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Old 06-10-2013, 04:32 AM
kynnjo kynnjo is offline
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Default Exchange of expectations in derivation of bias-variance decomposition

[I originally posted this question to the wrong sub-forum. My apologies.]

In the derivation of the bias-variance decomposition (on p. 63), there is a step in which the taking of expectation wrt \mathcal{D} and wrt \mathbf{x} are exchanged.

It's not clear to me that these two expectations commute: the choice of \mathbf{x} depends on the choice of \mathcal{D}, and viceversa.

I would appreciate a clarification on this point.

Thanks in advance,

kj

P.S. Pardon the "raw LaTeX" above. Is there a better way to include mathematical notation in these posts.
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Old 06-10-2013, 07:10 AM
Elroch Elroch is offline
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Default Re: Exchange of expectations in derivation of bias-variance decomposition

With regard to LaTex, you just need to surround your LaTex code with (math) and (/math), except that those brackets need to be square.

With regard to the exchange of expectations, this can always be done with probability distributions. The condition that makes it easier than swapping the order of integrations and sums in general is that probability densities are always positive. Problems with swapping the order of integrals and sums only occur when you have conditional convergence, with infinite positive and negative contributions cancelling out in a way which is order-dependent.

[EDIT: thanks to Yaser for being more precise than me. In this case it's not merely that probability distributions are positive that matters, it's also that the error function being integrated is non-negative. Without this, the change of order could be invalid if the function was pathological. It would be safe if the function was Lebesgue measurable].
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Old 06-10-2013, 12:55 PM
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yaser yaser is offline
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Default Re: Exchange of expectations in derivation of bias-variance decomposition

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Originally Posted by kynnjo View Post
[I originally posted this question to the wrong sub-forum. My apologies.]

In the derivation of the bias-variance decomposition (on p. 63), there is a step in which the taking of expectation wrt \mathcal{D} and wrt \mathbf{x} are exchanged.

It's not clear to me that these two expectations commute: the choice of \mathbf{x} depends on the choice of \mathcal{D}, and viceversa.

I would appreciate a clarification on this point.

Thanks in advance,

kj

P.S. Pardon the "raw LaTeX" above. Is there a better way to include mathematical notation in these posts.
It is not independence that allows us to change the order of integration. It is the fact that the integrand is always nonnegative. Think of it as a double summation. You are adding up the same set of numbers whether you start with one sum or the other. The problem arises when some of these numbers are positive and some are negative, since in that case you can play tricks with different orders of the summation to converge to different values. Look up "absolute convergence" versus "conditional convergence."
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