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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.

Aslan986
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    $T$ is actually $\chi^2(2n)$ distributed, since $\Gamma(k/2,2)=\chi^2(k)$. Does that give you the right answer? :) – MånsT Jun 08 '12 at 09:19
  • Ok, I wrote it wrong. I actually considered $\chi^2(2n)$, but the answer is wrong :( ... – Aslan986 Jun 08 '12 at 09:24
  • Having read your solution more carefully, I think that you got the distribution of $Y$ wrong. It should be $\Gamma(n,1/\lambda)$; see http://en.wikipedia.org/wiki/Gamma_distribution#Special_cases – MånsT Jun 08 '12 at 09:43
  • You are right. I followed Italian Wikipedia which is wrong ( http://it.wikipedia.org/wiki/Distribuzione_esponenziale#Distribuzioni ). I'f you want to write an answer i will accept it. Thank you very much. – Aslan986 Jun 08 '12 at 10:08
  • 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 :) – MånsT Jun 08 '12 at 10:22

3 Answers3

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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.)

Community
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BruceET
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  • Hi BruceET. Never is too late for a minor correction :). The exponential MGF is $\dfrac{\lambda}{\lambda -t}$ and the MGF for the Gamma$(n,\lambda)$ is $\left(\dfrac{\lambda}{\lambda -t}\right)^n$. In both cases the MGFs formulas presented have 1 instead of $\lambda$ in the denominator. The final result for $\bar X$ PDF is correct. – bluemaster Sep 15 '22 at 19:20
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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.

MånsT
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  • I believe this is incorrect. See other Answer. – BruceET Feb 20 '17 at 10:09
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    To any passing through readers: there are two ways to parameterize the Gamma distribution, either with a rate (similar to exponential rate) or a scale (similar to exponential mean) and they are related by $\text{scale}=\frac1{\text{rate}}$. The OP in the question uses $\lambda$ as a rate parameter and so it will be inherited by the Gamma as a rate parameter as well in this instance. – jdods Dec 15 '21 at 12:35
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Considering these properties:

  1. $c \times \text{Gamma} (\alpha, \lambda) \sim \text{Gamma} (n, \lambda/c)$ for $c>0$
  2. $X+Y \sim \text{Gamma} (\alpha+\beta, \lambda)$ when $X \sim \text{Gamma} (\alpha, \lambda)$ and $Y \sim \text{Gamma} (\beta, \lambda)$ are independent
  3. $\chi^2_m \sim \text{Gamma} (\frac{m}{2}, \frac{1}{2}) $

we have the following

$$ \color{blue}{2 n\lambda \bar{X} \sim \chi^2_{2n}}$$ $$ \color{blue}{n\bar{X} \sim \text{Gamma} (n, \lambda)}$$ $$ \color{blue}{\bar{X} \sim \text{Gamma} (n, n\lambda)}.$$

Amir
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