2
$\begingroup$

Let A_r be a real-valued normal random variable whose mean is an integer A.

Let A_i be the rounding of A_r such that A_r = A_i + e, where e is an uniformly distributed random variable taking the value from -0.5 to 0.5.

Now taking variance,

Var(A_r)=Var(A_i)+var(e)+cov(A_i,e)=Var(A_i)+1/12+cov(A_i,e);

What would be the cov(A_i,e)? Will it be less than -1/12 so that Var(A_r) is always less than Var(A_i)?

$\endgroup$
5
  • $\begingroup$ It can go either way. For a family of examples, see stats.stackexchange.com/a/35138/919. $\endgroup$
    – whuber
    Dec 10, 2019 at 15:11
  • $\begingroup$ @whuber Thanks for the info. $\endgroup$ Dec 10, 2019 at 15:19
  • $\begingroup$ You're welcome. Please note that the rounding error is not usually uniformly distributed, although in the case of a Normal random variable $A$ it may be remarkably close to uniform: see the discussion of method 8 at stats.stackexchange.com/a/117711/919. This raises questions concerning what you are trying to ask: is your question actually about rounding or is it about adding uniformly-distributed noise to a variable? $\endgroup$
    – whuber
    Dec 10, 2019 at 15:39
  • $\begingroup$ @whuber Thanks. It is about rounding in my question. I assumed rounding a normal random variable would generate error which is uniformly distributed. $\endgroup$ Dec 12, 2019 at 2:59
  • $\begingroup$ It's a fair assumption but not quite correct. At best we can say the distribution of errors will be symmetric if and only if the mean of the Normal variable is an integral multiple of $1/2.$ Since all uniform distributions on intervals are symmetric, this rules out that possibility in general. As an extreme example consider a Normal distribution with a mean of $1/4$ and such a tiny standard deviation that almost all its probability lies between $0$ and $1/2.$ Most of the rounding errors will be in the interval $[-1/2,0],$ showing they cannot be uniformly distributed on $[-1/2,1/2].$ $\endgroup$
    – whuber
    Dec 12, 2019 at 16:23

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.