1
$\begingroup$

I am a bit confused by the min and max values of hypergeometric random variable. Suppose that we have a bag with b blue marbles and r red marbles. we choose k marbles at random (without replacement) and Let X be the number of blue marbles in your sample. So, the range is:

RX={max(0,k−r),...,min(k,b)}.

Would be great if somebody explain me what max(0,k−r) and min(k,b) exactly means?

$\endgroup$
  • $\begingroup$ I don't understand why you were downvoted. Someone seems to be arbitrarily downvoting a lot of new posts without a good reason. $\endgroup$ – Glen_b -Reinstate Monica Apr 16 '17 at 10:22
2
$\begingroup$

$\max(0,k-r)$ describes the smallest possible number of blue marbles you draw. Clearly you can't have fewer than $0$ blue marbles, but if there are fewer red marbles available to be drawn in the bag than the number of marbles you draw, you can't have $0$ blue marbles (e.g. if there are two red marbles and you draw 3 marbles, at least one must be blue). So if $k-r$ is positive, $X$ can't be smaller than that.

Consequently whichever is the bigger of $0$ and $k-r$ is the smallest $X$ can be. Now $\max$ is the maximum function, which just returns the biggest of its arguments, so $\max(0,k-r)$ is simply how you express that lower limit mathematically.

For the upper limit, the number of blue marbles you draw cannot exceed the number of marbles you draw, nor the number of blue marbles that are there to be drawn. That is, the most $X$ can be is the smaller of $k$ and $b$, which you write as $\min(k,b)$.

$\endgroup$
  • $\begingroup$ "max is the maximum function, which just returns the biggest of its arguments", this is exactly what I was looking for :-), thank you very much for your reply $\endgroup$ – user6013305 Apr 16 '17 at 15:43
  • $\begingroup$ Some (hopefully useful) links: The formal definition of the maximum is here; the explanation that the $\max$ abbreviation means that maximum function is here (first entry in the "m"s). $\endgroup$ – Glen_b -Reinstate Monica Apr 17 '17 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.