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I was going through the concept of probability density functions and had a small confusion about the notation that a pdf can take values greater than one.I found this How can a probability density be greater than one and integrate to one useful but the answers involved pdf's that are defined over a closed interval.

Is it correct to say that if a continuous random variable $X$ can take any value between $(-\infty,\infty)$, then the maximum value that its pdf can take is 1?

For example if $X\sim\mathcal{N}(0,\sigma^2)$, then $f_X(x)\leq1$ for every possible value of $x\in\mathbb{R}$.

In other words, pdfs that are defined over limited domain can take values greater than 1 but if they can take any possible value then their pdfs will never exceed value 1.

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No, that is not true. If a random variable can take any value on the real line, but it is exceedingly likely that said variable will be between $0$ and $0.1$, then the pdf will have values that are, if not over, then at least close to $10$ in that small interval.

As another example, if $X\sim\mathcal{N}(0,\sigma^2)$ with $\sigma$ very small, then $f_X(0)$ will be very large, and you can get it as large as you want by choosing $\sigma$ ever closer to $0$. For instance, $X\sim\mathcal{N}(0,0.1^2)$ has $f_X(0) = 3.99$, and for each time you divide $\sigma$ by $10$, $f_X(0)$ is multiplied by $10$.

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  • $\begingroup$ Thank you for answering it. However, I am not sure if scaling by factor of 10 in your last example is correct. since $f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp(-x^2/2\sigma^2)$, we do get a scaling factor of 10 because of $\frac{1}{\sigma_{new}\sqrt{2\pi}}, \sigma_{new}=\frac{\sigma}{10}$ part but what about the $\exp$ part. It will also change with $\sigma_{new}$ $\endgroup$ – NAASI Jun 21 '16 at 23:32
  • $\begingroup$ But you did answer the original question..... $\endgroup$ – NAASI Jun 21 '16 at 23:34
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    $\begingroup$ @NAASI Yes, the exponential part will also change, but since we're evaluating at $x = 0$, it will end up becoming $\exp(0) = 1$ anyways. It is, however, the changing exponential part that allows the total integral to become $1$ even though the value at $x = 0$ increases (the graph of $f_X(x)$ becomes narrower from the exponential part and taller from the non-exponential part at "the same pace", in the sense that the total area remains the same). $\endgroup$ – Arthur Jun 21 '16 at 23:51

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