Is $P(U=u)=1/x$ well defined when $x$ is the values a discrete random variable takes












2












$begingroup$


If $X$ is a discrete random variable with the following probability distribution function
$P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.



Now let $U$ be a discrete uniform random variable such that :
$P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)



I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
How can the probability that a random variable $U$ take the value $u$ be random?



If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?



Thanks






probability-theory probability-distributions







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    2












    $begingroup$


    If $X$ is a discrete random variable with the following probability distribution function
    $P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.



    Now let $U$ be a discrete uniform random variable such that :
    $P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)



    I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
    How can the probability that a random variable $U$ take the value $u$ be random?



    If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?



    Thanks






    probability-theory probability-distributions







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      If $X$ is a discrete random variable with the following probability distribution function
      $P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.



      Now let $U$ be a discrete uniform random variable such that :
      $P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)



      I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
      How can the probability that a random variable $U$ take the value $u$ be random?



      If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?



      Thanks






      probability-theory probability-distributions







      share|cite|improve this question









      $endgroup$




      If $X$ is a discrete random variable with the following probability distribution function
      $P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.



      Now let $U$ be a discrete uniform random variable such that :
      $P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)



      I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
      How can the probability that a random variable $U$ take the value $u$ be random?



      If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?



      Thanks






      probability-theory probability-distributions




      probability-theory probability-distributions






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      asked Jan 7 at 20:49









      user3503589user3503589

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          $begingroup$

          I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$



          Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.



          To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
          and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$






          share|cite|improve this answer











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            $begingroup$

            I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$



            Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.



            To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
            and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$






            share|cite|improve this answer











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              2












              $begingroup$

              I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$



              Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.



              To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
              and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$



                Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.



                To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
                and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$






                share|cite|improve this answer











                $endgroup$



                I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$



                Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.



                To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
                and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 21:37

























                answered Jan 7 at 21:18









                lulululu

                39.8k24778




                39.8k24778






























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