Suppose that the possible values of X are x1 x2 x3 and PX x1

Suppose that the possible values of X are x1, x2, x3 and
P(X =x1) = 3

Solution

let p(x_i)=p_i where i=1,2,3

p₁=3*p₂=4*p₃

as we konw

p₁+p₂+p₃=1

hence

p₁*(1+1/3+!/4)=1

p₁=12/19

p₂=4/19

p₃=3/19

for a=E(x),you get minimum error

because

E(a-x)^2=E [ {((Y - E [Y ]) + (E [Y ] - a))} ^2 ]

            = E [ (Y - E [Y ])^2 ] + E h (E [Y ] - a)^ 2 ]+ 2