2
$\begingroup$

In section 4.4 of 'Introduction for Machine Learning' by Ethem Alpaydin the following example of estimating a prior density us given:

For example let us say that we are told that [the random variable] $\theta$ is approximately normal and with 90 percent confidence, $\theta$ lies between 5 and 9, symmetrically around 7. Then we can write $p(\theta)$ [(the prior density)] to be normal with mean 7 and because: \begin{align} P\{-1.64 < \frac{\theta - \mu}{\sigma} < 1.64\} &= 0.9\\ P\{\mu -1.64\sigma < \theta < \mu + 1.64\sigma\} &= 0.9 \end{align} we take $1.64\sigma = 2$ and use $\sigma = 2/1.64$. We can thus assume $p(\theta) = \mathcal{N}(7, (\frac{2}{1.64})^2)$.

Based on the information given I think that $p(\theta)$ looks like this:

Normal distribution described above

However I do not fully understand the equalities form the example. Since the confidence interval is 90% the standard error of the mean is 1.64. But how do you go from that knowledge to the equalities in the example. Why is the z-score used in the first equality? And why do they take 1.64$\sigma = 2$, I assume that that has something to do with the fact that $9 - 7 = 7 -5 = 2$.

Please note that I know that I can find the standard deviation by solving the following equation:

\begin{equation} \int_5^9 \frac{1}{\sqrt{2\pi}\sigma} \exp{\left(\frac{-x-7}{2x\sigma^2}\right)}dx = 0.9 \end{equation}

But the method used in the example seems a lot simpler than solving the equation ($\sigma = 1.21951 \approx \frac{2}{1.64}$) above.

$\endgroup$

1 Answer 1

1
$\begingroup$

This is simply the formula for the z-score:

$z=\frac{x-\mu}{\sigma}$

90% of the area under the curve lies between z=-1.64 and z=+1.64. So on your diagram, that corresponds to 5 and 9.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.