Q. 7.2 - A First Course in Probability

Question

7.2. Suppose that $X$ is a continuous random variable with

density function $f$. Show that $E[IX-a\mid ]$ is minimized

when $a$ is equal to the median of $F$.

Hint: Write

$E[I X-a l]=|x-a| f\left(x\right) d x$

Now break up the integral into the regions where $x<a$

and where $x>a$, and differentiate.

Step-by-Step Solution

Answer

Differentiate $E[IX-a\mid ]$ respective to $a$ is proved as minimized equal to the median.

Step 1: Given Information

$X$ is a continuous random variable with density function with the formula using $E[X-a ]=|x-a| f\left(x\right) d x$.

Step 2: Explanation

We have that,

$E\left(\right|X-a\left|\right)={\int }_{\mathrm{ℝ}}|x-a|f\left(x\right)dx={\int }_{x<a}|x-a|f\left(x\right)dx+{\int }_{x\ge a}|x-a|f\left(x\right)dx$

$={\int }_{x<a}(a-x)f\left(x\right)dx+{\int }_{x\ge a}(x-a)f\left(x\right)dx$

Using the differentiation respective to $a$ and setting it equal to zero, we have that

$\frac{d}{da}E\left(\right|X-a\left|\right)={\int }_{x<a}\frac{d}{da}(a-x)f\left(x\right)dx+{\int }_{x\ge a}\frac{d}{da}(x-a)f\left(x\right)dx={\int }_{x<a}f\left(x\right)dx-{\int }_{x\ge a}f\left(x\right)dx=0$

Step 3: Explanation

so the minimum is obtained for $a$ such that

${\int }_{x<a}f\left(x\right)dx={\int }_{x\ge a}f\left(x\right)dx$

and it is the definition of the median of the distribution. Hence, we have proved the claimed.

Step 4: Final answer

Differentiate $E[IX-a\mid ]$ respective to $a$ is proved as minimized equal to the median.