# Finding a HPD Interval for unknown population mean

I'm doing a question from a Bayesian stats textbook for study and am a bit confused with the result. Pretty sure I've messed something up but I'm not sure exactly what is going on here so any help is much appreciated.

QUESTION: A random sample of $$25$$ observations is obtained from a normal distribution with $$\mu$$ unknown and a known variance of $$\sigma^2=1$$. The sample mean is found to be $$0.29$$. Given that $$\mu>0$$, use a flat prior to determine the posterior. Plot this posterior and determine a $$95\%$$ highest posterior density (HPD) interval for $$\mu$$.

My attempt: The bayesian posterior here (with sample mean $$y=0.29$$, and utilising a flat prior) is given by $$\mu|\mathbf{y}, \sigma^2 \sim N\bigg(\overline{y},\frac{\sigma^2 }{n}\bigg)$$

We known $$n=25$$ and $$\sigma^2=1$$, so the posterior is

$$\mu|\mathbf{y} \sim N\bigg(0.29,\frac{1}{25}\bigg) = N\big(0.29,0.04\big).$$

Then the plot of the distribution looks like this: Plot of posterior

We see that the posterior is symmetric and unimodal, and so the $$95\%$$ HPD interval should just be the regular $$95\%$$ CI of $$(0.2116, 0.3684)$$, right? Something seems wrong here because I don't see why the question would be asking for a HPD interval if this was actually the case...

1. I amended your notation of $$\mu|\mathbf{y}$$; I understood what you meant but your notation was wrong.

2. the result you found is incorrect:

$$\overline{y}-1.96\times \frac{\sigma}{\sqrt{n}}\leq\mu\leq \overline{y}-1.96\times \frac{\sigma}{\sqrt{n}}$$

$$0.29-1.96\times \frac{1}{5}\leq\mu\leq 0.29-1.96\times \frac{1}{5}$$

...you wrongly considered $$\frac{1}{25}$$ instead of $$\frac{1}{\sqrt{25}}$$

Finally: Bayesian CI can be calculated in several ways, Centered Intervals or HPD.

The fact that, in this particular case, they are the same it is not significant.