Question

Suppose X is a random variable such that E(X) and E(X² ) both exist, and are finite. Consider the function f(c) of a real number c given by

f(c) = E[(X ? c)² ].

(a) (2 pts.) Find this function f(c) when X ? Bin(3, 1/2). Among the ’zoo’ of functions that you know about, what kind of function is it?

(b) (8 pts.) Find the value of c which MINIMIZES the function f(c). Hint: expand out the (X ? c)² . Which parts are constant, which change with c?

Answer 1

a) f(c)=E(X-c)^2

=E(X^2-2Xc+c^2)=E(X^2)-2c*E(X)+E(c^2).=E(X^2)-2c*E(X)

=E[X(X-1)+X] -2c*E(X)=E[X(X-1)]+E(X) -2c*E(X).

X~Binomial (3,1/2).

E(X)=3*1/2=1.5

V(X)=3*1/2*(1-1/2)=3/4=0.75.

Then, E(X(X-1))+E(X)=V(X)+{E(X)}^2

=0.75+1.5^2=3.

f(c)=3-(2*c*1.5)=3-3c=3(1-c).

f(c)=Mean Squared Deviation about c.

b)

f(c)=E(X-c) ^2

=E[(X-E(X))+(E(X)-c)] ^2

=E(X-E(X) )^2+(E(X) -c)^2+2*E(X-E(X))*E(E(X)-c).

=V(X)+(E(X) -c)^2+0.

V(X) is a positive quantity and is free of c.

So f(c) is minimum iff (E(X)-c)^2 is minimum, that is 0.

So,iff E(X)=c then, f(c) is minimum.

Here ,f(c) is minimum iff c=1.5.