Question

Let the random variable Yn have a distribution that is b(n, p).

(a) Prove that Yn/n converges in probability to p. This result is one form of the weak law of large numbers.

(b) Prove that 1 − Yn/n converges in probability to 1 − p.

(c) Prove that (Yn/n)(1 − Yn/n) converges in probability to p(1 − p).

Answer 1

Explanation:

Let the random variable
`Y_n` have a distribution thst id b(n, p)

a)

`Pr[[(Y_n)/(n)-E[(Y_n)/(n)]\geq E]\leq(1)/(E^2)var[(Y_n)/(n)]\\\\=Pr[((Y_n)/(n)-P)\geq E]\leq (1)/(n^2E^2)np(1-p)=0`

Hence, it is obtained that

`Pr[(1-(Y_n)/(n)-(1-p))\geq E]\leq (1)/(n^2E^2)np(1-p)=0`

Hence, it is obtained that
`i-(Y_n^p)/(n)=1-p`

C)

As it was shown that
`(Y_n^p)/(n)=p` obtained the following

`[(Y_n)/(n)]^2^p=p^2`

Difference the above

`(Y_n)/(n)-((Y_n)/(n))^2^p=p-p^2`

Hence the following expression has been obtained.

`(Y_n)/(n)(1-(Y_n)/(n))^p=p(1-p)`