Stuck on counting question.

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HaloWarrior

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I don't see how the expectation relates to finding the numbers

Assume two numbers are drawn from a magicians hat at random with replacement from the set{1, 2, . . . , n} and let Z be the maximum of the two numbers. Find E[Z].
 
@Dennis, Please read the OP again.

HaloWarrior said:
Assume two numbers are drawn from a magicians hat at random with replacement from the set{1, 2, . . . , n} and let Z be the maximum of the two numbers. Find E[Z].
Click to expand...
Do you realize that there are n different values for \(\displaystyle \mathcal{Z}~?\)
BUT there are also \(\displaystyle n^2\) possible elementary events (ordered pairs).
Also, \(\displaystyle \mathcal{Z}=1\) in only one way; \(\displaystyle \mathcal{Z}=n\) in \(\displaystyle 2n-1\) ways. Why or How?
Thus \(\displaystyle P(\mathcal{Z}=1)=\dfrac{1}{n^2}\), & \(\displaystyle P(\mathcal{Z}=n)=\dfrac{2n-1}{n^2}\)

Now can you find \(\displaystyle E(\mathcal{Z})~?\)
 
Last edited:
pka said:
@Dennis, Please read the OP again.


Do you realize that there are n different values for \(\displaystyle \mathcal{Z}~?\)
BUT there are also \(\displaystyle n^2\) possible elementary events (ordered pairs).
Also, \(\displaystyle \mathcal{Z}=1\) in only one way; \(\displaystyle \mathcal{Z}=n\) in \(\displaystyle 2n-1\) ways. Why or How?
Thus \(\displaystyle P(\mathcal{Z}=1)=\dfrac{1}{n^2}\), & \(\displaystyle P(\mathcal{Z}=n)=\dfrac{2n-1}{n^2}\)

Now can you find \(\displaystyle E(\mathcal{Z})~?\)
Click to expand...

Up to now I have only done expectations with PMF tables where we simply multiply an observed value with a defined probability.
I can't see how you would apply this here ( I am probably being an idiot).
 
pka said:
Do you realize that there are n different values for \(\displaystyle \mathcal{Z}~?\)
BUT there are also \(\displaystyle n^2\) possible elementary events (ordered pairs).
Also, \(\displaystyle \mathcal{Z}=1\) in only one way; \(\displaystyle \mathcal{Z}=n\) in \(\displaystyle 2n-1\) ways. Why or How?
Thus \(\displaystyle P(\mathcal{Z}=1)=\dfrac{1}{n^2}\), & \(\displaystyle P(\mathcal{Z}=n)=\dfrac{2n-1}{n^2}\)
Click to expand...

HaloWarrior said:
Up to now I have only done expectations with PMF tables where we simply multiply an observed value with a defined probability. I can't see how you would apply this here ( I am probably being an idiot).
Click to expand...

It should be your job to construct the "PMF tables (and) simply multiply an observed value with a defined probability.
After all I gave you the first and last values. So here is what must be done:
\(\displaystyle \Large{\displaystyle{E({\mathcal{ Z}}) = \sum\limits_{k = 1}^n {k \cdot \frac{{2k - 1}}{{{n^2}}}}}}\)
 
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