Question

2. Let X be a continuous r.v. with pdf fx() for which EX exists. We want to find the real number a that minimizes EX-al over all real a. Determine an equation involving the CDF Fx(-) that can be solved to obtain a*.

Answer 1

From the definition of expectation, we know that

$E(X)=\int_{-\infty}^{\infty}xf_{X}(x)dx$

Hence, using the definition of modulus function we have  

E (X -a) ー00 4

Differentiate with respect to we get

E(X - al)

Let $a^*$ be the value which minimizes

Ефе-а) !

Then

Oo Pr(X <a*) = Pr(X> a *)

Now, we know that $F_{X}(x)=\Pr(X\leq x)=\int_{-\infty}^{x} f_{X}(x)dx$

Hence, by (A) the required equation is

LFxa