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Given a random number $X$, from popular websites it seems to me that the expected value of $[X]$ is the same as the expected value of $X$. Where $[X]$ denotes the statistical (or banker) integer rounding of $X$. However, I tried to search a proof for this statement but could not find any material.

Can anyone give me a hand on this ?

whuber
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Phong Le
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    In some situations this form of rounding can introduce bias, so you would need to add some assumptions to get a situation in which it necessarily has none. For example, I expect you will need to be more specific about what you mean by "a random number". – Glen_b Oct 18 '20 at 07:14
  • Let's say $X$ belongs to a set of symmetric distribution (like Laplace, Gauss). If you have any relevant materials regarding the statistical bias of rounding, then I am appreciated if you can share them with me. – Phong Le Oct 18 '20 at 12:36
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    One reason you cannot find a proof is because the statement is not true. Consider, for instance, any *symmetrical* Normal distribution with mean $2.5$ and tiny standard deviation. Because most of its probability will lie in the interval $[1,3]$ and all numbers in that interval round to $2,$ virtually all the probability of the rounded value will be at $2,$ whence the mean of the rounded value will be (extremely) close to $2,$ which is far less than the mean of the original variable (compared to its standard deviation). That's a strong bias. – whuber Oct 18 '20 at 15:51
  • @whuber: No the statistical (or banker) rounding is difference, if x = 2.6 it can be rounded to 3, and x = 2.1 it will be rounded to 2, so intuitively the expected value after rounding is till 2.5. I heard something related to Sheppard's corrections, but not sure what are all conditions so that E[X] = E [ [X] ] – Phong Le Oct 18 '20 at 16:36
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    Let's slightly modify the example, then: consider a Normal distribution with a mean of 2.25 and tiny standard deviation. Almost all its values round to 2, no matter what your rounding convention might be. The bias is there and it remains large. The point is that the very best result about bias you can hope for will be an approximate one and the accuracy of the approximation will depend on the variance of the distribution. – whuber Oct 18 '20 at 16:40
  • You can also easily show that rounding to the nearest 10 (for example) introduces bias for any data where [Benford's law](https://en.wikipedia.org/wiki/Benford%27s_law) holds. – Eoin Oct 18 '20 at 18:02

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