Addition of Independent Normal Random Variables

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Addition of Independent Normal Random Variables

• Theorem 1
• Let X and Y be two independent random variables
  with the N(?, ?2) distribution. Then,
  is N(2?, 2?2) .
• Proof

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Subtraction of Independent Normal Random Variables

• Theorem 2
• Let X and Y be two independent random variables
  with the N(?, ?2) distribution. Then,
  is N(0, 2?2) .
• Proof
• Similar to the proof of Theorem 1.

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Generalization of Theorem 1

• Let X1, X2, , Xn be n mutually independent
  normal random variables with means µ1, µ2, ,
  µn and variances , respectively.
• Then, is
  .

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Chi-Square Distribution with High Degree of
Freedom

• Assume that X1, X2, , Xk are k independent
  random variables and each Xi is N(0, 1).
• Then, the random variable Z defined by
  is called a chi-square random variable
  with degree of freedom k and is denoted by
  .

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Addition of Chi-Square Distributions

• A
• A

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Example of Chi-Square Distribution with Degree of
Freedom 2

• Let X,Y be two independent standard normal random
  variables, and Z X2 Y2.

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Example of Chi-Square Distribution with Degree of
Freedom 3

• Let X,Y,Z be three independent standard normal
  random variables, and W X2 Y2 Z2.

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• Distribution Function of is

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• Note that,
• The surface area of a sphere in the
  k-dimensional vector space is

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Moment-Generating Function of the
Distribution

• The moment-generating function of a distribution
  with p.d.f f(x) is defined to be
• Note that

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• Theorem The moment-generating function
• Mk(z) of is
• Proof since the p.d.f of is only
  defined in 0,8)
• we now consider two cases
• (1) k2h
• (2) k2h1

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• Case(1)

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• By applying the same technique repetitively we
  get
• we can prove that for both k2h and k2h1
  cases,

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• implies that

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Estimation of the Expected Value of a Normal
Distribution

• Let X be a normal random variable with unknown µ
  and s2.
• Assume that we take n random samples of X and
  want to estimate µ ands2 of X based on the
  samples.
• Let denote an estimation of µ.
• Then, the likelihood function of n samples
• is

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continues

• Therefore, is a maximum likelihood
  estimator of µ.

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continues

• Let X1, X2, Xn be the random variables
  corresponding to sampling X n times.
• Since is called an
  unbiased estimator of µ.
• Furthermore, since ,the
  confidence interval of µ approaches 0 , as n ?8,
  provided that is used as the
  estimator of µ.
• The confidence interval of µ is

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Estimation of the Variance of a Normal
Distribution
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That is the confidence interval of S2 approaches
0 as n?8, provided that S2 is used as the
estimator of s2
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An Important Observation
There are more general situation in which a
degree of freedom of chi-square distributions is
lost for each parameter estimated.
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Test of Statistical Hypotheses

• The goal is to prove that a hypothesis H0 does
  not hold in the statistical sense.
• We first assume that H0 holds and design a
  statistical experiment based on the assumption.
• The probability that we reject H0 when H0
  actually holds is called the significance level
  of the test. Typically, we use ? to denote the
  significance level.

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An Example of Statistical Tests

• We want to prove that a coin is biased.
• We make the following hypothesis The coin is
  unbiased.
• We design an experiment that is to toss the coin
  n times and we claim the coin is biased, i.e.
  rejecting the hypothesis, if we observe either
  one side is up k times or more, where k gt ½ n.

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continues

• Let X be the random variable corresponding the
  number of times that one particular side is up in
  n tosses.
• Under the hypothesis The coin is unbiased,
  approaches N(0, 1).
• The significance level of the test is

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continues

• Assume that we want to achieve a significance
  level of 0.05 and we toss the coin 100 times.
  Since ,
• Assume that we want to achieve the same level of
  significance and we toss the coin 1000
  times.Then,

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Test of Equality of Several Means

• Assume that we conduct k experiments and all the
  outcomes of the k experiments are normally
  distributed with a common variance. Our concern
  now is whether these k normal distributions,
  N(?1,?2), N(?2,?2),, N(?k,?2), have a common
  mean, i.e. ?1 ?2 ?k.

