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We all know that probability of a single point should be equal to zero. I am a bit confused , then why $f(0)=\frac{1}{\sqrt{2\pi}}$, where $f(x) = \frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}$ is PDF for standard normal distribution.

Thanks

Ahsan Iqbal
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  • For a continuous distribution, you get non-zero results if you ask for the probability that your variable is in some range. For instance, the probability that a normal variable is in the interval $[0,1]$ has a non-zero value. – lulu Oct 13 '21 at 14:30
  • https://math.stackexchange.com/questions/210630/what-does-the-value-of-a-probability-density-function-pdf-at-some-x-indicate https://math.stackexchange.com/questions/3849959/if-gx-is-my-density-function-then-gx-0-is-simply-px-x-0 https://math.stackexchange.com/questions/1720053/how-can-a-probability-density-function-pdf-be-greater-than-1 https://math.stackexchange.com/questions/4263617/the-probability-of-any-particular-value-of-a-continuous-distribution-occurring-i https://math.stackexchange.com/questions/180283 – leonbloy Oct 13 '21 at 14:42

1 Answers1

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Because $f_X(x_0)$ does not show a probability but only a probability density.

Observe that $f(x_0)$ can also be greater than 1

take as an example $X\sim N(0;0.1)$

In this case you have $f_X(0)\approx 1.26$

The probability is calculated using an integral. The fact that for continuous distribution the probability of any point is zero is showed in the following way

$$\mathbb{P}[X=x_0]=\int_{x_0}^{x_0}f(x)dx=0$$

tommik
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