Answer:
Step-by-step explanation:
Let the random variable [tex]Y_n[/tex] have a distribution thst id b(n, p)
a)
[tex]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[/tex]
Hence, it is obtained that [tex]\frac{Y_n^P}{n}=P
B)
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})[/tex][tex]Pr[(1-\frac{Y_n}{n}-(1-p))\geq E]\leq \frac{1}{n^2E^2}np(1-p)=0[/tex]
Hence, it is obtained that [tex]i-\frac{Y_n^p}{n}=1-p[/tex]
C)
As it was shown that [tex]\frac{Y_n^p}{n}=p[/tex] obtained the following
[tex][\frac{Y_n}{n}]^2^p=p^2[/tex]
Difference the above
[tex]\frac{Y_n}{n}-(\frac{Y_n}{n})^2^p=p-p^2[/tex]
Hence the following expression has been obtained.
[tex]\frac{Y_n}{n}(1-\frac{Y_n}{n})^p=p(1-p)[/tex]
