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3.5.12 Suppose there are two urns. Urn I contains 100 chips: 30 are labelled 1, 40...

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3.5.12 Suppose there are two urns. Urn I contains 100 chips: 30are labelled 1, 40 are labelled 2, and 30 are labelled 3. Urn 2contains 100 chips: 20 are labelled 1, 50 are labelled 2, and 30are labelled 3. A coin is tossed and if a head is observed, then achip is randomly drawn from urn 1, otherwise a chip is randomlydrawn from urn 2. The value Y on the chip is recorded. If anoccurrence of a head on the coin is denoted by X = 1, a tail by X =0, and X ? Bernoulli(3/4), then determine E(X | Y), E(Y | X), E(Y),and E(X)

Answer & Explanation Solved by verified expert
3.8 Ratings (665 Votes)

P(H)=3/4=0.75

if head appears then X=1 and urn 1 is chosen

P(label 1 from urn 1)=0.30

P(label 2 from urn 1)=0.40

P(label 3 from urn )=0.30

and P(label 1 from urn 1 and head)=0.75*0.30=0.225

similarly.....we will do other calculations

-----------------------------

joint distribution table obtained is

X=0 X=1
Y=1 0.05 0.225 0.275
Y=2 0.125 0.3 0.425
Y=3 0.075 0.225 0.3
total 0.25 0.75 1
X 0 1 total
P(X=x) 0.25 0.75 1
E(x)=X*P(X) 0 0.75 0.75

E(X)=0.75

----------------------------------------

Y P(Y) E(Y)=Y*P(Y)
1 0.275 0.275
2 0.425 0.85
3 0.3 0.9
total 1 2.025

E(Y)=2.025

---------------------------------------

Y|X

X=0 X=1 P(Y|X=0) P(Y|X=1) Y*P(Y|X=0) Y*P(Y|X=1)
Y=1 0.05 0.225 0.2 0.3 0 0
Y=2 0.125 0.3 0.5 0.4 0.5 0.4
Y=3 0.075 0.225 0.3 0.3 0.6 0.6
total 0.25 0.75 1 1 1.1 1

so, E(Y | X)=?Y*P(Y|X=0) +? Y*P(Y|X=1) = 1.1+1=2.1

---------------------------------------------------

X|Y

X=0 X=1
Y=1 0.05 0.225 0.275
Y=2 0.125 0.3 0.425
Y=3 0.075 0.225 0.3
P(X|Y=1) 0.1818182 0.818182 1
P(X|Y=2) 0.2941176 0.705882 1
P(X|Y=3) 0.25 0.75 1
X*P(X|Y=1) 0 0.818182 0.818182
X*P(X|Y=2) 0 0.705882 0.705882
X*P(X|Y=3) 0 0.75 1.524064

E(X | Y) = ?X*P(X|Y=1)+?X*P(X|Y=2)+?X*P(X|Y=3) = 0.818181818+0.705882353+1.524064171 = 3.0481
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