An introduction to inverse transform sampling
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- Опубликовано: 14 май 2018
- Explains how to independently sample from a distribution using inverse transform sampling.
This video is part of a lecture course which closely follows the material covered in the book, "A Student's Guide to Bayesian Statistics", published by Sage, which is available to order on Amazon here: www.amazon.co.uk/Students-Gui...
For more information on all things Bayesian, have a look at: ben-lambert.com/bayesian/. The playlist for the lecture course is here: • A Student's Guide to B...
You have no idea how long i looked for an explanation of this for an uni assignment thank you so much.
Thank you so much! Very nice series for Bayesian Statistics! I watch them every day.
What an amazing video.! Never thought I could understand this concept within 11 minutes
Thanks Ben, you are always great in explaining concepts. I am happy :-)
Thank you so much Ben. you do not know how much I was looking for this.
Great explanation Thanks
Thank you so much for this, Ben!
Hello Ben, huge fan here :) Thank you for your videos they are awesome.
Wonderful explanation. Thank you so much :)
great explanation, thanks!
Why do we have to "go through" the CDF computation to generate a normally distributed variable from a uniformly distributed one? One could have a hunch that since we ultimately want that the resulting variable follows a desired PDE, it seems it should be enough to invert the desired PDE and compute back the variables from uniformly distributed variables. There are some proofs on wikipedia but I can't get any intuition about it. Thanks!
you mentioned that calculating the CDF in higher dimensions is not feasible. Is that because it requires us the calculate the integral?
Good question - yep, that’s right; calculating the CDF via integration typically isn’t possible. Best, Ben
Then, why is important the uniform pdf?. I mean you could sample directly from one distribution to another just by putting the value returned from the CDF of the first pdf as input to the inverse CDF of pdf you want to arrive at. Am I wrong?
thanks!
I was searching like a mad lad where was this originally published - does anybody know? It's probably Smirnov, but I can't find anything.
How come sometimes I see the inverse transformation of an exponential distribution written like X=(-1/lamda)*log(u) ? Am I missing something, I’ve seen it an answer for the actuary CT6 2018 paper question 1, I’m confused please shed some light on it
Lamda is the rate parameter for an exponential distribution, in Ben's example the rate parameter is one so it kinda just "disappeared", but the PDF of an exponential distribution is -l*e^-lx (l is lamda), and you can verify yourself that the inverse CDF ends up being X=(-1/l)*log(1-u). As for the 1-U vs U, U is a uniform random variable taken from 0-1 so sampling from 0-1 and taking 1 minus a sample from 0 to 1 is the same thing.
(1-u) and u have the same distribution in this case
Is CDF same as Probability Density Function (PDF)?
No the CDF is the integral over the PDF that should equal 1.
No, as @Michel said. It's the cumulative distribution function.
If You want to increment the visits I believe You have to write full your speaked. In this way everyone will translate and understand better with help of a translater 😉
Bye and thanks ☺
it has to be ln not log