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Suppose that $X_1, ldots, X_n oversetidsim F$. Is there a function $h$ such that $h(X_1), ldots, h(X_n) oversetiidsim G$?



The 2019 Stack Overflow Developer Survey Results Are InSuppose $X_1,ldots,X_n$ are independent random variables. Find $rho(S_k,S_n)$.Suppose that $X_1, X_2, …, X_n$ are i.i.d. random variables such that $X_1sim N(mu, 0.5)$ and $n = 100$.calculating $P(X_n=max(X_1,X_2,ldots, X_n))$$X_1, dots, X_n$ are independent random variables. Suppose $M = min(X_1, X_2, dots, X_n)$Do we not have that $X_1, ldots, X_n sim F$ iff $X_1, ldots, X_n sim f$?If $X_1 perp X_2 perp ldots perp Z$, where $X_1, ldots, X_n$ are iid, is $fracX_1Z, ldots ,fracX_1Z$ are independent of $Z$?Suppose that $X_1, X_2$ are iid random variables with the same CDF, is it true that $P(X_2>X_1) = X_1$?Suppose $X_1, ldots, X_n$ are iid standard Cauchy random variables, does $frac1nsum_i=1^n}X_i$ converge in probability or almost surely?If $X_1,ldots,X_n$ are independent, does $mathbbP{X_n>maxX_1,ldots,X_n-1=mathbbP{X_n>X_1)cdotsmathbbPX_n>X_n-1$ hold?










0












$begingroup$


Suppose that we have $n$ dependent random variables distributed as:



$$
X_1, ldots, X_n oversetidsim F
$$



Is there a function $h$ such that:



$$
h(X_1), ldots, h(X_n) oversetiidsim G
$$



for a new distribution $G$ where each of the $h(X_i)$ are now independent as well? I thought about the following:



$$
h(x) = 0cdot x + epsilon
$$



where $epsilon sim N(0,1)$. However, this relies on a random variable, which I am not sure if it would be valid.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Is $G$ given to you?
    $endgroup$
    – Minus One-Twelfth
    Mar 31 at 1:34















0












$begingroup$


Suppose that we have $n$ dependent random variables distributed as:



$$
X_1, ldots, X_n oversetidsim F
$$



Is there a function $h$ such that:



$$
h(X_1), ldots, h(X_n) oversetiidsim G
$$



for a new distribution $G$ where each of the $h(X_i)$ are now independent as well? I thought about the following:



$$
h(x) = 0cdot x + epsilon
$$



where $epsilon sim N(0,1)$. However, this relies on a random variable, which I am not sure if it would be valid.










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Is $G$ given to you?
    $endgroup$
    – Minus One-Twelfth
    Mar 31 at 1:34













0












0








0





$begingroup$


Suppose that we have $n$ dependent random variables distributed as:



$$
X_1, ldots, X_n oversetidsim F
$$



Is there a function $h$ such that:



$$
h(X_1), ldots, h(X_n) oversetiidsim G
$$



for a new distribution $G$ where each of the $h(X_i)$ are now independent as well? I thought about the following:



$$
h(x) = 0cdot x + epsilon
$$



where $epsilon sim N(0,1)$. However, this relies on a random variable, which I am not sure if it would be valid.










share|cite|improve this question









$endgroup$




Suppose that we have $n$ dependent random variables distributed as:



$$
X_1, ldots, X_n oversetidsim F
$$



Is there a function $h$ such that:



$$
h(X_1), ldots, h(X_n) oversetiidsim G
$$



for a new distribution $G$ where each of the $h(X_i)$ are now independent as well? I thought about the following:



$$
h(x) = 0cdot x + epsilon
$$



where $epsilon sim N(0,1)$. However, this relies on a random variable, which I am not sure if it would be valid.







probability statistics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 30 at 23:41









user321627user321627

998515




998515







  • 1




    $begingroup$
    Is $G$ given to you?
    $endgroup$
    – Minus One-Twelfth
    Mar 31 at 1:34












  • 1




    $begingroup$
    Is $G$ given to you?
    $endgroup$
    – Minus One-Twelfth
    Mar 31 at 1:34







1




1




$begingroup$
Is $G$ given to you?
$endgroup$
– Minus One-Twelfth
Mar 31 at 1:34




$begingroup$
Is $G$ given to you?
$endgroup$
– Minus One-Twelfth
Mar 31 at 1:34










1 Answer
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$begingroup$

$h= 1$ is such a function. There are cases where $h$ is necessarily constant: e.g., $X_1=X_2$ with standard normal distribution.






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    2












    $begingroup$

    $h= 1$ is such a function. There are cases where $h$ is necessarily constant: e.g., $X_1=X_2$ with standard normal distribution.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      $h= 1$ is such a function. There are cases where $h$ is necessarily constant: e.g., $X_1=X_2$ with standard normal distribution.






      share|cite|improve this answer









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        2





        $begingroup$

        $h= 1$ is such a function. There are cases where $h$ is necessarily constant: e.g., $X_1=X_2$ with standard normal distribution.






        share|cite|improve this answer









        $endgroup$



        $h= 1$ is such a function. There are cases where $h$ is necessarily constant: e.g., $X_1=X_2$ with standard normal distribution.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 31 at 0:00









        Kavi Rama MurthyKavi Rama Murthy

        74.1k53270




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