Let $X_1,X_2,...,X_n$ be i.i.d r.v's with common p.d.f.

$$ \mbox f(x)=\frac{x^5e^{-x/\theta}}{5!\theta^6} $$

where $\theta$ > 0. Show that the Neyman-Pearson lemma produces a test of $H_0: \theta=\theta_0$ against $H_1: \theta=\theta_1 (\theta_1>\theta_0)$ with critical region of the form $\sum^n_{i=1}x_i≥c$ where $i=1,2,...n,x_i$ is the observation on $X_i$ and $c$ is some constant.

Use the result "if $X_i$ are i.i.d with $X_i\sim Ga(\alpha,\beta)$, then $\sum^n_{i=1}X_i\sim Ga(n\alpha,\beta)$" to identify the distribution of $\sum^n_{i=1}X_i$. Hence find the equation which must be solved to find $c$.

My attempt:

$$ \frac{L(\theta_0)}{L(\theta_1)} =\frac{\prod^n_{i=1}\frac{x_i^5e^{-x_i/\theta_0}}{5!\theta_0^6}}{\prod^n_{i=1}\frac{x_i^5e^{-x_i/\theta_1}}{5!\theta_1^6}} $$ Which I think simplifies to: $$ =\prod^n_{i=1}\frac{e^{-x_i/\theta_0}\theta_1^6}{e^{-x_i/\theta_1}\theta_0^6} $$

$$ =(\frac{\theta_1}{\theta_0})^{6n}e^{-\sum x_i(\frac{1}{\theta_0}-\frac{1}{\theta_1})} $$

$$ =(\frac{\theta_1}{\theta_0})^{6n}e^{-\sum x_i(\frac{1}{\theta_0}-\frac{1}{\theta_1})} $$

$$ (\frac{\theta_1}{\theta_0})^{6n}e^{-\sum x_i(\frac{1}{\theta_0}-\frac{1}{\theta_1})} <c $$

Edit: attempted continuation assuming I'm right (which is a dangerous assumption)

$$ 6n log(\frac{\theta_1}{\theta_0})-\sum x_i(\frac{1}{\theta_0}-\frac{1}{\theta_1}) <c $$

$$ \sum x_i > \frac{6n log(\frac{\theta_1}{\theta_0})-c}{\frac{1}{\theta_0}-\frac{1}{\theta_1}} $$

$$ \sum x_i > c' $$

  • 2
    $\begingroup$ taking logs of a product turns that product into a sum. $\endgroup$ – Eric Peterson May 17 '14 at 19:25
  • 1
    $\begingroup$ Routine bookwork questions are treated a little differently on the site. Please read the self-study tag wiki info and add the tag. You do show your attempt, and where you got stuck, so you may not need to edit your question further. Hopefully Eric's suggestion allows you to proceed. If you think you have a good answer, you should post it (as an answer to your question). You might consider using $g(\mathbf{x})=e^{\log(g(\mathbf{x}))}$ when making use of his suggestion. $\endgroup$ – Glen_b May 18 '14 at 1:11
  • $\begingroup$ Is this right? $$ =(\frac{\theta_1}{\theta_0})^{6n}\frac{e^{-\sum x_i/\theta_0}}{e^{-\sum x_i/\theta_1}} $$ $\endgroup$ – user123965 May 18 '14 at 9:42
  • $\begingroup$ your expression in logs at the end of your post is wrong (it's s simple error). If you fix it, you may find your progress easier. $\endgroup$ – Glen_b May 19 '14 at 4:42
  • $\begingroup$ Is that better? $\endgroup$ – user123965 May 19 '14 at 17:44

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