If we have a box that contains 1000000 balls numbered from 1 to 1000000.
if we take "K" balls without replacement. let X be the maximum number and y be the minimum number in the K balls.
What is the expected value E(x-y) as function of "K".
If we have a box that contains 1000000 balls numbered from 1 to 1000000.
if we take "K" balls without replacement. let X be the maximum number and y be the minimum number in the K balls.
What is the expected value E(x-y) as function of "K".
@billobillo
You will never understand this question unless you yourself sit and work with the numbers.
May I suggest that you cut the question down to 100 balls an let K=8.
List the smallest eight numbers. What is $\displaystyle x=?~,~y=?~,~x-y=?$
List the larest eight numbers. What is $\displaystyle x=?~,~y=?~,~x-y=?$
List any eight numbers. What is $\displaystyle x=?~,~y=?~,~x-y=?$
Now what is the maximum for $\displaystyle x$?
Now what is the minimum for $\displaystyle y$?
Now what is $\displaystyle ?\le x-y\le ?$?
The think about number partitions?
When you have done that. Tell us what you find.
The title is a bit misleading...
Try it Moo, but maybe not start with K=8...
if K=2, you draw 2 balls out of 100, what is the expected value of the bigger numbered ball?
Now you draw 3 balls...
Then you should get it by now.
prove by using principle of symmetry
The set $\displaystyle \{1,2,3,4,5,6,7,8\}$ has $\displaystyle x=8,~y=1,~\&~x-y=7$
The set $\displaystyle \{93,94,95,96,97,98,99,100\}$ has $\displaystyle x=100,~y=93,~\&~x-y=7$.
The set $\displaystyle \{1,24,37,45,50,61,87,100\}$ has $\displaystyle x=100,~y=1,~\&~x-y=99$
Clearly $\displaystyle 7\le~x-y~\le 99$
Now $\displaystyle \mathcal{P}(x-y=99)=\dfrac{\binom{98}{6}}{\binom{100}{8}}$.
Thanks for your replies, before posting this topic, I worked on a sample of 7 balls with K=3 and I was able to find E(x) and E(y) as function of K but not E(x-y),
I'm not a math specialist but I guess that x and y are not independent events, any suggestions?