This is a question regarding the central limit theorem. In my model, I have five sources of disturbances, each following a particular distribution. I sample the data from each and sum to determine the final disturbance. Will the distribution of the final disturbances be normal?

I read this sentence " In large problems, we can invoke the central limit theorem for all but the first few completion times and use normal distributions as close approximations to the convolutions we need. " -Kenneth Baker I am unable to justify it. Since only the means of the large sample follow normal distribution how can the random variable themselves form normal distribution when sampled? This is a question on application of central limit theorem!

  • $\begingroup$ What are the distributions for the individual disturbances? You have some disturbance $\epsilon_i$ that is the sum of 5 other disturbances (eg. $\epsilon_i = w_i + u_i + v_i + \eta_i + \phi_i$)? $\endgroup$ Nov 5, 2016 at 4:37
  • $\begingroup$ @MatthewGunn yes. disturbance is sum of 5 other disturbances. the distributions of the individuals are gamma, exponential, normal etc. (how does it matter?; does the answer depend on the distributions?) $\endgroup$
    – kosmos
    Nov 5, 2016 at 4:39
  • $\begingroup$ I advise you do a QQPlot of your disturbance term and see how far it is from being normally distributed. $\endgroup$ Nov 5, 2016 at 5:20

1 Answer 1


As I understand the question:

  • You have 5 independent random variables: eg. $u$, $v$, $w$, $x$, $y$.
  • Random variable $z$ is the sum: $z = u + v + w + x + y$.
  • You want to know if $z$ is normally distributed.

The answer is it depends (but from what you said, probably not)

Summing 5 independent random variables does not magically make the sum normally distributed.

  • If $u$, $v$, $w$, $x$ and $y$ all are normally distributed, then their sum is normal.
  • If any of $u$, $v$, etc... are not normally distributed:
    • The sum is not going to be normal either (unless you're in a very peculiar case).
    • As a practical matter, summing them will move things closer to the normal distribution, in some sense.

You can't invoke a classic central limit here because you only have 5 random variables. You can't answer this question in some abstract, general way. The answer will depend on what's the exact distribution of your disturbance terms. Why not do a QQPlot and see what it looks like?


It could be really quite off (eg. this is a sum of a beta(.4..4), 3*gamma(1,.1), and 3 normals with stdev .05). enter image description here

Or it could sorta be moving in the right direction... (eg. this is a sum of 5 IID uniform random variables) enter image description here But it's still off... enter image description here

  • $\begingroup$ "As a practical matter, summing them will move things closer to the normal distribution, in some sense"... Could you please explain this? $\endgroup$
    – kosmos
    Nov 5, 2016 at 5:39
  • 1
    $\begingroup$ @kosmos The central limit is about $n \rightarrow \infty$ for some normalized sum of $n$ random variables. $n=5$ isn't $\infty$, but you might, in some sense, be moving in the right direction. All I can say is play around with some simulations. Note also that you're not normalizing anything either, and if the next disturbance term you add is huge, it can make things look incredibly worse if the disturbance term is both non-normal and larger than the others. $\endgroup$ Nov 5, 2016 at 5:44

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