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How would I even begin proving this? I've been thinking about this for an hour now with no luck.

I attempted to get E(Z) and go from there, but I can't seem to prove it

  • Try rearranging each sum and see both sums have the same terms. – Jonathaniui Mar 02 '17 at 20:40
  • Hint: $P(i)=P(Z=i)=P(Z≥i)-P(Z≥i+1)$. Thus, say, $1\times P(1)+2\times P(2)=P(Z≥1)-P(Z≥2) + 2P(Z≥2)-2P(Z≥3)=P(Z≥1)+P(Z≥2)-2P(Z≥3)$ and so on. – lulu Mar 02 '17 at 20:43

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