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
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, $$P(Y=0\,|\,X=x)=\frac 1{2^x}$$