The expression is:
$=\sum_{i=1}^{M}[(\frac{1}{L})^N\frac{1}{M+1}] $
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1$\begingroup$ What happens to $P(X|L)=1/L$ when $L=0$? $\endgroup$– gunesCommented Jul 30, 2020 at 14:02
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$\begingroup$ That's why I think I'm doing it wrong. $\endgroup$– user292800Commented Jul 30, 2020 at 14:04
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1$\begingroup$ Yes, you're doing calculations wrong, but this is about the definitions. It can't be calculated this way. $\endgroup$– gunesCommented Jul 30, 2020 at 14:04
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$\begingroup$ Isn't U(X|A,B) defined as 1/(B-A+1) so U(X|1,L) = 1/L? $\endgroup$– user292800Commented Jul 30, 2020 at 14:06
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$\begingroup$ Ok, but my original question remains unanswered. What happens when $L=0$? $U(X|1,L=0)$ doesn't make sense. $\endgroup$– gunesCommented Jul 30, 2020 at 14:07
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1 Answer
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If $P(X|L=0)$ is assumed to be $0$, you can remove $i=0$ case from the summation. With some cleaning, we have
$$\sum_{i=1}^{M}\left[\prod_{i=j}^N P(X_j|L=i)\right]P(L=i)=\frac{1}{M+1}\sum_{i=1}^{M}\frac{1}{i^N}$$
This is a p-series and doesn't have a closed form formula.
