We've got a random sample of iid $X_1,\dots,X_n$. We're testing the mean of $X \sim \mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. The size of the test $\alpha=0.05$.

$H_0: \mu=0$

$H_1: \mu=v$

By the Neyman-Pearson lemma the Most Powerful test is $\phi(X) = \mathbf{1}_A$ where the set $A =\{ x: \prod_{i=1}^n \frac{f(\mu_1,\sigma^2)}{f(\mu_0,\sigma^2)} > k \} $

simplyfying we can reduce the test to: \begin{equation} \frac{1}{n}\sum_{i=1}^n X_i > \frac{2\sigma^2(\log k+\mu_1^2 n)}{n \mu_1} \end{equation}

where $\mu_1 = v$. Calculating the critical value $k$ we evaluate: $$ \begin{align} \alpha=\mathbb{E}[\phi(X) |H_0] &= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty}\phi(x) e^{-\frac{x^2}{2\sigma^2}} dx\\ &=\frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty}\mathbf{1}_A e^{-\frac{x^2}{2\sigma^2}} dx\\ &= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{k}^{\infty}e^{-\frac{(x-\mu_1)^2}{2\sigma^2}} dx\\ &= \frac{1}{\sqrt{2 \pi }} \int_{\frac{k-\mu_1}{\sigma}}^{\frac{\infty-\mu_1}{\sigma}}e^{-\frac{x^2}{2}} dx\\ &=1-\Phi\Bigg(\frac{k-\mu_1}{\sigma} \Bigg) \end{align} $$

therefore we can derive $k=\mu_1 + \sigma \Phi^{-1}(1-\alpha)$

Is my approach correct, or did I mess up the calculation of the critical value?


You messed up the calculation of the critical value. First of all It is true that the test statistic you have is reduced to

$$T(X)=\frac{1}{n}\sum_{i=1}^n X_i>\gamma,\quad\quad \mu>0$$

Let $T(X)$ be a random variable $Y$. Then,

$Y\sim{\mathcal{N}}(0,\frac{\sigma^2}{n})$ if $\mathcal{H_0}$ is correct

$Y\sim{\mathcal{N}}(\mu,\frac{\sigma^2}{n})$ if $\mathcal{H_1}$ is correct

Now we simply have $$\alpha=P(Y>\gamma|\mathcal{H_0})=Q\left(\frac{\gamma}{\sqrt{\sigma^2/n}}\right)$$

As you can see $\alpha$ is independent of $\mu$. In a similar way you can calculate the detection probability as $$\beta=P(Y>\gamma|H_1)=Q\left(\frac{\gamma-\mu}{\sqrt{\sigma^2/n}}\right)$$

From here, since $Q$ is monotonically increasing as $1-Q$ is a CDF, it accepts an inverse form and eventually the critical value can be found as


  • $\begingroup$ So you mean that my PDF for the expectation was incorrect? $\endgroup$
    – shimee
    May 31 '13 at 15:53
  • $\begingroup$ Yes exactly. False alarm rate in this problem must be independent of $\mu$. Check the relevant parts in your solution. $\endgroup$ May 31 '13 at 16:22
  • $\begingroup$ I guess your approach makes sense. Right, you have a point that the distribution of the test statistic differs from that of the population. Alright. $\endgroup$
    – shimee
    May 31 '13 at 19:54

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