If $X$ is a normal random variable with mean $0$ and variance $1$, then the pdf of $Y=X^2$ is $f_Y(y)=(2\pi y)^{-\frac{1}{2}}e^{-\frac{y}{2}}$ (for $y\ge0$). But $f_Y(y)$ is not integrable at $0$, since it has a divergence. So we can't use it to compute probabilities in any interval of the form $[0,a]$ (for $a>0$). Am I wrong? If so, where am I wrong? And if not, then what is the "reason" that the pdf is behaving this way?
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1$\begingroup$ No, it is integrable at $0.$ $\endgroup$– zhw.Nov 7, 2015 at 20:19
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Yes, I was wrong. The pdf of $Y$ is indeed integrable. A function with a singularity might indeed be integrable, as in the example: $$\int_0^a{y^{-\frac{1}{2}}dy}=2a^{\frac{1}{2}}$$
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$\begingroup$ Indeed, the integral obtained by dropping the exponential factor from the integrand provides an upper bound on the cdf, since the exponential factor reduces the integrand in practice. This proves integrability of the pdf. $\endgroup$– J.G.Nov 7, 2015 at 22:15