Statistics 630 - Assignment 7

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Statistics 630 - Assignment 7
(due Thursday, March 21, 2019, 8am CST)
Name ________________________________________________
Email Address ___________________________________________________________
Statistics 630 - Assignment 7

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Problem 4.2.2
{P(Xn-Y)>=£}={P(Xn=Yn-Y)>= £}=0 as n͢ ȹ for all £>0
Hence the sequence {Xn} converges in probability Y.
Therefore, the two distributions Xn and Y do not degenerate.
Problem 4.2.10
Let Zn be a sequence of the squares of independent random variables, each
having the same mean μ Zn =X1+..............+Xn
M= 1/n(X1+...........+Xn) Whereby M is the sample mean.
The weak law of large numbers states L: if {Zn} is identically independent
distribution or is independent with constant mean and bounded variance,
then the sample mean M =1/n(X1+.........+Xn) converges in probability to E(Zn
).
Wirth variance v less than or equal to ȹ, then for {£>0.limn=ȹP (Zn-
m)>=£}=0
Therefore,
E(Zn)=m
Var(Zn)= 1/n2{var(X1)+var(X2)+.......+var(Xn)}
<=1/n2(v+V+....v)
=1/n2(nv)
=v/n
Problem 4.2.12
E(Mn)=1/5
By law of large numbers, P(1/5-£<Mn<1/5+£) increases from 1/5 as n͢ 20
P(1/5-£<Mn<1/5+£) ͢ 1 as n͢ ȹ
P(1/5-£<Mn<1/5+£) ͢ 0 as n͢ 1

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P(1/5-£<Mn<1/5+£) =0.198
P(1/5-£<Mn<1/5+£) =0.21
Mn=0.204
Hence for a large number n, the average value Mn is very close to 1/5.
Problem 4.4.4
P(Wn<=w)={0 (1+x/n) otherwise 0<x<1
Let W,W1,W2,....be random variables.
The sequence {wn converges in the distribution to W if for all w£R1.
Hence P(W=w)=0 ,limn͢ ȹP(Wn=w)=P(W=w)
Thus Wn ͢ D W
Problem 4.4.6
The variables in the sequence each has a mean of μ and variance σ2
We can use the sample mean to approximate mean μ
And the the central Limit theorem:
=liln=ȹP(Mn-3σ/√n<u<Mn+3σ/√n)
Using Table D,
ɸ(3)-ɸ(-3)=0.9987-(1-0.9987)
=0.9974
0.9974>-4470 Hence P(∑i=1900Zi>=-4470.
4.4.12
Exp(16,0.5) is the exponential distribution of service time at n=16,ʎ=0.5
Y1,Y2,...is a random sample of service time.
E(Y)=enʎ