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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?

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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.

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

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