REVIEW Stochastic convergences Markov' s inequality Assum-glyL≥ t for all y∈R Py IYL> ci< gop Proof. Assum-that Y is a continuous r v. with pdf f 1 L. D-fin - e ylg lyL> ch and A2e ylg lyL< c).Th-n Eo IYli e/glyLf lyIdyh/ g1yLf 1yLdy ≥/ g lylf lyLy f lyly e cPolyL> ci Th-stat-d r-sult follows from this Chebyshevs inequality -t y b-ar v. with E lYLe u and Var lYLe o Pdy >1< PqY-Hl PLY -uL Ely Example 5 c with probability v n with probability n PdZn-c≥iePc ThREVIEW 3 Stochastic convergences Markov’s inequality Assume g (y) ≥ 0 for all y ∈ R P [g (Y ) ≥ c] ≤ g [Y ] c . Proof. Assume that Y is a continuous r.v. with pdf f (·). Define A1 = {y|g (y) ≥ c} and A2 = {y|g (y) < c} . Then E [g (Y )] =  A1 g (y) f (y) dy +  A2 g (y) f (y) dy ≥  A1 g (y) f (y) dy ≥  A1 cf (y) dy = cP [g (Y ) ≥ c] . The stated result follows from this. Chebyshev’s inequality Let Y be a r.v. with E (Y ) = µ and V ar (Y ) = σ 2 . Then P [|Y − µ| ≥ r] ≤ σ 2 r 2 . Proof. P [|Y − µ| ≥ r] = P  (Y − µ) 2 ≥ r 2  ≤ E (Y − µ) 2 r 2 = σ 2 r 2 . Example 5 Zn =
c with probability 1 − 1 n n with probability 1 n P [|Zn − c| ≥ ε] = P [Zn = n] = 1 n → 0. Thus Zn P→ c