Expectation of conditional normal random variablesconditional expectations for independent random...

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Expectation of conditional normal random variables


conditional expectations for independent random variablesExpectation of random variablesShow Independence of two random variables $X$ and$Y$Conditional expectation of independent random variablesHow to find the CDF of the sum of independent uniformly distributed random variables?Conditional probability for two normal distributed variables.Conditional Expectation as a random variable of independent rendom variablesConditional Expectation of random sum of independent random variables (when $N$ and $X_i$ are dependent)Conditional Expectation of Random Variables.A specific question in conditional expectation with mixed discrete and continuous random variables













-1












$begingroup$


I need help with a question in probability.
If X,Y are independent N(0,1) random variables. What is E(X-Y| X>Y)



All help would be much helpful.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Welcome to MSE. Please read this text about how to ask a good question.
    $endgroup$
    – José Carlos Santos
    Mar 15 at 8:31






  • 1




    $begingroup$
    $Z=X-Y$ is itself normal. Can you compute $E(Z|Z>0)$?.
    $endgroup$
    – Kavi Rama Murthy
    Mar 15 at 8:43












  • $begingroup$
    I tried to preform that transformation using the transformation theorem setting X=U Z=X-Y. However, I got stuck when trying to get the marginal distribution of Z.
    $endgroup$
    – Two-norm
    Mar 15 at 9:29












  • $begingroup$
    Please include your work, whatever you have done, in the post itself. The fact that $X-Y$ is again normal follows from a general result.
    $endgroup$
    – StubbornAtom
    Mar 15 at 10:23
















-1












$begingroup$


I need help with a question in probability.
If X,Y are independent N(0,1) random variables. What is E(X-Y| X>Y)



All help would be much helpful.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Welcome to MSE. Please read this text about how to ask a good question.
    $endgroup$
    – José Carlos Santos
    Mar 15 at 8:31






  • 1




    $begingroup$
    $Z=X-Y$ is itself normal. Can you compute $E(Z|Z>0)$?.
    $endgroup$
    – Kavi Rama Murthy
    Mar 15 at 8:43












  • $begingroup$
    I tried to preform that transformation using the transformation theorem setting X=U Z=X-Y. However, I got stuck when trying to get the marginal distribution of Z.
    $endgroup$
    – Two-norm
    Mar 15 at 9:29












  • $begingroup$
    Please include your work, whatever you have done, in the post itself. The fact that $X-Y$ is again normal follows from a general result.
    $endgroup$
    – StubbornAtom
    Mar 15 at 10:23














-1












-1








-1





$begingroup$


I need help with a question in probability.
If X,Y are independent N(0,1) random variables. What is E(X-Y| X>Y)



All help would be much helpful.










share|cite|improve this question









$endgroup$




I need help with a question in probability.
If X,Y are independent N(0,1) random variables. What is E(X-Y| X>Y)



All help would be much helpful.







probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 15 at 8:27









Two-normTwo-norm

1




1








  • 1




    $begingroup$
    Welcome to MSE. Please read this text about how to ask a good question.
    $endgroup$
    – José Carlos Santos
    Mar 15 at 8:31






  • 1




    $begingroup$
    $Z=X-Y$ is itself normal. Can you compute $E(Z|Z>0)$?.
    $endgroup$
    – Kavi Rama Murthy
    Mar 15 at 8:43












  • $begingroup$
    I tried to preform that transformation using the transformation theorem setting X=U Z=X-Y. However, I got stuck when trying to get the marginal distribution of Z.
    $endgroup$
    – Two-norm
    Mar 15 at 9:29












  • $begingroup$
    Please include your work, whatever you have done, in the post itself. The fact that $X-Y$ is again normal follows from a general result.
    $endgroup$
    – StubbornAtom
    Mar 15 at 10:23














  • 1




    $begingroup$
    Welcome to MSE. Please read this text about how to ask a good question.
    $endgroup$
    – José Carlos Santos
    Mar 15 at 8:31






  • 1




    $begingroup$
    $Z=X-Y$ is itself normal. Can you compute $E(Z|Z>0)$?.
    $endgroup$
    – Kavi Rama Murthy
    Mar 15 at 8:43












  • $begingroup$
    I tried to preform that transformation using the transformation theorem setting X=U Z=X-Y. However, I got stuck when trying to get the marginal distribution of Z.
    $endgroup$
    – Two-norm
    Mar 15 at 9:29












  • $begingroup$
    Please include your work, whatever you have done, in the post itself. The fact that $X-Y$ is again normal follows from a general result.
    $endgroup$
    – StubbornAtom
    Mar 15 at 10:23








1




1




$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Mar 15 at 8:31




$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Mar 15 at 8:31




1




1




$begingroup$
$Z=X-Y$ is itself normal. Can you compute $E(Z|Z>0)$?.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 8:43






$begingroup$
$Z=X-Y$ is itself normal. Can you compute $E(Z|Z>0)$?.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 8:43














$begingroup$
I tried to preform that transformation using the transformation theorem setting X=U Z=X-Y. However, I got stuck when trying to get the marginal distribution of Z.
$endgroup$
– Two-norm
Mar 15 at 9:29






$begingroup$
I tried to preform that transformation using the transformation theorem setting X=U Z=X-Y. However, I got stuck when trying to get the marginal distribution of Z.
$endgroup$
– Two-norm
Mar 15 at 9:29














$begingroup$
Please include your work, whatever you have done, in the post itself. The fact that $X-Y$ is again normal follows from a general result.
$endgroup$
– StubbornAtom
Mar 15 at 10:23




$begingroup$
Please include your work, whatever you have done, in the post itself. The fact that $X-Y$ is again normal follows from a general result.
$endgroup$
– StubbornAtom
Mar 15 at 10:23










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