Mixing
Density function of conditional random variable without knowing joint distrbution.
$begingroup$
I have Z and X which are two random variables with density:
$f_Z(z) = 3(1-z)^2mathbb{1}_{[0,1]}(z)$
$f_X(x) = 6x(1-x)mathbb{1}_{[0,1]}(x)$
I want to find $f_{Z vert X = x}(z)$, but to do that I have to calculate the joint density which is unknown. (The two random variables are not independent). Is there another way? If not, how do I calculate the joint density?
EDIT: The random variable Z was defined as: $$ Z = XU $$ with $f_U(u) = mathbb{1}_{[0,1]}(u)$
probability conditional-expectation
asked Jan 15 at 19:59
qcc101qcc101
622213
$endgroup$
add a comment |
$begingroup$
I have Z and X which are two random variables with density:
$f_Z(z) = 3(1-z)^2mathbb{1}_{[0,1]}(z)$
$f_X(x) = 6x(1-x)mathbb{1}_{[0,1]}(x)$
I want to find $f_{Z vert X = x}(z)$, but to do that I have to calculate the joint density which is unknown. (The two random variables are not independent). Is there another way? If not, how do I calculate the joint density?
EDIT: The random variable Z was defined as: $$ Z = XU $$ with $f_U(u) = mathbb{1}_{[0,1]}(u)$
probability conditional-expectation
asked Jan 15 at 19:59
qcc101qcc101
622213
$endgroup$
1
$begingroup$
Without knowing anything about the dependence, it doesn't seem that there is enough information to solve the problem.
$endgroup$
– Greg Martin
Jan 15 at 20:12
$begingroup$
@GregMartin I see, I will edit and add more information.
$endgroup$
– qcc101
Jan 15 at 20:16
1
$begingroup$
Again you have to tell us the relationship between $X$ and $U$, otherwise we cannot be certain. Assume you means they are independent. Then the question is easy - you do not need to do any tedious derivation - the conditional distribution of $Z$ given $X = x$ is simply $xU sim text{Uniform}(0, x)$.
$endgroup$
– BGM
Jan 16 at 3:23
$begingroup$
Yes sorry about that, they indeed are independent.
$endgroup$
– qcc101
Jan 16 at 6:31
add a comment |
$begingroup$
I have Z and X which are two random variables with density:
$f_Z(z) = 3(1-z)^2mathbb{1}_{[0,1]}(z)$
$f_X(x) = 6x(1-x)mathbb{1}_{[0,1]}(x)$
I want to find $f_{Z vert X = x}(z)$, but to do that I have to calculate the joint density which is unknown. (The two random variables are not independent). Is there another way? If not, how do I calculate the joint density?
EDIT: The random variable Z was defined as: $$ Z = XU $$ with $f_U(u) = mathbb{1}_{[0,1]}(u)$
probability conditional-expectation
asked Jan 15 at 19:59
qcc101qcc101
622213
$endgroup$
I have Z and X which are two random variables with density:
$f_Z(z) = 3(1-z)^2mathbb{1}_{[0,1]}(z)$
$f_X(x) = 6x(1-x)mathbb{1}_{[0,1]}(x)$
I want to find $f_{Z vert X = x}(z)$, but to do that I have to calculate the joint density which is unknown. (The two random variables are not independent). Is there another way? If not, how do I calculate the joint density?
EDIT: The random variable Z was defined as: $$ Z = XU $$ with $f_U(u) = mathbb{1}_{[0,1]}(u)$
probability conditional-expectation
probability conditional-expectation
asked Jan 15 at 19:59
qcc101qcc101
622213
asked Jan 15 at 19:59
qcc101qcc101
622213
edited Jan 15 at 20:18
qcc101
asked Jan 15 at 19:59
qcc101qcc101
622213
asked Jan 15 at 19:59
qcc101qcc101
622213
asked Jan 15 at 19:59
qcc101qcc101
622213
622213
1
$begingroup$
Without knowing anything about the dependence, it doesn't seem that there is enough information to solve the problem.
