## Taylor expansion of a probability density function

To further improve my understanding of probabilities I am reading Data Analysis: A Bayesian Tutorial – which I thoroughly recommend for anyone with basic mathematical knowledge but wanting to grasp the key concepts in this area.

But I have a query about one of the concepts it introduces: a Taylor expansion of a probability density function around its maximum to give a confidence interval (this stuff matters for the day job as well as the computer science).

We have a probability density function $P(x)$ (in this context this is a posterior pdf from Bayes‘s equation, but that is not important, as far as I can tell, to the maths).

The maximum of $P(x) = P(x_0)$ and $\frac{d}{dx}P(x_0) = 0$.

To get a smoother function we look at $log_e(P) = L$. As logarithms are a strictly monotonic function $\frac{d}{dx}L(x_0) = 0$.

So expanding $L$, according to the book gives this: $L = L(x_0) + \frac{1}{2}\frac{d^2L}{dx^2}_{x_0}(x - x_0)^2 + \ldots +$ (we ignore higher terms).

I am not sure why the derivatives are the coefficients of the expansion, but I can read up on that later, but given that I understand why there is no $\frac{dL}{dx}$ term as $\frac{dL}{dx}_{x_0} = 0$.

OK … well this is the power of blogging as a means of clarifying thought: because just as I was about to ask my question – why isn’t the first term dependent on $(x - x_0)$ – I realised the answer. The first term is, in fact the zeroth term of the expansion and so the dependency on $(x - x_0)$ is in fact a dependency on $(x - x_0)^0 = 1$.

2. 