Minimize expected value of the absolute difference

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To minimize E[|X - b|] for a continuous random variable X with median m, it is essential to analyze the expression E[|X - b|] = E[|X - m|] + 2 ∫(x - b) f(x) dx from b to m. The integral indicates that the sign of (x - b)f(x) is positive for x ≥ b, leading to a positive integral. Consequently, to minimize the expected value, setting b equal to the median m results in E[|X - b|] being minimized, as the integral evaluates to zero. The factor of 2 in the expression arises during the development of the right side but does not affect the minimum when b equals m. This confirms that the optimal choice for minimizing the expected absolute difference is b = m.
cielo
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Homework Statement


Let X be a continuous random variable with median m.
Minimize E[|X - b|] as a function of b. Hint: Show that E[|X - b|] = E[|X - m|] + 2 \int (x - b) f(x) dx , where the integral is from b to m.

Homework Equations





The Attempt at a Solution


I wanted to try a solution but I even don't know how to determine whether it is minimum or not. Please help.
 
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treat is as a function of b
f(b) = E(|X-b|)

how would you find the minima of f w.r.t. b?
 
cielo said:

Homework Statement


Let X be a continuous random variable with median m.
Minimize E[|X - b|] as a function of b. Hint: Show that E[|X - b|] = E[|X - m|] + 2 \int (x - b) f(x) dx , where the integral is from b to m.

Homework Equations





The Attempt at a Solution


I wanted to try a solution but I even don't know how to determine whether it is minimum or not. Please help.

Can you show the expression given above? If so, what do you know about the sign of

<br /> \int_b^m (x-b) f(x) \, dx<br />

Finally, note that the right-side is a function of the number b. What happens when you evaluate at that function at a (cleverly) chosen value?
 
statdad said:
Can you show the expression given above? If so, what do you know about the sign of

<br /> \int_b^m (x-b) f(x) \, dx<br />

Finally, note that the right-side is a function of the number b. What happens when you evaluate at that function at a (cleverly) chosen value?

I am having a hard time how to evaluate this integral because the function f(x) is not given.
<br /> \int_b^m (x-b) f(x) \, dx<br />
 
Aaaaah, that's the point: if you were given a specific f(x), any result you obtained would apply only to that particular function , not in general. This exercise is meant to give a general result.

Hint: In the integral

<br /> \int_a^m (x-b) f(x) \, dx<br />

the interval of integration consists of values x \ge b, so

i) What is the sign of (x-b)f(x) over the interval?
ii) What does this say about the sign of the integral of (x-b)f(x)?
iii) Using your answers to `i' and `ii'', if you want to choose b to minimize

<br /> E(|X-m|) + 2 \int_b^m (x-b) f(x) \, dx<br />

as a function of b, what choice does it?
 
statdad said:
Aaaaah, that's the point: if you were given a specific f(x), any result you obtained would apply only to that particular function , not in general. This exercise is meant to give a general result.

Hint: In the integral

<br /> \int_a^m (x-b) f(x) \, dx<br />

the interval of integration consists of values x \ge b, so

i) What is the sign of (x-b)f(x) over the interval?
ii) What does this say about the sign of the integral of (x-b)f(x)?
iii) Using your answers to `i' and `ii'', if you want to choose b to minimize

<br /> E(|X-m|) + 2 \int_b^m (x-b) f(x) \, dx<br />

as a function of b, what choice does it?


okay, let me see if I figure this out right.
i) The sign of (x-b)f(x) over the interval is positive.
ii) The sign of the integral of (x-b)f(x) is positive.
iii) Substituting b to m in the expression <br /> E(|X-m|) + 2 \int_b^m (x-b) f(x) \, dx<br />

results to <br /> E(|X-b|) + 2 \int_b^b (x-b) f(x) \, dx<br />
which becomes E(|X - b|) + 2*(0)
and finally to E(|X - b|)
...so E[|X - b|] is minimized when b = m.
I hope this time I finally got it right with your guidance.. =)
 
However, I'm wondering what is the use of 2 when the integral is just zero?
 
The 2 comes in when the expression on the right is developed.
 
Stat Dad -- Thank you for your guidance on this problem -- I was terribly lost and had a very similar question!
 

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