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We have 10 cards numbered from 1 to 10. We pick two cards among them. What is the expected value of the product of these two cards with and without replacement?

My thought: Let X and Y be the number on the two card For with replacement, X and Y are independent, so E(X*Y)=E(X)*E(Y)=30.25

For without replacement, since X and Y are dependent, I feel it is the same, but could not think a nice way to prove it. Thanks!

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Note that with replacement, $P(X=n,Y=k)=\frac{1}{90}$ for $n\neq k$, so the measure is still uniform. This makes the calculation not all that different since we just have to account for the contribution of the diagonal terms where $n=k$. In other words,

$$ 90*E(X,Y)=\sum_{1\leq k\neq n\leq 10} nk=\sum_{1\leq k,n\leq 10} nk-\sum_{j=1}^{10} j^2=100*\left(\frac{11}{2}\right)^2-\frac{10\cdot11\cdot 21}{6} $$ and from here, I guess you can take it.

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  • $\begingroup$ Thanks! so E(X,Y) for without replacement is 88/3, so is smaller than with replacement $\endgroup$
    – Ellie.P
    Oct 11 '20 at 8:57

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