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

查看解析

登录即可查看完整答案

类似问题

Following the question above, what is the conditional expectation of 𝑋 1 given 𝑋 2 = 1 ?

If the conditional variance for tomorrow's return is 0.0004, 𝑉 𝑎 𝑅 𝑡 ( 𝑅 𝑡 + 1 ) = ( 0.02 ) 2 = 0.0004 , then the conditional expectation for tomorrow's return is 2% or -2%.

A fair coin is tossed 2 times. The random variable X is the number of Heads in two tosses and Y is the number of Heads on the first toss.The conditional expectation E(X|Y) is given by

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单