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I am confused at how the normal distribution's PDF capable of calculating a density for a single variable. I understand that the CDF probability of an exact continuous random variable $X$ is 0. Therefore, to calculate probability of $X$, we may define a range such that probability of $X$ is $P(a < X < b)$. It appears this range is usually referred to as the interval (please correct me if I am wrong).

PDF for normal distribution is $\frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu)^2}{2 \sigma^2}}$ so if we assume $x=1$, $\mu=0$ and $\sigma=1$ the result from these parameters is 0.2419707 density using dnorm in R. How is the PDF capable of coming to this conclusion as we do not specify a interval?

indrid
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Cumulative distribution function (CDF): $$F_X(x) = \Pr[X \le x],$$ the probability of observing a value of the random variable $X$ less than some given value $x$.

Probability density function (PDF): $$f_X(x) = F_X'(x) = \lim_{\Delta x \to 0} \frac{1}{\Delta x}\Pr[x < X < x + \Delta x],$$ the instantaneous rate of change of the CDF at $x$, or the limit of the ratio of the probability that $X$ is observed in the interval $(x,x + \Delta x)$ divided by the length $\Delta x$ of that interval, as the width of the interval tends to $0$. More formally we can define the PDF in terms of the Radon-Nikodym derivative, but the above definition suffices for the typical continuous random variables encountered in elementary probability. Loosely speaking, the PDF measures in some sense a normalized likelihood of observing $X$ to be a particular value. It is not a probability.

heropup
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