0

I have a random variable $X$ with PDF $f_X(x)=\frac{e^{-x}}{(1+e^{-x})^2}, \ x\in(-\infty,\infty)$. When I apply the transformation $Y=X^2$, I obtain Y with PDF $f_Y(y)=\frac{1}{2\sqrt{y}}\left(\frac{e^{\sqrt{y}}}{(1+e^{\sqrt{y}})^2}+\frac{e^{-\sqrt{y}}}{(1+e^{-\sqrt{y}})^2}\right) \ y\in[0,\infty)$. I checked that this integrates to one, however what strikes me as strange is that the PDF is undefined at zero. What does this mean? Is this a valid PDF? Do all transformations result in valid PDFs given that the starting PDF was valid?

Undertherainbow
  • 760
  • I am not a theorist, so others can perhaps comment more precisely. My understanding is that it is quite OK for the PDF to have infinite values, see here and here for examples. In my mind, the CDF is the more "fundamental" object and the PDF is merely its derivative, if the derivative exists. – antkam Sep 23 '19 at 18:57
  • Further, if the PDF is continuous and finite at $x_0$, then $P( [x_0 - \delta, x_0 + \delta] ) \to 2 f(x_0) \delta$, i.e. the prob shrinks linearly with $\delta$. Any time it shrinks slower than linearly, it will have infinite PDF, I think. I haven't checked your math, but I am curious if $P(Y \le \delta)$ in your case shrinks as $\propto \sqrt{\delta}$... – antkam Sep 23 '19 at 19:01

0 Answers0