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As the title suggests, suppose we form a set $A$ by selecting each element in $\{1,2,3,\dots,N\}$ with probability $\delta$ independently.

What is the expected size of $|A|$? The book I am reading claims it is $\delta N$.

By definition of expectation, this should be equal to $\sum_{n=0}^N n$ $N\choose n$ $\delta^n (1-\delta)^{N-n}$. I'm struggling to show this is equal to $\delta N$.

trynalearn
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    Hint: Let $E(n)$ be the expected size when drawing from a set of size $n$. Can you show that $E(n)=\delta+E(n-1)$? – Michael Burr Jun 29 '22 at 14:49
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    You can take a factor $N\delta$ outside the sum (both being constants) and think about what that does to the individual summands. – Mark Bennet Jun 29 '22 at 14:50

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