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March 4th, 2014, 10:27 AM   #1
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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 . Compare c with the value of b that minimizes .

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 , then taking ?

Thanks for any help on this
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March 4th, 2014, 12:45 PM   #2
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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).
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March 4th, 2014, 11:00 PM   #3
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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
exists, then the sum is called the mathematical expectation or expected value of the function , and it is denoted by

In the problem I posted, what is ?

By definition, will always ? since it is a p.m.f.?
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March 4th, 2014, 11:22 PM   #4
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Re: Help understanding/putting together this type of problem

I am sorry, in that last part, I am asking if
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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 .
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.
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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
exists, then the sum is called the mathematical expectation or expected value of the function , and it is denoted by

In the problem I posted, what is ?

By definition, will always ? 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|).
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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 .
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.
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March 5th, 2014, 01:24 PM   #8
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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.
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March 5th, 2014, 06:10 PM   #9
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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.

, so

Which will be



In general, how do I find the value of that minimizes this w/ out the value being given?

thanks for all the help
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March 5th, 2014, 11:36 PM   #10
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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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let x be a random variable with support
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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 E(|x-C|)

,

argue that c=5 is the value that minimizes

,

let x be the random variable with support {1,2,3,5,15,25,50}, each point which has the same probablilty 1/7.

,

value of 5/15/25

,

let x be a random support c=5 minimizes compare

,

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 E(|x-C|) answers

,

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 E(/x c/) compare this to the value of b that minimizes E /(x b)^2/

,

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 E(|x-c|). compare this to a value b that minimizes E(|x-b|^2)

,

Let X be a random variable with support {1,2,3,5,15,25,50}, each point which has the same probability 1/7 Argue that C=5 is the value that minimizes E(|x-C|). Compare this to the value of b that minimizes E[(x-b)^2].-solution

,

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 E(|x-c|). Compare this to the value of b that minimizes E[(x-b)^2]

,

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 mini- mizes h(c) = E( |X − c| ). Compare c with the value of b that minimizes g(b) = E[(X − b)2]

,

let x be a random variable with support {1,2,3,5,15,25,50} each of which has the same probability 1/7. argue that c=5 is the value that minimizes

,

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 E(|x-c |) compare this to the value of b that minimizes E[(x-b)^2]

,

1,2,3,5,15,25,50

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