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• One application of this type of statistical tests
  is to determine whether the students in several
  schools have similar academic performance.
• The hypothesis of the test is . ?1 ?2 ?k.

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• Let ni denote the number of smaples that we take
  from distribution N(?i,?2).
• As a result, we have the following radom
  variables
• X11, X12,, X1n1 samples from N(?1,?2).
• X21, X22,, X2n2 samples from N(?2,?2).
• Xk1, Xk2,, Xknk samples from N(?k,?2).

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Chi-Square Test of Independence for 2x2
Contingence Tables

• Assume a doctor wants to determine whether a new
  treatment can further improve the condition of
  liver cancer patients. Following is the data the
  doctor has collected after a certain period of
  clinical trials.

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continues

• The improvement rate when the new treatment is
  applied is 0.85 and the rate is 0.77 when the
  conventional treatment is not applied. So, we
  observe difference. However, is the difference
  statistically significant?
• To conduct the statistical test, we set the
  following hypothesis H0 The effectiveness of
  the new treatment is the same as that of the
  conventional treatment.

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continues

• Under H0, the two rows of the 2x2 contingence
  table corresponds to two independent binomial
  experiments, denoted by X1 and X2 , respectively.
• Define parameters as follows.

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• Let Y1, Y2, , Yn1 be n1 samples taken from a
  normal distribution N( p, p(1-p) ).
• Then, is

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• Therefore, the distribution of Z1 approaches that
  of . Similarly, let be
  samples taken from a normal distribution N( p,
  p(1-p)). Then the distribution of Z2 approaches
  that of
• Since and are
  samples from a statisctical test of two means,
• is ,
  where

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• Since the distributions of Z1 and Z2 approach
  those of and , respectively,
• approaches .
• is an estimator of the mean of Yi and Wj,
  which is p. Therefore, if we use
  as the estimator,
• then we have
  approaches

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• According to our previous observation,

we have
In conclusion,
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continues

• Applying the data in our example to the equation
  that we just desired, we get
• Therefore, we have over 97.5 confidence that the
  new treatment is more effective than the
  conventional treatment.

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continues

• On the other hand, if the number of patients that
  have been involved in the clinical trials is
  reduced by one half, then we get
• Therefore, we have less than 90 confidence when
  claiming that the new treatment is more effective
  than the conventional treatment.

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A Remark on the Chi-Square Test of independence

• The chi-square test of independence only tell us
  whether two factors are dependent or not. It does
  not tell us whether they are positively
  correlated or negatively correlated.
• For example, the following data set gives us
  exactly identical chi-square value as our
  previous example.

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Measure of Correlation

• Correlation (A,B) P(AB) / P(A)P(B)
  P(AB)P(B) P(BA)P(A).
• Correlation(a,b) 1 implies that A and B are
  independent.
• Correlation(a,b) gt 1 implies that A and B are
  positively correlated.
• Correlation(a,b) lt 1 implies that A and B are
  negatively correlated.

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Generalizing the Chi-Square Test of Independence

• Given a 3?2 contingency table as follows.
• Let Z1, Z2 and Z3 be the three random variables
  corresponding to the experiments defined by the
  three rows in the contingency table.

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• Then

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The Chi-Square Statistic for Multinomial
Experiments

• Given the 3?2 contingency table .
• We can regard the table as the outcomes of 2
  independent multinomial experiments.

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• Therefore, we derive the following fact about a
  multinomial experiments as follows

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Contingence Table with m Rows and n Columns

• The chi-square statistichas degree of freedom
  (m-1)(n-1).

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Determining the Degree of Freedom of a 2-D
contingency Table

• Once the counts in cells marked by ? are
  determined. Then, the remaining counts are
  determined.

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Determining the Degree of Freedom on 3-D
contingency Table

• (rs 1)(t 1) (r 1)(s 1) rst r s
  t 2