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)

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

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