Maximum Likelihood Question 8 An urn contains 200 marbles. An unknown number of
ID: 3069061 • Letter: M
Question
Maximum Likelihood Question 8 An urn contains 200 marbles. An unknown number of the marbles in the urn are black. 100 draws are made from the urn. It is not known whether the draws are made with or without replacement. This procedure is repeated three times. The count of black marbles observed is: 15, 12,17. The probability distribution generating the observations is either binomial (draw with replacement) or hypergeometric (draw without replacement). See https://en.wikipedia.org/wiki/Hypergeometric distribution for properties of the hypergeometric distribution. 8A Identify the single maximum likelihood parameter p(s) -B/(B+W) for the binomial At that that parameter, what is the log likelihood of observing 15,12,17? ANSWER 8B Identify the single maximum likelihood parameter p[s on first draw)-B/(B+W) for the hypergeometric distribution. At that parameter, what is the log likelihood of observing 15,12,17? ANSWER 8C Which is more probable: the observed draws were done with replacement, or without replacement? ANSWERExplanation / Answer
the urn has 200 marbles. Some of which are black and rest are white. 100 marbles are drawn 3 times.
the count of black marbles are 15,12,17 respectively.
8a) assuming that the draws are done with replacement.
then let X denotes the number of black marbles in the 100 drawn marbles
then X~Bin(100,p) where p is the probability of drawing a black marble out of 200 marbles
so p=number of black marbles/(number of black marbles+number of white marbles)=B/(B+W)
for first draw maximum likelihood estimate of p is 15/100
for the second draw the maximum likelihood estimate of p is 12/100
for the third draw it is 17/100
hence one single maximum likelihood estimate of p is (0.15+0.12+0.17)/3=0.1467 [answer]
assuming the 3 repeatations are independent.
hence the likelihood of getting 15,12,17 black marbles are
P[X=15]*P[X=12]*P[X=17]
=[100C150.146715(1-0.1467)100-15]*[100C120.146712(1-0.1467)100-12]*[100C170.146717(1-0.1467)100-17]
=0.1106*0.0903*0.0858
=0.0008569
so log likelihood is log(0.0008569)=-7.062189281 [answer]
8b) let the draws are done by without replacement.
let X be number of black marbles in 100 drawn marbles
then X~hypergeometric(200,100,p) where p is the probability of black marble is first draw.
for first draw maximum likelihood estimate of p is 15/100
for the second draw the maximum likelihood estimate of p is 12/100
for the third draw it is 17/100
hence one single maximum likelihood estimate of p is (0.15+0.12+0.17)/3=0.1467 [answer]
assuming the 3 repeatations are independent.
hence the likelihood of getting 15,12,17 black marbles are
P[X=15]*P[X=12]*P[X=17]
=[(200*0.1467C15*200*(1-0.1467)C100-15)/200C100]*[(200*0.1467C12*200*(1-0.1467)C100-12)/200C100]*[(200*0.1467C17*200*(1-0.1467)C100-17)/200C100]
=0.155748325*0.097190128*0.097190128
=0.001471186
so loglikelihood is log(0.001471186)=-6.521686149 [answer]
8c) since the log likelihood of drawing without replacement is larger than that of with replacment hence more probable was that the draws were done without replacement