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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)?

Thiruppathirajan
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  • It can go either way. For a family of examples, see https://stats.stackexchange.com/a/35138/919. – whuber Dec 10 '19 at 15:11
  • @whuber Thanks for the info. – Thiruppathirajan Dec 10 '19 at 15:19
  • 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 https://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? – whuber Dec 10 '19 at 15:39
  • @whuber Thanks. It is about rounding in my question. I assumed rounding a normal random variable would generate error which is uniformly distributed. – Thiruppathirajan Dec 12 '19 at 02:59
  • 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 $[-1/2,1/2].$ – whuber Dec 12 '19 at 16:23

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