Conditional variance problem on INTRODUCTION TO PROBABILITY(2nd edition)
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How can I get results in the red box?
conditional-probability variance
asked Nov 26 at 6:09
GoingMyWay
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How can I get results in the red box?
conditional-probability variance
asked Nov 26 at 6:09
GoingMyWay
1609
add a comment |
up vote
0
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How can I get results in the red box?
conditional-probability variance
asked Nov 26 at 6:09
GoingMyWay
1609
How can I get results in the red box?
conditional-probability variance
conditional-probability variance
asked Nov 26 at 6:09
GoingMyWay
1609
asked Nov 26 at 6:09
GoingMyWay
1609
edited Nov 26 at 6:19
asked Nov 26 at 6:09
GoingMyWay
1609
asked Nov 26 at 6:09
GoingMyWay
1609
asked Nov 26 at 6:09
GoingMyWay
1609
1609
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add a comment |
1 Answer
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It is a well-known fact that the variance of a uniformly distributed random variable on the interval $(a, b)$ is given by $(b - a)^{2}/12$. This can be shown using the Law of the Unconscious Statistician.
So,
$$text{var}(X mid Y = 1) = frac{(1)^{2}}{12} = frac{1}{12}$$
$$text{var}(X mid Y = 2) = frac{(2)^{2}}{12} = frac{4}{12} $$
answered Nov 26 at 6:32
Ekesh
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It is a well-known fact that the variance of a uniformly distributed random variable on the interval $(a, b)$ is given by $(b - a)^{2}/12$. This can be shown using the Law of the Unconscious Statistician.
So,
$$text{var}(X mid Y = 1) = frac{(1)^{2}}{12} = frac{1}{12}$$
$$text{var}(X mid Y = 2) = frac{(2)^{2}}{12} = frac{4}{12} $$
answered Nov 26 at 6:32
Ekesh
4855
add a comment |
up vote
0
down vote
accepted
It is a well-known fact that the variance of a uniformly distributed random variable on the interval $(a, b)$ is given by $(b - a)^{2}/12$. This can be shown using the Law of the Unconscious Statistician.
So,
$$text{var}(X mid Y = 1) = frac{(1)^{2}}{12} = frac{1}{12}$$
$$text{var}(X mid Y = 2) = frac{(2)^{2}}{12} = frac{4}{12} $$
answered Nov 26 at 6:32
Ekesh
4855
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It is a well-known fact that the variance of a uniformly distributed random variable on the interval $(a, b)$ is given by $(b - a)^{2}/12$. This can be shown using the Law of the Unconscious Statistician.
So,
$$text{var}(X mid Y = 1) = frac{(1)^{2}}{12} = frac{1}{12}$$
$$text{var}(X mid Y = 2) = frac{(2)^{2}}{12} = frac{4}{12} $$
answered Nov 26 at 6:32
Ekesh
4855
It is a well-known fact that the variance of a uniformly distributed random variable on the interval $(a, b)$ is given by $(b - a)^{2}/12$. This can be shown using the Law of the Unconscious Statistician.
So,
$$text{var}(X mid Y = 1) = frac{(1)^{2}}{12} = frac{1}{12}$$
$$text{var}(X mid Y = 2) = frac{(2)^{2}}{12} = frac{4}{12} $$
answered Nov 26 at 6:32
Ekesh
4855
answered Nov 26 at 6:32
Ekesh
4855
answered Nov 26 at 6:32
Ekesh
4855
answered Nov 26 at 6:32
Ekesh
4855
4855
add a comment |
add a comment |
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