# 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?

## 1 Answer

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.

• Thanks, J.G.! This is a quite neat solution! Great stuff! – Ivan Svetunkov Apr 4 at 22:45