Section 4.3-Ch3

[Actuary@22]

Hi
I am not clear with the derivation of Var(mew bar -mew).
1.When we move from step 2 to step 3 so where did the N^2 go?
2.Also how does the law of large numbers imply that V/N tends to E(Vi)?
Pleas explain.
Thank you.

 

[Andrew Martin]

Hello

1.

\( E[V_1 + ... + V_N ] = E[V_1] + ... + E[V_N ] = N * E[V_i] \)

So:

\( (E[V_1 + ... + V_N ])^2 = N^2 * E[V_i] ^2 \)

2.

Think about taking a sample from some distribution (assume it has finite mean mu and some varianace sig^2). We can calculate the sample mean as sum(xi) / n. The random variable form of this expression is sum(Xi) / n. From the CLT, the distribution of this is (approximately) N(mu, sig^2 / n). Now as n gets larger and larger this distribution gets tighter and tighter around mu, the true population mean. So the probability of getting an outcome a particular distance away from mu gets smaller and smaller as n gets larger. In the limit, we talk about this as sum(Xi) / n converging in probability to the mean mu.

This is the same thing with V / N as V = sum(Vi) and N is n in my example above.

The LLN says exactly this - that averages from large samples should be close to the expected value and tend to get closer as the sample size increases.

Hope this helps

Andy