Density function of conditional random variable without knowing joint distrbution.












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 20:18







qcc101

















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














  • 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










0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074885%2fdensity-function-of-conditional-random-variable-without-knowing-joint-distrbutio%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074885%2fdensity-function-of-conditional-random-variable-without-knowing-joint-distrbutio%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more