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$X_i$ are exponential(\lamda) distribution and identically independent distribution.

Y = $\sum_{i=1}^n$$X_i$

$X_i$ is an unbaised estimator of \lamda.

Y is a sufficient estiamtor of \lamda.

solve E[$X_1$|Y]


I know that Y is Gamma distribution // gamma(1,1/\lamda)

but i can't solve this problem.

how to solve this problem?

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closed as off-topic by mrtaurho, Leucippus, Cesareo, José Carlos Santos, YiFan Mar 18 at 10:20

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  • $\begingroup$ $Y$ is a sufficient statistic of $\lambda$ rather than an estimator. Its shape parameter is $n$ rather than the $1$ you suggest $\endgroup$ – Henry Mar 17 at 15:04
  • $\begingroup$ thanks! i wrote it wrong!! $\endgroup$ – JeonSeokJun Mar 17 at 15:10
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Hint:

Consider $$\mathbb E\left[X_1 \mid Y\right] = E\left[X_j \mid Y\right] = \frac1n\sum_{i=1}^n \mathbb E\left[X_i \mid Y\right]= \frac1n\mathbb E\left[\sum_{i=1}^nX_i \;\Bigg| \; Y\right]$$

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  • $\begingroup$ thanks a lot! but i don't know that i understand well. E[\X_1\+\X_2\+...+\X_n\|Y]=E[\X_1\|Y]+E[\X_2\|Y]+...E[\X_n\|Y] required that X and Y are independent? $\endgroup$ – JeonSeokJun Mar 17 at 15:02
  • $\begingroup$ @JeonSeokJun You did not quite finish that comment, but I would guess you intended $\mathbb E[Y \mid Y]$. If you did, then that would be equal to $Y$ $\endgroup$ – Henry Mar 17 at 15:05
  • $\begingroup$ @JeonSeokJun Now you have finished the comment - independence is not required and indeed is not present here. Your expression is true because of linearity of expectation $\endgroup$ – Henry Mar 17 at 17:27

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