$endgroup$
– Greg Martin
Jan 15 at 20:12
$begingroup$
@GregMartin I see, I will edit and add more information.
$endgroup$
– qcc101
Jan 15 at 20:16
1
$begingroup$
Again you have to tell us the relationship between $X$ and $U$, otherwise we cannot be certain. Assume you means they are independent. Then the question is easy - you do not need to do any tedious derivation - the conditional distribution of $Z$ given $X = x$ is simply $xU sim text{Uniform}(0, x)$.
$endgroup$
– BGM
Jan 16 at 3:23
$begingroup$
Yes sorry about that, they indeed are independent.
$endgroup$
– qcc101
Jan 16 at 6:31
add a comment |
1
$begingroup$
Without knowing anything about the dependence, it doesn't seem that there is enough information to solve the problem.
$endgroup$
– Greg Martin
Jan 15 at 20:12
$begingroup$
@GregMartin I see, I will edit and add more information.
$endgroup$
– qcc101
Jan 15 at 20:16
1
$begingroup$
Again you have to tell us the relationship between $X$ and $U$, otherwise we cannot be certain. Assume you means they are independent. Then the question is easy - you do not need to do any tedious derivation - the conditional distribution of $Z$ given $X = x$ is simply $xU sim text{Uniform}(0, x)$.
$endgroup$
– BGM
Jan 16 at 3:23
$begingroup$
Yes sorry about that, they indeed are independent.
$endgroup$
– qcc101
Jan 16 at 6:31
1
1
$begingroup$
Without knowing anything about the dependence, it doesn't seem that there is enough information to solve the problem.
$endgroup$
– Greg Martin
Jan 15 at 20:12
$begingroup$
Without knowing anything about the dependence, it doesn't seem that there is enough information to solve the problem.
$endgroup$
– Greg Martin
Jan 15 at 20:12
$begingroup$
@GregMartin I see, I will edit and add more information.
$endgroup$
– qcc101
Jan 15 at 20:16
$begingroup$
@GregMartin I see, I will edit and add more information.
$endgroup$
– qcc101
Jan 15 at 20:16
1
1
$begingroup$
Again you have to tell us the relationship between $X$ and $U$, otherwise we cannot be certain. Assume you means they are independent. Then the question is easy - you do not need to do any tedious derivation - the conditional distribution of $Z$ given $X = x$ is simply $xU sim text{Uniform}(0, x)$.
$endgroup$
– BGM
Jan 16 at 3:23
$begingroup$
Again you have to tell us the relationship between $X$ and $U$, otherwise we cannot be certain. Assume you means they are independent. Then the question is easy - you do not need to do any tedious derivation - the conditional distribution of $Z$ given $X = x$ is simply $xU sim text{Uniform}(0, x)$.
$endgroup$
– BGM
Jan 16 at 3:23
$begingroup$
Yes sorry about that, they indeed are independent.
$endgroup$
– qcc101
Jan 16 at 6:31
$begingroup$
Yes sorry about that, they indeed are independent.
$endgroup$
– qcc101
Jan 16 at 6:31
add a comment |
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1
$begingroup$
Without knowing anything about the dependence, it doesn't seem that there is enough information to solve the problem.
$endgroup$
– Greg Martin
Jan 15 at 20:12
$begingroup$
@GregMartin I see, I will edit and add more information.
$endgroup$
– qcc101
Jan 15 at 20:16
1
$begingroup$
Again you have to tell us the relationship between $X$ and $U$, otherwise we cannot be certain. Assume you means they are independent. Then the question is easy - you do not need to do any tedious derivation - the conditional distribution of $Z$ given $X = x$ is simply $xU sim text{Uniform}(0, x)$.
$endgroup$
– BGM
Jan 16 at 3:23
$begingroup$
Yes sorry about that, they indeed are independent.
$endgroup$
– qcc101
Jan 16 at 6:31