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I've been solving problems from the book by DeGroot and Schervish and I can't understand why m is the upper limit of integration in the solution to this problem. Why not the lower one?

Here is the problem:

Suppose that a random variable X has a continuous distribution for which the p.d.f. f is as follows:

$f(x) = 2x$ for $ 0< x <1, 0 $ otherwise

Determine the value of $d$ that minimizes $E(|X − d|)$.

Here is the solution:

$ \int_0^m 2x \, dx=0.5 $

Thank you very much in advance.

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    ahaha;) cool! mathematics doesnt like confusion although it has this inside of his blood..2012-08-18

3 Answers 3

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The expected value is given by (assuming $d$ is in $[0,1]$): $E = \int_0^1{2x|x-d|}dx = \int_0^d{2x(d-x)}dx + \int_d^1{2x(x-d)}dx$ $ = -\frac{2}{3} + d - \frac{2d^3}{3}$ Derive by $d$ and compare to zero: $E' = 1 - 2d^2 = 0 \rightarrow d = \frac{1}{\sqrt{2}}$

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    @MichaelHardy yes you are right. WHILE I posted my answer I did it based on nbubis's answer. I didnt check it by myself. I only focused on the second part of the question which say minimum! although it seemed to be maximum based on his answer. But you are completely right.2012-08-18
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The value of $d$ that that minimizes $E(|X-d|)$ is the median value of $X$, i.e. the value $m$ such that $ \Pr(X\le m) = 0.5 = \Pr(X\ge m). $

Thus for continuous distirbutions it is the value of $m$ for which $ \int_{-\infty}^m f_X(x)\,dx = 0.5 = \int_m^\infty f_X(x)\,dx. $

In your case, you want $ \int_0^m 2x\,dx = 0.5 = \int_m^1 2x\,dx. $ So $m^2 -0^2 = 0.5 = 1^2-m^2$. Consequently $m= \sqrt{0.5}= \sqrt{2}/2$.

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I will note one important thing about the question. You will get $E^{'}[X]=1-2d^2$ as @nbubis wrote down already. When you find $E^{''}[X]=-4d$ and since $d\in[0,1]$ you have $E^{''}[X]<0$ which indicates that you have a maximum instead of minimum as you stated in the question EDIT: Result was based on $E^{'}[X]=1-2d^2$ Since it is $E^{'}[X]=2d^2-1,$ there is a minimum. I didnt solve the first derivative by myself just copied.

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    This is nonsense. You get a minimum, not a maximum. The derivative should be $2d^2-1$, not $1-2d^2$.2012-08-18