Deriving the q-Gaussian PDF

Question

Ok it may sound a bit too simple but I am quite confused here. While studying generalized entropic forms, in my case that of $S_q$ or in another words the Tsallis Entropy, I reach a point where I have to derive the maximal distribution that corresponds to $S_q$.

In order for that to be done, one has to impose some constraints and follow the Lagrange parameters method. In this particular case the constraints required would be:

\begin{align} & \int_{0}^{\infty}p(x)dx=1 \quad \text{(Normalization Constraint)} \\ & \langle x_q \rangle=\int_{0}^{\infty}xP(x)dx=X_q \quad \text{(q-mean value)} \end{align} where $P(x)$ is called the Escort Distribution and is defined as: \begin{align} P(x)=\frac{[p(x)]^q}{\int_{0}^{\infty}[p(k)]^qdk} \end{align} Now we define the quantity: \begin{equation} Φ(x;p;q)=\frac{1-\int_{0}^{\infty}[p(x)]^qdx}{q-1}-α \int_{0}^{\infty}p(x)dx-\beta_q \frac{\int_{0}^{\infty}x[p(x)]^qdx}{\int_{0}^{\infty}[p(x)]^qdx} \end{equation} and demand that $\partial{Φ}/\partial{p}=0$. By solving that, one ends up with the pdf: \begin{equation} p_{opt}(x)=\frac{e_q^{-\beta_q(x-X_q)}}{\int_{0}^{\infty} e_q^{-\beta_q (x'-X_q)}dx'} \end{equation} where $e_q^x$ are the q-expodentials. This $p_{opt}(x)$ is also known as q-Gaussian pdf.

My problem in deriving the pdf, is that I am not able to see how to calculate the quantity:

\begin{equation} \frac{\partial{}}{\partial{p}}\left( \frac{\int_{0}^{\infty}x[p(x)]^qdx}{\int_{0}^{\infty}[p(x)]^qdx}\right) \end{equation}

Perhaps treating it like a function of the form $h(x)=f(x)/g(x)$?

Am I making a mistake thinking of it in this way? Because I am not able to reach the final formula of the pdf. Also, I am not able to find any paper where the derivation of $p_{opt}$ is worked out.

I would really appreciate your help. Thank you!

Answer 1

EDIT: This is only an answer to the first version of the asker's question.

This looks like physicists' notation, where p and x are codependent variables. If you want to derive this by $p$, you need to formulate the entire term as being dependent on only $p$ instead of $x$. If $p$ is dependent on $x$, you need to rewrite the dependency to make $x$ dependent on $p$, i.e. you need to convert your function $p(x)$ to an implicit or explicit function $x(p)$, use the integral substitution rules to reformulate $dx$ to $dp$, then derive

$$Φ(x(p);p;q) =\frac{\int_{p(0)}^{p(\infty)}x(p)p^qx'(p)dp}{\int_{p(0)}^{p(\infty)}p^qx'(p)dp}.$$

Answer 2

The chain rule for derivatives looks like this for physicists:

$${\partial{\Phi}\over\partial{x}} = {\partial{\Phi}\over\partial{p}} {\partial{p}\over\partial{x}}$$

This means that ${\partial{\Phi}\over\partial{x}} = 0$ is a necessary condition for ${\partial{\Phi}\over\partial{p}} = 0$. I assume that ${\partial{\Phi}\over\partial{x}} = 0$ is the differential equation you are looking for.