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Questions and Comments for:
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'''[[Generation of N-dimensional normally distributed random numbers from two categories with different priors|Generation of normally distributed random numbers from two categories with different priors]]'''
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A [https://www.projectrhea.org/learning/slectures.php slecture] by Jonghoon Jin
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Please leave me comment below if you have any questions, if you notice any errors or if you would like to discuss a topic further.
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=Questions and Comments=
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This slecture was reviewed by Khalid Tahboub:
 
This slecture was reviewed by Khalid Tahboub:
  
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4) I think it might give a nice demonstration if you plot the histogram of U and X to show that this method really functions in the desired way.
 
4) I think it might give a nice demonstration if you plot the histogram of U and X to show that this method really functions in the desired way.
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Back to '''[[Generation of N-dimensional normally distributed random numbers from two categories with different priors|Generation of normally distributed random numbers from two categories with different priors]]'''

Revision as of 16:49, 3 May 2014

Questions and Comments for: Generation of normally distributed random numbers from two categories with different priors

A slecture by Jonghoon Jin


Please leave me comment below if you have any questions, if you notice any errors or if you would like to discuss a topic further.


Questions and Comments

This slecture was reviewed by Khalid Tahboub:

Great job! few minor remarks:

1) I think the first equation should be

$ F^{-1}(u)=inf\{ x|F(x)\geq u, \quad u\in [0, 1] \} $

instead of

$ F^{-1}(u)=inf\{ x|F(x)\leq u, \quad u\in [0, 1] \} $
2) How we reach
$ X <- F^{-1}(U)\quad $
from
$ F^{-1}(u)=inf\{ x|F(x)\geq u, \quad u\in [0, 1] \} $
is not very clear to me


3)I think the equation

$ F(x) = \int_{-\infty}^x \lambda exp(-\lambda x') dx' = \int_0^x \lambda exp(-\lambda x') dx' = [-exp(-\lambda x')]_0^x = 1-exp(-\lambda x) \leq u $

should be instead

$ F(x) = \int_{-\infty}^x \lambda exp(-\lambda x') dx' = \int_0^x \lambda exp(-\lambda x') dx' = [-exp(-\lambda x')]_0^x = 1-exp(-\lambda x) \geq u $

4) I think it might give a nice demonstration if you plot the histogram of U and X to show that this method really functions in the desired way.



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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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