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I was reading up on Conditional expectation and Sigma field and saw this statement which I dun really understand.

The coarsest $\sigma$-field for which $\mathbb E(X\mid Y)$ is a random variable is the $\sigma$-field generated by the $Y$ that is denoted by by $\sigma(Y)$

I don't really understand this statement. Need some explanation.

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$\mathbb E(X\mid Y)$ is in particular, by definition, a real valued random variable which is $\sigma(Y)$-measurable.

Without further restrictions, we cannot assert that it is measurable with respect to a strictly smaller $\sigma$-algebra (for example take $Y=X$).

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  • $\begingroup$ the smallest $\sigma$-field is $\sigma$(Y), it can't be any smaller. am i right to say this? $\endgroup$
    – lakshmen
    Commented Sep 29, 2013 at 12:56
  • $\begingroup$ In general yes, but for example when $X$ is constant, it may be a smaller one. $\endgroup$ Commented Sep 29, 2013 at 13:02
  • $\begingroup$ "for example when X is constant, it may be a smaller one." can explain this further? $\endgroup$
    – lakshmen
    Commented Sep 29, 2013 at 13:05
  • $\begingroup$ In this case, $\mathbb E(X,\mid Y)$ is constant (equal to $\mathbb E(X)$), hence it is also measurable with respect to $\{\emptyset,\Omega\}$. $\endgroup$ Commented Sep 29, 2013 at 13:07
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    $\begingroup$ Each constant random variable is independent of any other one, but here the computation can be carried elementarily. $\endgroup$ Commented Sep 29, 2013 at 13:12

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