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I please ask someone to check if my calculations are right.

I have $X_1, ..., X_n$ from a $\mathcal{E}(\lambda): f(x, \lambda) = \lambda e^{-\lambda x}$.

I have to find the $k$ such that $P(\bar{X} \le k) = \alpha$, where $\bar{X}$ is the sample mean; i did: $Y=\sum_{i = 1}^{n} X_i$ $Y \sim \Gamma (n, \lambda)$ $\bar{X} = \frac{1}{n} Y \sim \Gamma(n, \frac{\lambda}{n})$ $T = 2\frac{n}{\lambda} \bar{X} \sim \Gamma(\frac{2n}{2}, 2) \stackrel{d}{=}\chi^2 (2n) $ $P(\bar{X} \le k) = P(T \le k' = 2\frac{n}{\lambda} k) = \alpha $

Then i can find the value of $k'$ from the table, and finally find $k$. I'm i missing something? (I can't reach the result stated by the book).

Thank you very much.

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    This is a common mistake: the different parametrisations used for the exponential distribution often cause confusion. I hope that you corrected the error in Italian Wikipedia :)2012-06-08

2 Answers 2

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My comment from above, so that the question can be marked as answered:

Having read your solution more carefully, I think that you got the distribution of Y wrong. It should be Γ(n,1/λ); see Wikipedia.

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    I believe this is incorrect. See other Answer.2017-02-20
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Correction.

In discussing this question, I have discovered errors here.

Specifically if $n$ observations are sampled at random from $\mathsf{Exp}(\text{rate} = \lambda),$ as shown in the Question above, then $T \sim \mathsf{Gamma}(\text{shape}=n,\, \text{rate}=\lambda).$

The proof is that the MGF of $X_i$ is $M_X(t) = \frac{\lambda}{1-t},$ so the MGF of $T$ is $M_T(t) = (\frac{\lambda}{1-t})^n,$ which is the MGF of $\mathsf{Gamma}(\text{shape}=n,\, \text{rate}=\lambda).$

Consequently, $\bar X \sim \mathsf{Gamma}(n, n\lambda).$ (This relationship is illustrated in the link.)