# Distribution of a probability density function

I have probability density function of the log normal distribution $$f_X(x, \mu, \sigma^2)$$, where $$x$$ is the random variable, $$\mu$$ is the location and $$\sigma^2$$ is the scale parameters of the distribution. For various sets of reasons I have realised that in order to solve a problem that I have, I might need to calculate the expectation of the PDF with respect to $$x$$. So this is not a $$E(x)$$, but it is $$E(f_X(x, \mu, \sigma^2))$$ that I need. In a way, I am saying that the PDF itself is a random variable. I would be able to calculate the expectation myself if I knew the distribution of the PDF of the log normal distribution. Unfortunately, I cannot find any information on this.

So, my question: is there a theorem, or any property, or anything else that would connect PDF of exponential family (log normal in my case) with the distribution of the PDF?

You can compute the mean of any function of $$X$$, provided said mean exists. In particular, $$\Bbb Ef_X(X)=\int_{\Bbb R}f_X^2(x) dx$$. In the lognormal case $$f_X=\frac{1}{\sigma x\sqrt{2\pi}}\exp-\frac{(\ln x-\mu)^2}{2\sigma^2}$$ on the support $$[0,\,\infty)$$. We'll use $$z:=\frac{\ln x-\mu}{\sigma}$$ to compute $$Ef_X^2(X)=\int_0^\infty\frac{1}{2\pi\sigma^2 x^2}\exp-\frac{(\ln x-\mu)^2}{\sigma^2}dx=\int_{\Bbb R}\frac{1}{2\pi\sigma}\exp (-\mu-\sigma z-z^2)dz.$$Now we'll substitute $$w=z+\frac{\sigma}{2}$$, so the mean is$$\int_{\Bbb R}\frac{1}{2\pi\sigma}\exp\left(\frac{\sigma^2}{4}-\mu-w^2\right)dw=\frac{1}{2\sigma\sqrt{\pi}}\exp\left(\frac{\sigma^2}{4}-\mu\right).$$Now double-check my arithmetic.