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I've got a one major problem using the Neyman-Pearson lemma.

We're testing hypotheses $H_0: \theta \le \theta_0$ vs. $H_1: \theta > \theta_0$. Our $$f(x,\theta) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$$

We use the Neyman-Pearson lemma and get the following expression for the critical value of the test: $$\phi(X) = \mathbf{1}_{\{f(x,\theta_1) > k f(x,\theta_0)\}}(x) + \gamma \mathbf{1}_{\{f(x,\theta_1) = k f(x,\theta_0)\}}(x)$$

Now I can get some expression in terms of $\frac{1}{n} \sum_{i=1}^n X_i > g(k)$ for the test, but I need a value for $k$ from the expression for $\phi(X)$.

How would I compute $$\mathbb{E}[\phi(X) |H_0] =\alpha$$ so that I can solve for $k$?

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Note that $\frac{1}{n} \sum_{i=1}^n X_i > g(k) \implies \sum_{i=1}^n X_i >n g(k)=c(\text{say})$

Consider your $\phi(X)=\begin{cases}1 & \text{if}\,\,\sum x_i>c \\ 0 & \text{otherwise} \end{cases}$.

Since $\sum X_i$ is coninuous R.V. we have $P[\sum X_i=c]=0$.

Then $c$ is determined from $$\begin{align} \alpha &= \mathbb{E}[\phi(X) |H_0] \\ &=1\cdot P_{H_0}[\sum X_i>c]\\&=P_{H_0}[\dfrac{2\sum X_i}{\theta}>\dfrac{2c}{\theta_0}]\\&=P[\chi^2_{2n}>\dfrac{2c}{\theta_0}] \end{align}$$

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  • $\begingroup$ Does the distribution of $X$ make any difference in the conversion to chi-squared? $\endgroup$
    – shimee
    Jun 19 '13 at 19:07
  • $\begingroup$ @shimee: Yes it will. $\endgroup$
    – A.D
    Jun 19 '13 at 19:12

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