have a distribution thst id b(n, p)![Pr[[\frac{Y_n}{n}-E[\frac{Y_n}{n}]\geq E]\leq\frac{1}{E^2}var[\frac{Y_n}{n}]\\\\=Pr[(\frac{Y_n}{n}-P)\geq E]\leq \frac{1}{n^2E^2}np(1-p)=0](/tpl/images/0432/7085/7815a.png)
![\frac{Y_n^P}{n}=PB)As mentioned in a), use chebyshevs inequality.[tex]Pr(\frac{Y_n}{n}-E(\frac{Y_n}{n})\geq E)\leq \frac{1}{E^2}var(\frac{Y_n}{n})](/tpl/images/0432/7085/4934a.png)
![Pr[(1-\frac{Y_n}{n}-(1-p))\geq E]\leq \frac{1}{n^2E^2}np(1-p)=0](/tpl/images/0432/7085/34352.png)
obtained the following ![[\frac{Y_n}{n}]^2^p=p^2](/tpl/images/0432/7085/34cad.png)
Let the random variable yn have a distribution that is b(n, p).
(a) prove that yn/n con...
Mathematics, 25.12.2019 07:31 Amyra2003
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).
Answers: 2
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