Let Yi=v¯¯¯(v¯¯¯TXi) , or equivalently v¯¯¯(v¯¯¯⋅Xi)=(v¯¯¯⋅Xi)v¯¯¯ , where v¯¯¯=v∥v∥ is the same as in part (c). We will compare the asymptotic distribution of X¯¯¯¯n you obtain in part (d) to the asymptotic distribution of Y¯¯¯¯n where Y¯¯¯¯n=∑niYin . What is the expectation E[Yi] of Yi ?

step-by-step explanation:

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0

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