# Homework Help: Minimize expected value of the absolute difference

1. Aug 21, 2010

### cielo

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

2. Aug 21, 2010

### lanedance

treat is as a function of b
f(b) = E(|X-b|)

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

3. Aug 21, 2010

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

$$\int_b^m (x-b) f(x) \, dx$$

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?

4. Sep 16, 2010

### cielo

I am having a hard time how to evaluate this integral because the function f(x) is not given.
$$\int_b^m (x-b) f(x) \, dx$$

5. Sep 16, 2010

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

$$\int_a^m (x-b) f(x) \, dx$$

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

$$E(|X-m|) + 2 \int_b^m (x-b) f(x) \, dx$$

as a function of $$b$$, what choice does it?

6. Sep 19, 2010

### cielo

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 $$E(|X-m|) + 2 \int_b^m (x-b) f(x) \, dx$$

results to $$E(|X-b|) + 2 \int_b^b (x-b) f(x) \, dx$$
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.. =)

7. Sep 20, 2010

### cielo

However, I'm wondering what is the use of 2 when the integral is just zero?

8. Sep 20, 2010