Assignment On Bernoulli Distribution | Questions & Answers


Added on  2019-09-30

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AnswersQ1. a.i.XiA has a Bernoulli distribution with parameter pA or in other words the distribution is bonomial with parameters unity and success probability pA.ii.By definition if a random variable Y takes the values y1,y2.. with corresponding probabilities p1,p2.., then E(Y)=i=1yipi. Here XiA takes the values 1 and 0 with probabilities pA and 1pA. Then E(XiA)=1xpA+0x(1pA)=pAE(XiA2)=12xpA+02x(1pA)=pAHence Var(XiA)=E(XiA2)[E(XiA)]2=pA(1pA)b. i.If WA denotes the random variable representing the total number of games wonby player A, then WA=i=1nAXiA.ii.If we assume that XiA are independently distributed and probability of wining a game remains fixed at pA for player A , then from the additive property of binomial distribution, WA is distributed as Binomial with parameters nA and pA.iii.Results used: For any k random variables Yi,i=1,2,..,k with finite expectation, thenE(i=1kYi)=i=1kE(Yi).If these k random variables are independent with finite variances, thenVar(i=1kYi)=i=1kVar(Yi).Then for the given problem, k=nA and E(i=1kXiA)=i=1kE(XiA)=nApA using (bi) above. AlsoVar(i=1kXiA)=i=1kVar(XiA)=nApA(1pA) using (bi) above.

c.Null hypothesis H0:pA=pB, Alternative hypothesis Ha:pA>pB as the higher the winning chance, the better is the player.i. Define h(pA,pB)=pApB. If p¿k is the maximum likelihood estimator of pk based on nkobservations then p¿k=Wk/nk. Since Wk is distributed as Binomial with parameters nk and pk, it follows from the normal approximation to binomial (p¿kpk) is approximatelyN(0,pk(1pk)/nk), independently for each k=A,B. Thus (p¿Ap¿BpA+pB) is approximatelyN(0,pA(1pA)/nA+pB(1pB)/nB) and hence h(p¿A,p¿B)h(pA,pB)pA(1pA)/nA+pB(1pB)/nB is approximately statndard normal. Replacing the unknown pk by p¿k, we get the Wald test stsistic for H0:h(pA,pB)=0 asT={h(p¿A,p¿B)0p¿A(1p¿A)/nA+p¿B(1p¿B)/nB}2=¿¿It can be shown that under the null hypothesis T is approximately a chi square random variable with one degree of freedom. Thus Wald test rejects the null if¿¿Since Player A wins 19 games out of 27 games played and player B wins 28 games out of 54games played, we have WA=19,nA=27 and WB=28,nB=54 and hence p¿A=19/27 andp¿B=28/54.Thus observed value of T comes out as¿¿Since χ1,.052=3.841459 and observed T is less than this, we fail to reject the null at 5% significance level. Therefore, the players A and B have significantly equal ability in winning casino games.Q2. a. i.In many sistuations, a point estimator does not provide enough information about the population parameter. However, a confidence interval gives two limits (based on data), within which the true value is expected to lie with high chance. Therefore, we get an interval within which the true value is expected rather than putting a single value (i.e.pint estimate) for the true parameter.

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