In any real experiment, Eventhough we conduct experiment with utmost accuracy some random errors are bound to occur.
This is the Gaussian distribution which defines the probability of getting $x$ in any experiment with mean $x_0$ and standard deviation $\sigma$.
Considering this we get that,
since this is probalility then $P(x_0)\le1$ or,
$$\sigma \ge \frac{1}{\sqrt{2\pi}}$$
What this means is that however nicely an experiment is done the $\sigma$ can never be less than $\frac{1}{\sqrt{2\pi}}$
So is there a fundamental limit to accuracy or I have been wrong in this derivation?
Please Explain?

  • $\begingroup$ the probability of getting any single value x is always zero for any continuous distribution. Note that $p(x_0)$ is the density function at $x_0$ and is not the probability $P[x = \{x_0\}]$ $\endgroup$ – user144410 May 20 '18 at 8:25
  • $\begingroup$ Your expression is a probability density function for a continuous random variable. Pointwise probability is always zero; you have to specify an interval to talk about probability $\endgroup$ – daruma May 20 '18 at 8:25
  • $\begingroup$ $\sigma$ is just a parameter which we're free to choose (well, in reality it's determined by whatever source of randomness you're investigating, but mathematically it's just a parameter). You can't possibly bound it below like that, it can be any positive real number. This is a bit like saying "Let $R$ be a rectangle of width $w$ and height $h$" and then proving that $w>\frac 12$ or something. $\endgroup$ – Jack M May 20 '18 at 8:34
  • $\begingroup$ You might be interested in the topic of error analysis as discussed, for example, in the book "An Introduction to Error Analysis" by John R. Taylor. $\endgroup$ – awkward May 20 '18 at 12:22

$P(x)$ is not the probability of getting $x.$ The probability of getting $x$ is zero for any real number. (How would we even know whether we got $x$ or not in a real experiment? It's not like our instruments ever output an infinite number of digits).

$P(x)$ is a probability density function. What it tells you is that for an interval $[a,b],$ the probability that $x$ lies in $[a,b]$ is $$ \int_a^bP(x)dx.$$

As $P(x)$ is not a probability, there is no particular reason it needs to be less than one. In fact, you can see it actually has units of $1/x,$ so it doesn't even make sense to think about its relationship to $1.$

In principle, $\sigma$ can be anything (it also has units of $x$ so it would make no sense for it to be bounded by a dimensionless number). $\sigma$ is just the precision of the measurement. It's possible there's a fundamental limit on this (for physics reasons) but there's no limit set by probability theory.

  • $\begingroup$ for any interval the probability must be less than 1 $\endgroup$ – Abhishek May 20 '18 at 8:28
  • $\begingroup$ @Dr.Math Yeah, but that doesn't imply $P(x)<1.$ What if $b-a = 0.001$ and $P(x) \approx 100$ on that interval? Then the probability of being in the interval is $\int_a^b P(x)dx \approx 0.1,$ which is totally fine. But $P(x)$ is much greater than one. (Also, again, my point about units. What does $P(x)$ greater than one even mean?) $\endgroup$ – spaceisdarkgreen May 20 '18 at 8:30

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