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?


1 Answer 1


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, 2013 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, 2013 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, 2013 at 19:54

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