Statistics 630 - Assignment 7

This assignment includes solved problems on convergence in probability, weak law of large numbers, distribution convergence, exponential distribution, covariance, F distribution, strong law of large numbers, parameterization, and estimation.

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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

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= e16(0.5)
=e80
P(2.5<Ῡ<2.5) =P(2.5-0.5<e80-0.5<2.5+0.5)
P(2/16<(e80-0.5)/16<3/16)
P(0.125<(e80-0.5) <0.1875)
When n=36;
E(Y)=enʎ
=e36(0.5)
=e18
P(2.5<Ῡ<2.5) =P(2.5-0.5<e18-0.5<2.5+0.5)
=P(2/√36<(e18-0.5)/√36<3/√36)
When n=100;
E(Ῡ)=e100(0.5)
=e50
P(2.5<Ῡ<2.5) =P(2.5-0.5<e50-0.5<2.5+0.5)
P(2/√100<(e50-0.5)/√100<3√100)
┌(x)=∫ȹ0e-ttx-1dt
Problem 4.4.14
a) E(X) =α/ʎ
=5/1/10
=50.
b) P(40<Ῡ<40)=P(40-0.5<Ῡ<50+0.5)
P(39.5/√6<Ῡ/√6<50.5/√6)

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c) P(50<Ῡ<50) =P(50-0.5<Ῡ<50+0.5)
=P(49.5/√6<49.5/√6<50.5/√6)
d) P(40-50<Ῡ<40-50) =P(-10<Ῡ<10)
=P(-10/√6<Ῡ<10/√6)
Problem 4.4.17
A sequence of {X20} of 20 random numbers converges in probability to X if
limn͢ ȹ P(|X20-X|<=£) =0
E(M20) =4
By law of large numbers, P(4-£<M20<4+£) ͢ 1 as n approaches to infinity.
Hence when N becomes large to 10^4, the average value M20 is very close to
4.
E(X20)=P(2.5<=M20<=3.3)
=p(2.5-0.5<=4-0.5<=3.3+0.5)
=p(2/√20<=3.5/√20<=3.8/√20)
Problem 4.6.1.
a) Cov(U,V)=∑∑aibj cov(XiXJ)
=7*5(5X1+(c-5)X2)
=35(5X1+(c-5)X2)
=[175 +(35c-175)][x1x2]
b) 35(5x1+(c-5)X2)=0
35C-175=0
C=5
Problem 4.6.3
Cov(U,V)=∑∑aibjcov(XiXJ)

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=28cov(xiXJ)
=28C=0
C=0
Hence c1=c2=c3=c4=c5=0
Problem 4.6.6
Cov(U,V)=∑∑aibjCov(XiXJ)
3n/n=c
C=3.
Problem 4.6.7
Cov(u,v)= ∑∑abcov(xy)
C+1=0
C=-1
Problem 4.6.8
Cov(u,v)=∑∑abcov(XY) =0 if X and Y are independent,
=40+c=0
C=-40.
Problem 4.6.10
X12 is a normal distribution is with mean 0 and variance σ2
X21 +X22 I a linear combination of two independent distribution with mean μ
and variance σ2.
X1/(X20+X30+X40)3 is an F distribution with n degrees of freedom. It
converges to a standard normal distribution as n tends to infinity.᷉^^͠
3X10/√(X20+X30+X40] is an F distribution with n and m degrees of freedom.

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Problem 4.6.11
a) K=1,n=0
b) y+1=0.05
y=-0.95.
c) the values of b and c are true. F distribution has got 2 degrees of freedom.
d) w+1+1+1=0.05
w+3+0.05
w=-2.95.
Problem 4.6.12
a) D converges almost to 1 as n tends to infinity. This obeys the strong
law of large numbers. Hence e(Y1) =1.
b) C also converges almost to 1 but greater than D as n tends to infinities
c) Z converges to 0 as n increases.
d) P(C<D)=P(C-£<Mn<C+£)
=P(C/√n<Mn/√n<C/√n)
e) P(U>633.25) =P(U-0.5>Mn>633.25+0.5)
=(U/√n>(Mn-0.5)/√n>633.75/√n)
Problem 5.3.11
You are asked to parameterize to identify to the source of the sample.

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The model Ψ =ln{θ/(1-θ)} is given.
The statistical model {Pθ:θ£ῼ } is identified, whereby;
Θ is the parameter of the model,ῼ is the parameter space for all values of θ
[0,1]
The probability function for the kth sample is fθ(xj)=θxi(1-θ)1-xi
Problem 5.4.11.
a) FX(hjj)=∫-ȹhihX(x)dx.
At N(3,2)
=1[X2/2]32
=4.5-2
=2.5 Units.
b) >y=rnorm(1000, mean=3,sd=sqrt(2));
>plot.ecdf(y);
>x=seq(-5,10, length=1);
>lines(x,pnorm(x,sd=sqrt(2)),col=2);
c) >y=rnorm(1000, mean=3,sd=sqrt(2));
>plot.ecdf(y);
>x=seq (-5,10, length=0.1);
>line(x,pnorm(x,mean=3,sd=sqrt(2)),col=2)

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d) The area of histogram (d) is slightly smaller than that of (c). This is
because of higher frequency in (c) as compared to (b) hence the
accuracy levels.
e) High number of observations may lead to a lot of errors, hence limits
the frequency of intervals when testing a big population.
Problem 5.5.5
>Y=rnorm(1000,mean=0.075,sd=sqrt(2));
>plot.ecdf(y);
>x=seq(-1.4,2.1,length=0.05);
>lines(x,pnorm(x,mean=3,sd=sqrt(2)),col=2);
>median(1.0, −1.2 ,0.4, 1.3, −0.3,-1.4,0.4,-0.5,-0.2,-1.3,0.0,-1.0,-
1.3,,2.0,1.0,0.9,0.4,2.1,0.0,-1.3);
>0.2
>Mean(1.0, −1.2 ,0.4, 1.3, −0.3,-1.4,0.4,-0.5,-0.2,-1.3,0.0,-1.0,-
1.3,,2.0,1.0,0.9,0.4,2.1,0.0,-1.3);
>0.075
Problem 5.5.20
a) >rand(50,mean=4,sd=sqrt(2));
>x=seq();
>x0.9=(x,pnorm,α=0.01)
>
b) The statistic estimate is higher than the normal distribution estimate.
c) I would prefer (b).Random estimates do fluctuates on recalculation
hence errors in the results.

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