Random variable with finite expectation and unbounded variance
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Example of Random variable with finite expectation
∑∞k=11k3=ζ(3)∑k=1∞1k3=ζ(3)
This means ∑∞k=11ζ(3)k3=1∑k=1∞1ζ(3)k3=1 would be a great probability distribution.
Let P(X=k)=1ζ(3)k3P(X=k)=1ζ(3)k3 on k=1,2,…k=1,2,…
Now E[X]=∑∞k=1kP(X=k)=∑∞k=11ζ(3)k2=ζ(2)ζ(3)<∞E[X]=∑k=1∞kP(X=k)=∑k=1∞1ζ(3)k2=ζ(2)ζ(3)<∞
But E[X2]=∑∞k=1k2P(X=k)=∑∞k=11ζ(3)kE[X2]=∑k=1∞k2P(X=k)=∑k=1∞1ζ(3)k diverges
And thus, the variance of XX is infinite
∑∞k=11k3=ζ(3)∑k=1∞1k3=ζ(3)
This means ∑∞k=11ζ(3)k3=1∑k=1∞1ζ(3)k3=1 would be a great probability distribution.
Let P(X=k)=1ζ(3)k3P(X=k)=1ζ(3)k3 on k=1,2,…k=1,2,…
Now E[X]=∑∞k=1kP(X=k)=∑∞k=11ζ(3)k2=ζ(2)ζ(3)<∞E[X]=∑k=1∞kP(X=k)=∑k=1∞1ζ(3)k2=ζ(2)ζ(3)<∞
But E[X2]=∑∞k=1k2P(X=k)=∑∞k=11ζ(3)kE[X2]=∑k=1∞k2P(X=k)=∑k=1∞1ζ(3)k diverges
And thus, the variance of XX is infinite
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Answer: Random variable with finite expectation and unbounded variance.
Step-by-step explanation:
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