This applies to natural numbers [itex]n[/itex] and [itex]N[/itex] where [itex]n<N[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

We have [itex]N[/itex] balls representing numbers [itex]1,2,...,N[/itex].

We randomly choose [itex]n[/itex] of those balls which happen to represent numbers [itex]{k_1},{k_2},...,{k_n}[/itex].

We then define a random variable [itex]X = {k_1} + {k_2} + ... + {k_n}[/itex].

What is the mean and variation of [itex]X[/itex]?

Well there are [itex]\left( {\begin{array}{*{20}{c}}N\\n\end{array}} \right)[/itex] equally likely combinations and every one of them brings [itex]n[/itex] summands. So we get [itex]\left( {\begin{array}{*{20}{c}}

N\\

n

\end{array}} \right) \cdot n[/itex] summands overall. Since there is no bias towards any particular number it means that every number is added [itex]\left( {\begin{array}{*{20}{c}}

N\\

n

\end{array}} \right) \cdot \frac{n}{N}[/itex] times hence our mean:

[itex]E\left( X \right) = \frac{{\left( {\begin{array}{*{20}{c}}

N\\

n

\end{array}} \right) \cdot \frac{n}{N} \cdot \left( {1 + 2 + ... + N} \right)}}{{\left( {\begin{array}{*{20}{c}}

N\\

n

\end{array}} \right)}} = \frac{n}{N} \cdot \frac{{N\left( {N + 1} \right)}}{2} = \frac{{n\left( {N + 1} \right)}}{2}[/itex].

Unfortunately variance seems to be an entirely different animal since I cannot just rip sums apart and add numbers in a different order. Any ideas?

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# What is Var(X) for my defined X?

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