# Chance of two Gaussian distributions being observations of the same phenomenom?

For an algorithm used for generation of a road map based upon position-samples, I am looking for a method of determining the probability of a sample belonging to an already discovered element of the map. For the sake of simplicity, the element on the map is represented as one point $v$ (a node in the graph), for which the position $\mu_v$ is estimated using the average of all measurements $M_v$ matched to this position. Each measurement comes from a GPS-source, which gives an approximate position ($\mu_m$) and accuracy ($\sigma_m$). The accuracy follows a Gaussian distribution. The distribution $\sigma_v$ for $v$ is assumed Gaussian, for ease of computation.

In the algorithm I use the chance of a measurement and the already estimated node on the map belonging together to determine if a measurement should be merged with the node, or should become a new node on the map. If the chance of belonging together is acceptable, the measurement will be added to the node. If this chance is below a certain threshold (probably 0.05), the measurement will be added to the map generating a new node in the graph.

My question is therefore:

Given $\mu_v$, $\sigma_v$, $\mu_m$, and $\sigma_m$:

What is the chance that $m$ and $v$ are observations of the same phenomenom?

-

Maybe I am not understanding your problem correctly. if that is the case, please let me know. Here is one approach I would consider. It is more like a hypothesis test.

Suppose we have $X_1,X_2,... X_n$ is a sample from $N(\mu_1, \sigma^2_1)$ and $Y_1,Y_2,... Y_n$ is a sample from $N(\mu_2, \sigma^2_2)$. So your $X$'s are your known location estimates, and $Y$'s are the new data points you got, you are considering merging

If your null hypothesis is $\mu_1=\mu_2$ and the alternative hypothesis is that the mean's are not equal. I think there are standard hypothesis testing procedures for this type of sample tests. (I am assuming you do not really care about variance) If you google something like "testing equality of means for two normal distributions unknown variance". I remember there is an explicit formula. It will be a t-test.

I think you can still use a t-test, if you are willing to make the assumption $\sigma_1=\sigma_2$. See attachment. The statistics displayed is distributed as a $T_{m+n-2}$ distribution under the null hypothesis. Hope this helps

-
Thanks for your insight. Welch's t-test is what I came to find. The only difficulty with this t-test is the Welch-Satterthwaite equation in it, which results in a division by zero when the population size is 1. This is the case when I provide one measurement to compare to the measurement-population making the node. For now, I can work around by replacing $N_i-1$ with max($N_i-1$, 1), but better solutions are welcome. – Roelof van den Berg Mar 21 '13 at 15:06
OH! your popurlation is 1, every time? How many people are there in your original population? (roughly) – Lost1 Mar 22 '13 at 12:34
@RoelofvandenBerg I added a formula. I think this should be what you want. You cannot assume $\sigma_1\neq\sigma_2$ if you only have only 1 sample from population 2, because you cannot estimate the variance. If you have only a few samples, the variance estimate is likely to be bad anyway. – Lost1 Mar 22 '13 at 12:47
I can estimate the variance quite well with one sample, because each sample is a sample of a GPS-position (with a position, and an error-value which is the standard deviation of the measurement). – Roelof van den Berg Apr 15 '13 at 13:48
@RoelofvandenBerg ah! – Lost1 Apr 15 '13 at 13:49

From the question:

Given $\mu_v$, $\sigma_v$, $\mu_m$, and $\sigma_m$:

What is the chance that $m$ and $v$ are observations of the same phenomenom?

Clarified by the comment:

I can estimate the variance quite well with one sample, because each sample is a sample of a GPS-position (with a position, and an error-value which is the standard deviation of the measurement).

I think what you're asking is what is the chance that these are two observations of the same position. First you need to define same. It's not equal because then the probability is zero. But you can ask what is the probability that the two observations are within $d$ of each other.

Now I'm not really qualified to answer--in fact I found this question looking for the answer myself. But after thinking about it, I think what you're looking for is just the probability that the difference of the distributions is less than $x$:

$$|\mathcal{N}(\mu_v - \mu_m, \sigma^2_v + \sigma^2_m)| < d$$

which is:

$$\frac{1}{2}\Big[\mbox{erf}\Big(\frac{d-\mu}{\sigma\sqrt{2}}\Big) - \mbox{erf}\Big(\frac{-d-\mu}{\sigma\sqrt{2}}\Big)\Big]$$ where $\mu=\mu_v - \mu_m$, $\sigma = \sigma^2_v + \sigma^2_m$

-