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 0.05 0.2 0.25
5 0.10 0.3 0.10

Using this information, please compute the unconditional expected value
𝐸
(
𝑋
)
and the conditional expected value
𝐸
(
𝑋
|
𝑌
=
2
)
. (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	0.05	0.2	0.25
	5	0.10	0.3	0.10

Using this information, please compute the unconditional expected value 
𝐸
(
𝑋
)
 and the conditional expected value 
𝐸
(
𝑋
|
𝑌
=
2
)
. (Please round your results to the 4th decimal place.)

BlackTom AI · Accepted Answer

We start by restating the problem and the given joint distribution so we can reason about the required expectations.

X takes values {4, 5} and Y takes values {1, 2, 3} with the joint probabilities:
- P(X=4, Y=1) = 0.05, P(X=4, Y=2) = 0.20, P(X=4, Y=3) = 0.25
- P(X=5, Y=1) = 0.10, P(X=5, Y=2) = 0.30, P(X=5, Y=3) = 0.10

Step 1: Compute the unconditional expectation E(X).
- First, find P(X=4) by summing the row for X=4: 0.05 + 0.20 + 0.25 = 0.50.
- Next, find P(X=5) by summing the row for X=5: 0.10 + 0.30 + 0.10 = 0.50.
- Then E(X) = 4*P(X=4) + 5*P(X=5) = 4*(0.50) + 5*(0.50) = 2.00 + 2.50 = 4.50.

Option-by-option analysis for E(X) and E(X|Y=2):

Option 1: E(X) = 2.2; E(X|Y=2) = 2.0
- The value 2.2 for E(X) is far from the correct 4.50, and 2.0 for E(X|Y=2) would imply a distribution dominated by X=2 which is not in the support. This option miscomputes both quantities, likely due to incorrect marginal or conditional computations.

Option 2: E(X) = 4.5; E(X|Y=2) = 4.6667
- The unconditional E(X) = 4.5 matches the correct calculation above, so this part is plausible. For E(X|Y=2), we compute using the Y=2 column: P(Y=2) = P(X=4,Y=2) + P(X=5,Y=2) = 0.20 + 0.30 = 0.50.
- The conditional probabilities are P(X=4|Y=2) = 0.20 / 0.50 = 0.40 and P(X=5|Y=2) = 0.30 / 0.50 = 0.60. Then E(X|Y=2) = 4*(0.40) + 5*(0.60) = 1.6 + 3.0 = 4.6, not 4.6667. So while the E(X) part is correct, the conditional expectation is incorrect here.

Option 3: E(X) = 4.5; E(X|Y=2) = 2.0
- Unconditional E(X) = 4.5 is correct as shown earlier, but E(X|Y=2) = 2.0 would imply a distribution with much weight near X=2, which is not part of X’s support. Given P(X=4|Y=2) = 0.4 and P(X=5|Y=2) = 0.6, the conditional expectation cannot be as low as 2.0. This option is inconsistent with the actual conditional distribution obtained from P(X,Y).

Option 4: E(X) = 4.5; E(X|Y=2) = 4.6
- This option has the correct unconditional expectation, 4.50, which aligns with P(X=4)=0.50 and P(X=5)=0.50. For the conditional part, with Y=2, we have P(X=4|Y=2) = 0.20 / (0.20 + 0.30) = 0.40 and P(X=5|Y=2) = 0.30 / (0.20 + 0.30) = 0.60. Therefore E(X|Y=2) = 4*(0.40) + 5*(0.60) = 1.6 + 3.0 = 4.6, which exactly matches this option’s value. This is the internally consistent choice based on the joint distribution.

Option 5: E(X) = 2.2; E(X|Y=2) = 4.6
- The conditional part E(X|Y=2) = 4.6 is correct, but the unconditional E(X) = 2.2 is incorrect, given the marginal of X sums to 1 with equal 0.50 mass on 4 and 5. The 2.2 value cannot arise from the given distribution.

Option 6: E(X) = 4.6667; E(X|Y=2) = 4.6667
- This would imply E(X) is 4.6667, which is inconsistent with P(X=4)=0.50 and P(X=5)=0.50, since that would give exactly 4.5, not 4.6667. The conditional 4.6667 would require a different conditional distribution than we have for Y=2, so this option is incorrect.

In summary, the only option that consistently matches both the unconditional expectation computed from the joint distribution and the conditional expectation E(X|Y=2) derived from the Y=2 column is the one with E(X) = 4.50 and E(X|Y=2) = 4.60.

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