This seems like it should be an easy question, at least the first part, but I can't think of the right way to do it or the search terms needed to google it.

Given a normal random variable $X$ with known distribution, what is the probability that an observation $r$ came from $X$? Is this even a well-posed question, or do I need to define some priors?

My end goal is, given a large set of random variables, $X_1 \dots X_n$, I want a quick way of assigning probabilities that a given measurement is from each random variable. Assuming I can reasonably define a closeness metric, $\rho_i$, for the measurement to each $X_i$ I can do $$ \begin{align} P(r \text{ is from} X_i) &= \frac{\rho_i}{\sum_{k=1}^n \rho_k} \end{align} $$

Bonus: What if the observation $r$ includes measurement noise (normally distributed)?

Double bonus: How do I extend this to bivariate normal distributions?

  • $\begingroup$ I know you can do this for a sample, but just one observation doesn't seem quite enough, imo, to see if the distribution matches (or in other words if it comes from a population with a given distribution). $\endgroup$
    – Dolma
    Apr 17, 2013 at 8:23
  • $\begingroup$ That question is actually slightly different conceptually in that it's asking if two distributions are the same, but I'm just trying to check if a point is from a known distribution. $\endgroup$
    – ryan0270
    Apr 17, 2013 at 9:05
  • $\begingroup$ Yes yes, I might have expressed myself badly. What I was trying to say was that (and someone correct me if I'm wrong) I think one point is just not enough to check if it comes from a given distribution. It could as well be from any other distribution because one point does not convey enough information about the population it might be from. I'm not entirely sure about his though ;) $\endgroup$
    – Dolma
    Apr 17, 2013 at 9:10
  • $\begingroup$ Ah, I understand now. I was a little bit unsure of that myself, which is why I asked if it was even a well-posed question. If that's the case, then, when requiring that it comes from 1 of the distributions, $X_1 \dots X_n$, it should make the problem well-posed, shouldn't it? $\endgroup$
    – ryan0270
    Apr 17, 2013 at 9:14
  • $\begingroup$ So the new problem: An observation r comes from one of the random variables $X_1....X_n$, what is the probability that it comes from $X_i$? $\endgroup$
    – bbnkttp
    Apr 17, 2013 at 9:45

1 Answer 1


If you have a number of random variables with known distribution and you know that a value $r$ is realised by exactly one variable. Then one can define the distribution of $Y$ ($Y=i$ if $r$ from $X_i$) as being:

$$P(Y=i)= \frac{f_i(r)}{\sum_{k=1}^n f_k(r)}$$

where $f_i(x)$ is the density function of $X_i$.

Intuitively working with densities makes sense, because if the $X_i$ are continuous and $f_i(r)> f_j(r)$ for a $i$ and $j$ then there is a $\delta$ such that $P(X_i \in (r-\delta,r+\delta)) > P(X_j \in (r-\delta,r+\delta))$. But densities are not probabilities, so caution is advised.

For the bonus: If you include noise which is normally distributed with mean 0 and variance $\sigma^2$, one can take $(r-\epsilon,r+\epsilon)$ with $\epsilon$ chosen so you have coverage with which you feel comfortable. Now you can use normal probabilities and state:

$$P(Y=i) =\frac{P(X_i \in (r-\epsilon,r+\epsilon))}{\sum_{k=1}^n P(X_ \in (r-\epsilon,r+\epsilon))}$$


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