Convergence in probability and in distribution 1

Question

0.0/2.0 points (graded)
Let (𝑇𝑛)𝑛≥1=𝑇1,𝑇2,… be a sequence of r.v.s such that

𝑇𝑛∼𝖴𝗇𝗂𝖿(5−12𝑛,5+12𝑛).

Given an arbitrary fixed number 0<𝛿<1, find the smallest number 𝑁 (in terms of 𝛿) such that 𝐏({|𝑇𝑛−5|>𝛿})=0 whenever 𝑛>𝑁.

Answer 1

For a random variable 𝑇𝑛 ~ Uniform(5-1/2𝑛, 5+1/2𝑛), we have 𝐄[𝑇𝑛] = (5-1/2𝑛 + 5+1/2𝑛)/2 = 5 and 𝐕[𝑇𝑛] = ((5+1/2𝑛 - (5-1/2𝑛))^2)/12 = 1/(4𝑛) for all 𝑛.

First, let's find the smallest number 𝑁 such that 𝐏(|𝑇𝑛-5|>𝛿) = 0 whenever 𝑛 > 𝑁.

Since the distribution of 𝑇𝑛 is symmetric around 5, we can rewrite the probability as 𝐏(𝑇𝑛 < 5-𝛿 or 𝑇𝑛 > 5+𝛿).

𝐏(𝑇𝑛 < 5-𝛿) + 𝐏(𝑇𝑛 > 5+𝛿) = 2𝐏(𝑇𝑛 > 5+𝛿).

To find the smallest 𝑁, we need to find the smallest 𝑛 such that 𝐏(𝑇𝑛 > 5+𝛿) = 0.

𝐏(𝑇𝑛 > 5+𝛿) = 1 - 𝐏(𝑇𝑛 ≤ 5+𝛿) = 1 - 𝐏(𝑇𝑛 ≤ 5+1/2𝑛+1/2𝛿) = 1 - (5+1/2𝑛+1/2𝛿 - (5-1/2𝑛))/(2/𝑛) = 1 - (1/2)/(1/𝑛) = 1 - (1/2)(𝑛/1) = 1 - 1/2𝑛.

Therefore, we need to solve the inequality 1 - 1/2𝑛 < 1, which simplifies to 1/2𝑛 > 0.

Solving for 𝑛, we get 𝑛 > 0.

Therefore, the smallest number 𝑁 is 1.