0

As the title suggested, if I draw two sample independently from the same continuous distribution, say uniform or normal, what is the probability those two samples have the same value? (I think I know how to do it when the distribution is discrete, but stuck at the continuous version)

Sam
  • 461
  • 1
    See the answer to https://math.stackexchange.com/questions/63794/probability-of-iid-random-variables-to-be-equal. – jakobdt Jan 05 '23 at 12:13
  • @jakobdt Actually the answer on the linked question only works if there is a density. – drhab Jan 05 '23 at 12:55
  • @drhab ...and the assumption here is that the distribution is continuous, hence it has a density. – jakobdt Jan 05 '23 at 14:20
  • @jakobdt No, not every continuous distribution has a density (wrt Lebesgue measure). If it has then the distribution is "absolutely continuous". – drhab Jan 05 '23 at 14:25
  • 1
    @drhab In my experience the terms "continuous distribution" and "absolutely continuous distribution" are used interchangeably (probably because many people do not know the difference). But I understand your point, and it is (as you do in your answer) good to note that even just continuity of one of the variables results in $P(X=Y)=0$. – jakobdt Jan 05 '23 at 14:34

1 Answers1

1

If $X,Y$ are independent random variables defined on the same probability space and $X$ has continuous distribution then: $$P(X=Y)=\int\int[x=y]dF_X(x)dF_Y(y)$$where function $[x=y]:\mathbb R^2\to\mathbb R$ takes value $1$ if $x=y$ and takes value $0$ otherwise.

Working that out we find:$$P(X=Y)=\int P(X=y)dF_Y(y)=0$$The second equality finds its cause in the continuïty of the distribution of $X$ which means exactly that $P(X=y)=0$ for every $y\in\mathbb R$.

Note that for this it is not necessary that both random variables have the same distribution. Moreover it is enough already if only one of them has continuous distribution.

drhab
  • 151,093
  • What if I use discrete distribution to approximate the continuous distribution? Then it depends on how accurate the approximation is? Something like as the granularity get is more and more refined, the probability becomes vanishingly small? – Sam Jan 05 '23 at 13:44
  • Yes, but what could be the reason for doing it on such cumbersome way? This especially if you can do it neatly on a direct and concise way. – drhab Jan 05 '23 at 14:31
  • Because in the scenario I'm interested in, the distribution may not be exactly continuous. But if I know the probability of that is $0$ for continuous case, then I can argue that in the limit it is approximately $0$ for discrete case. – Sam Jan 07 '23 at 05:17
  • Oh, I understand. Thank you for telling me. – drhab Jan 07 '23 at 07:46
  • no worries. your answer is very helpful so thank you too – Sam Jan 07 '23 at 08:09