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02ps-06 Heavytail Guassian Solution

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    The correct answer here is no.
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    Suppose we let "x" go to infinity,
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    then x-Mu2 for any fixed Mu would go to infinity.
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    So if x with a -infinity would go to zero--
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    because it is a constant--
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    Therefore, we know that in the limit of x goes to infinity--
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    this expression must be zero.
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    However, in this graph--it stays at alpha, and doesn't go to zero--
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    So, therefore, there can't be a valid Mu at sigma square.
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    If a deep improbability, you know that the area in the Gaussian
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    has to integrate into one,
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    and this area diverges, it is actually infinite in size,
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    so it's not even a valid execution.
Title:
02ps-06 Heavytail Guassian Solution
Description:

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Video Language:
English
Team:
Udacity
Project:
CS373 - Artificial Intelligence
Duration:
0:44
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