London School of Economics

• ST 202

• ST202 – LT Solutions to problem set 1 1. If A is the event of interest, it is easier to find the probability of Ac , i.e. no member of the population appearing more than once in the sample. Treating the order as important, we have P(A) = 1 − P(A c ) = 1 − |Ac | |Ω| = 1 − NPn Nn = 1 − N! (N − n)!Nn . The fraction in this expression can be written as N N N − 1 N . . . N − n + 1 N , which converges to 1 as N → ∞, so P(A) → 0. This confirms our intuition that, when sampling with replacement from a very large population, the probability of a member appearing more than once in the sample is negligible. 2. From the central limit theorem, if we define Zn = Y¯ n − µ p σ 2/n ⇐⇒ Y¯ n = µ + σ √ n Zn , we know that KZn (t) → t 2/2 as n → ∞. Rearranging, we find that KY¯n (t) = log E  e tY¯n  = log E  e t(µ+ √σ n Zn)  = µt + log E  e (tσ/√ n)Zn  = µt + KZn (tσ/√ n) → µt + σ 2 n t 2 2 as n → ∞. We deduce that Y¯ n d −−→ µ, a constant, because the variance of Y¯ n is σ 2/n, which converges to 0 as n → ∞. Furthermore, we have shown that convergence in distribution to a constant implies convergence in probability, so we conclude that Y¯ n p −−→ µ. 3. Probably the easiest approach is to prove that the first expression is equal to the second one, then proceed from there. We have Xn i=1 Yi(Yi − Y¯ ) = Xn i=1 (Y 2 i − YiY¯ ) = Xn i=1