I'm having trouble deriving the PDF for the following problem (from the book Doing Bayesian analysis with R and BUGS):
Consider a spinner of the kind often found with board games, but with a log10 scale from 1 to 100 (covering 0 to 360 deg). The spinner is fair, so any value is equally likely. What is the pdf at point $x$?
I know that the $x$ value at angle $t$ is given by
$$x = 100^\text{t/360}$$
Doing some transformation
$$x = \exp^{(t/360) \times \log(100)}$$
$$pdf(x) = pm(x)/dx$$
I also know
$$dt/dx = 360/(\log100 \times x)$$
Question
How to derive the answer below, using the information above?
Book answer
$$pdf(x) = 1/(2 \times ln(10) \times x)$$
Thanks.
self-study
tag to the question. $\endgroup$ – COOLSerdash Oct 7 '13 at 15:02