Probability

To put it simple, first,
let g(i) = [x(i) - u]^2
p(i) = P{ p = x(i) }

Then, g(i) can be divided into 2 subsets,

let g1(i) belongs to g(i) <= t , and, g2(i) belongs to g(i) > t ..... (a)

By definition,
Var(x) = Sum{ p(i) * g(i) } for all i = 0 ,1,2 ...... ,n
= Sum{p(i) * g1(i)} + Sum{p(i) * g2(i)}

since g(i) , g1(i) , g2(i) always >= 0,
then,

Var(x) >= Sum{p(i) * g2(i)}

Divide both sides by t, since t > 0, then
Var(x)/t >= Sum{p(i) * [g2(i) / t]}

by (a), since, g2(i) > t , then [g2(i) / t] > 1,
so, p(i) * [g2(i) / t] > p(i)

==> Var(x) / t > Sum{p(i) | g2}

by definition, Sum{p(i) | g2} = P[ g(i) > t ] , therefore,

Var(x) / t > P[ g(i) > t]

Let s^2 = Var(x)

Let t = k^2 * Var(x)
=> Var(x) / t = 1 / k^2
P[g > t] > Var(x) / t
P[g > (ks)^2] > 1 / k^2

then, (-1/k^2) < - P[g > (ks)^2]
==>
[1 - (1/k^2)] < {1 - P[g > (ks)^2]}
==>
[1 - (1/k^2)] <= P[g <= (ks)^2]
==>
[1 - (1/k^2)] <= P[-ks <= sqrt(g) <= ks]
==>
[1 - (1/k^2)] <= P[-ks <= (x-u) <= ks]

==>
[1 - (1/k^2)] <= P[u - ks <= x <= u + ks]