $g(b)= E[(X - b)^2]$. How do I argue that c minimizes h(c)? I have seen similar examples of these types of problems worked, just not quite understanding this type of problem. Will this involve $h(c)= x^2 - cx$, then taking $h'(c)$? Thanks for any help on this
 March 4th, 2014, 12:45 PM #2 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 Re: Help understanding/putting together this type of problem My approach: Step 1. Calculate h(c) for the 7 discrete values of X. Find the two lowest values (should be adjacent). Step 2. h(c) will be a nice (continuous and continuously differentiable) function in the interval between these values. Use elementary calculus to find minimum. I presume you know how to minimize g(b).
 March 4th, 2014, 11:00 PM #3 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Help understanding/putting together this type of problem I may need more help than I thought..... From the definition of this type of problem.... If f(x) is the p.m.f. of the random variable X of the discrete type w/ space S, and if the summation $\sum_{x \in s}u(x)f(x)$ exists, then the sum is called the mathematical expectation or expected value of the function $u(x)$, and it is denoted by $E[u(x)]$ In the problem I posted, what is $u(x)$? By definition, $f(x)$ will always $=1$? since it is a p.m.f.?
 March 4th, 2014, 11:22 PM #4 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Help understanding/putting together this type of problem I am sorry, in that last part, I am asking if $\sum f(x)$ $=1$
March 5th, 2014, 12:37 AM   #5
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Re: Help understanding/putting together this type of problem

Quote:
 Originally Posted by WWRtelescoping Let X be a random variable with support {1,2,3,5,15,25,50}, each point of which has the same probability, 1/7. Argue that c=5 is the value that minimizes $h(c)= E( |X-c|)$.
I reckon that:

a) For any odd-numbered set, c will be the middle term.
b) For any even-numbered set, c can be any number between (and including) the middle two terms.

a) To see this, consider the value of E(|X-c|) where c is the middle term and there are 2n+1 terms. If you increase c by a, you increase the value of E(|X-c|) by at least (n+1)a and reduce it by at most na. Similarly, if you reduce c.

In other words, moving away from the middle term always increases the expected value.

b) For an even set, remains constant from the n to the n+1 term and always increases outside of this range.

March 5th, 2014, 12:29 PM   #6
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Re: Help understanding/putting together this type of problem

Quote:
 Originally Posted by WWRtelescoping I may need more help than I thought..... From the definition of this type of problem.... If f(x) is the p.m.f. of the random variable X of the discrete type w/ space S, and if the summation $\sum_{x \in s}u(x)f(x)$ exists, then the sum is called the mathematical expectation or expected value of the function $u(x)$, and it is denoted by $E[u(x)]$ In the problem I posted, what is $u(x)$? By definition, $f(x)$ will always $=1$? since it is a p.m.f.?
u(x) = |x-c|, f(x) = 1/7.

To use my approach use the 7 values of X in turn for c to get E(|X-c|).

March 5th, 2014, 12:30 PM   #7
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Re: Help understanding/putting together this type of problem

Quote:
Originally Posted by Pero
Quote:
 Originally Posted by WWRtelescoping Let X be a random variable with support {1,2,3,5,15,25,50}, each point of which has the same probability, 1/7. Argue that c=5 is the value that minimizes $h(c)= E( |X-c|)$.
I reckon that:

a) For any odd-numbered set, c will be the middle term.
b) For any even-numbered set, c can be any number between (and including) the middle two terms.

a) To see this, consider the value of E(|X-c|) where c is the middle term and there are 2n+1 terms. If you increase c by a, you increase the value of E(|X-c|) by at least (n+1)a and reduce it by at most na. Similarly, if you reduce c.

In other words, moving away from the middle term always increases the expected value.

b) For an even set, remains constant from the n to the n+1 term and always increases outside of this range.
Your description is valid only of the discrete values are evenly spaced. Here they are not.

 March 5th, 2014, 01:24 PM #8 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Help understanding/putting together this type of problem If we take the set we have: 1, 2, 3, 5, 15, 25, 50. Start at c = 5 and increase by 2, say, so c = 7: 7 - 1 is +2 compared to 5 - 1 7 - 2 is +2 compared to 5 - 2 7 - 3 is +2 compared to 5 - 3 7 - 5 is +2 compared to 5 - 5 So, the sum of differences has increased by 4x2. This is actually regardless of the distribution or how large the increase is. On the other hand, the sum of differences of |X-c| for the higher terms will reduce by exactly 3x2. This is the most it could reduce by. If the increase in c is > 10, then it will reduce by less than this, as c will overshoot the next number 15. This argument depends only on the numbers being equally likely, but not evenly distributed.
 March 5th, 2014, 06:10 PM #9 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Help understanding/putting together this type of problem Sorry for dragging this out, just doing in baby steps to make sure I am understanding the set up. $E[u(x)]= \sum u(x)f(x)$ , so $E[|x-c|]= \sum |x-c|f(x)$ Which will be $|1-c|/7 + |2-c|/7 + |3-c|/7 + |5-c|/7 + |15-c|/7 + |25-c|/7 + |50-c|/7$ In general, how do I find the value of $c$ that minimizes this w/ out the value being given? thanks for all the help
 March 5th, 2014, 11:36 PM #10 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Help understanding/putting together this type of problem If there are an odd number of terms, c is always the middle term. If there an even number of terms, c can be any number between (and including) the middle two terms. If you don't see this, try it out with some examples. Start with 3 numbers and see what happens.

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