# conditional expectation of gamma distribution with alpha = 1 [closed]

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

Y = $$\sum_{i=1}^nX_i$$

$$X_i$$ 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?

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

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

Consider $$\mathbb E\left[X_1 \mid Y\right] = E\left[X_j \mid Y\right] = \frac1n\sum_{i=1}^n \mathbb E\left[X_i \mid Y\right]= \frac1n\mathbb E\left[\sum_{i=1}^nX_i \;\Bigg| \; Y\right]$$
• @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 $Y$ – Henry Mar 17 at 15:05