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Old 06-19-2015, 03:25 AM
iamds iamds is offline
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Default Noisy target - problem in understanding distribution

Noisy targets:

Quote:
" Indeed, we can formally
express any function f as a distribution P (y I x) by choosing P (y I x) to be
zero for all y except y = f (x) . Therefore, there is no loss of generality if we
consider the target to be a distribution rather than a function"
I am not able to understand how is no loss in generality ensured by considering target distribution and not target function ?
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Old 06-19-2015, 11:17 PM
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yaser yaser is offline
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Default Re: Noisy target - problem in understanding distribution

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Originally Posted by iamds View Post
I am not able to understand how is no loss in generality ensured by considering target distribution and not target function ?
If we can express a function as a distribution, then considering only distributions will not exclude functions, hence there would be no loss in generality. The fact that we can indeed express a function f({\bf x}) as a distribution P(y|{\bf x}) is based on using a "delta function" which allows distributions to put all the probability on a single value of y, thus making it effectively a function since y is uniquely determined by {\bf x}.
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Old 05-12-2016, 02:20 AM
waleed waleed is offline
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Default Re: Noisy target - problem in understanding distribution

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Originally Posted by yaser View Post
If we can express a function as a distribution, then considering only distributions will not exclude functions, hence there would be no loss in generality. The fact that we can indeed express a function f({\bf x}) as a distribution P(y|{\bf x}) is based on using a "delta function" which allows distributions to put all the probability on a single value of y, thus making it effectively a function since y is uniquely determined by {\bf x}.
thank you yaser
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