Consider two random variables:
𝑋
can take values
(
4
,
5
)
, and
𝑌
can take values
(
1
,
2
,
3
)
. The joint distribution of
𝑋
and
𝑌
is shown in the table below

𝑌

1 2 3

𝑋
4 0.20 0.25 0.3
5 0.05 0.10 0.1

Using this information, please compute the unconditional expected value
𝐸
(
𝑋
)
and the conditional expected value
𝐸
(
𝑋
|
𝑌
=
3
)
. (Please round your results to the 4th decimal place.)

Question

Consider two random variables: 
𝑋
 can take values 
(
4
,
5
)
, and 
𝑌
 can take values 
(
1
,
2
,
3
)
. The joint distribution of 
𝑋
 and 
𝑌
 is shown in the table below
	
𝑌

1	2	3

𝑋
	4	0.20	0.25	0.3
	5	0.05	0.10	0.1

Using this information, please compute the unconditional expected value 
𝐸
(
𝑋
)
 and the conditional expected value 
𝐸
(
𝑋
|
𝑌
=
3
)
. (Please round your results to the 4th decimal place.)

BlackTom AI · Accepted Answer

We start by unpacking the joint distribution to compute the quantities requested.
First, compute the unconditional E(X): sum over X values weighted by their marginal probabilities. For X = 4, the joint probabilities across Y are 0.20, 0.25, 0.30, which sum to 0.75. For X = 5, the joint probabilities across Y are 0.05, 0.10, 0.10, which sum to 0.25. Therefore the marginal probabilities are P(X=4)=0.75 and P(X=5)=0.25. Now multiply and add: E(X) = 4*(0.75) + 5*(0.25) = 3.00 + 1.25 = 4.25.

Next, compute the conditional expectation E(X | Y=3): we need the distribution of X given Y=3. The joint probabilities at Y=3 are P(X=4,Y=3)=0.30 and P(X=5,Y=3)=0.10, so P(Y=3) = 0.30 + 0.10 = 0.40. Thus P(X=4 | Y=3) = 0.30 / 0.40 = 0.75 and P(X=5 | Y=3) = 0.10 / 0.40 = 0.25. Then E(X | Y=3) = 4*(0.75) + 5*(0.25) = 3.00 + 1.25 = 4.25.

Now evaluate each option one by one:
- Option 1: E(X)=4.25; E(X|Y=3)=4.25. This matches the computed values above, using the marginal and conditional distributions derived from the table.
- Option 2: E(X)=2.15; E(X|Y=3)=4.25. The unconditional value 2.15 does not align with the marginal probabilities (0.75 for X=4 and 0.25 for X=5 yield 4.25). The second part happens to be correct but the first part is inconsistent with the data.
- Option 3: E(X)=4.25; E(X|Y=3)=2.2. The unconditional expectation is correct here, but the conditional expectation given Y=3 is not; using P(Y=3)=0.40 with P(X=4|Y=3)=0.75 and P(X=5|Y=3)=0.25 yields 4.25, not 2.2.
- Option 4: E(X)=2.15; E(X|Y=3)=2.2. Both numbers are inconsistent with the table; the unconditional 2.15 is not supported by the marginals, and the conditional 2.2 is also not what the conditional distribution gives.
- Option 5: E(X)=4.25; E(X|Y=3)=4.2. The unconditional value is correct, but the conditional value should be exactly 4.25 given the computed conditional probabilities; 4.2 would require slightly different probabilities which are not present in the table.
- Option 6: E(X)=4.2; E(X|Y=3)=4.2. Neither the unconditional nor the conditional results match the exact calculations from the joint table, where both are 4.25.

In summary, only the first option aligns with both the unconditional and conditional expectations derived directly from the provided joint distribution, while all other options misstate at least one of the two quantities.

BU.232.630.W1.SP25 Quiz 1 solutions

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