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Let $\textbf{X}=(X_1,\ldots,X_{n_1})$ be a random sample such that: $$X_i|\mu,\sigma^2\sim N(\mu,\sigma^2)$$ where $\sigma^2$ is known.

We assume a prior for $\mu$: $$\mu|\sigma^2\sim N\left(\mu_0,\frac{\sigma^2}{n_0}\right).$$

So we get that the posterior of $\mu|\textbf{X}$ is: $$\mu|\textbf{X}\sim N\left(\frac{n_0\mu_0+n_1\bar{X}}{n_0+n_1},\frac{\sigma^2}{n_0+n_1}\right)$$

Also, we assume the following hypothesis test:$$H_0:\mu=0 \quad \text{versus} \quad H_1:\mu>0$$

How can we find the value of $\bar{X}$ for which $P(\mu>0|\textbf{X})>\eta$ ?

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1 Answer 1

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Actually, I found out the answer:

We need to transform $$\mu|\textbf{X}\sim N\left(\frac{n_0\mu_0+n_1\bar{X}}{n_0+n_1},\frac{\sigma^2}{n_0+n_1}\right)$$ to $$Z\sim N(0,1)$$

Therefore, $$P(\mu>0|\textbf{X})=$$ $$P\left(\frac{\mu-\frac{n_0\mu_0+n_1\bar{X}}{n_0+n_1}}{\sqrt{\frac{\sigma^2}{n_0+n_1}}}>\frac{0-\frac{n_0\mu_0+n_1\bar{X}}{n_0+n_1}}{\sqrt{\frac{\sigma^2}{n_0+n_1}}}|\textbf{X}\right)=$$ $$P\left(Z>-\frac{n_0\mu_0+n_1\bar{X}}{\sigma\sqrt{n_0+n_1}}|\textbf{X}\right)=$$ $$1-P\left(Z\le-\frac{n_0\mu_0+n_1\bar{X}}{\sigma\sqrt{n_0+n_1}}|\textbf{X}\right)=$$ $$1-\Phi\left(-\frac{n_0\mu_0+n_1\bar{X}}{\sigma\sqrt{n_0+n_1}}\right)$$

So, $$1-\Phi\left(-\frac{n_0\mu_0+n_1\bar{X}}{\sigma\sqrt{n_0+n_1}}\right)>\eta\Leftrightarrow$$ $$\Phi\left(-\frac{n_0\mu_0+n_1\bar{X}}{\sigma\sqrt{n_0+n_1}}\right)<1-\eta\Leftrightarrow$$ $$-\frac{n_0\mu_0+n_1\bar{X}}{\sigma\sqrt{n_0+n_1}}<z_{1-\eta}\Leftrightarrow$$ $$\bar{X}>-\frac{\sigma\sqrt{n_0+n_1}z_{1-\eta}+n_0\mu_0}{n_1}$$

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