CDF (cumulative frequency) of multiple samples in summed normals?

Say I have some normally distributed data. I have an application where I compute the percentile (or cumulative frequency less than sample) for a particular sample using a CDF function along with the mean $\mu$ and standard deviation $\sigma$ of the samples.

so $$F_x(x) = \frac 12\left[1 + \text{erf} \left (\frac {x - \mu}{\sqrt{2 \sigma^2}}\right)\right]$$

Now I find myself in a situation where I want to determine the cumulative frequency of multiple samples across multiple data sets (finding something akin to an overall percentile of, say, three samples). Now assuming the variables are independent, I can sum the normals using

$$(\mu_\text{sum}, \sigma_\text{sum}) = (\mu_x + \mu_y + \mu_z), (\sqrt{σ^2_x + σ^2_y + σ^2_z})$$

Can I then sum the individual samples I care about and compare them to the new summed normal to compute a percentile of the three samples compared to the sum of the normals? Something tells me this doesn't work but I'd like to be sure. So I'm thinking something like computing the CDF using the sum of the samples I'm interested in:

$$F_x(x_x + x_y + x_z)$$

and using the $\mu$sum and $\sigma$sum in the CDF function above.

• Your notation is now confusing and contradictory. I suggest saying "Let $S=X+Y+Z$," replacing "$\text{sum}$" with "$s$" in subscripts and $F_x(x_x + x_y + x_z)$ with $F_S(x + y + z)$. If you agree that is what you're asking, then your question would make sense, and the short answer is "yes, that's what you do". Commented Oct 18, 2013 at 6:02

Your question implies that for independent random variables, $σ_\text{sum} = σ_x + σ_y + σ_z$. This is not the case.

The squares are additive: $σ^2_\text{sum} = σ^2_x + σ^2_y + σ^2_z$. So $σ_\text{sum} = \sqrt{σ^2_x + σ^2_y + σ^2_z}$.

However, otherwise you're correct - if you add three independent normal random variables, the distribution of the sum is normal with mean equal to the sum of their means and variance equal to the sum of their variances (indeed that applies to adding any number of terms).

Variances of correlated random variables are a little more complicated, but still straightforward. For correlated multivariate normals, you also still have normality.

• You're right, I've edited my question. The real question though, is can I use the new $\mu_\text{sum}$ and $\sigma_\text{sum}$ to compute a CDF using the sum of three specific samples. I don't think this works but I wanted to be sure. Commented Oct 18, 2013 at 5:37
• @mikepk are you asking whether you can use sample means and standard deviations to compute the population distribution function? Or are you trying to find the ECDF of the summed sample values? Or something else? What are you actually trying to achieve in the end? Commented Oct 18, 2013 at 5:40
• Thanks Glen, I'm probably not asking the right questions. I have a collection of three separate scores on three separate normally distributed ranges. Usually I compare the individual scores against all scores using a percentile (for which I'm using the CDF). I'd like to compare the collection of all three scores against all the other possible collections of those three scores. Effectively how does this collection compare to the population of all collections. Like I said in my original question, my stats memory is dim, but I don't think this works. Commented Oct 18, 2013 at 14:53
• Reading over the original question I think I've even confused myself :). So I have a single sample $x = 300$ that I want the CDF of, so I get a value back like $F_X(300) = 0.6$. So the probability of all random samples being <= 300 is less than or equal is 0.6. Now I have three samples lets say (300, 900, 100) from three different data sets (all normally distributed) and I want to do something similar, the probability of three samples being less than or equal to that particular collection of three samples. The more I think about this the more it doesn't quite make sense to me. Commented Oct 18, 2013 at 15:21
• @mikepk How do you come to know $\mu_X$ and $\sigma_X$? If you do know them and can assume independence, then I don't see any difficulty with doing what you suggest. Commented Oct 18, 2013 at 19:17