Can you please name a few.. i am new to this and would very much appreciate your help. Thanks in advance. :)

Your words mean a lot, thank you. And I wish you best of luck in your PhD!

Hey, that's super cool! I'm clearly more of a math person so I always love hearing when people have an actual application of some of these topics. Thanks!

Glad it was helpful!

Glad you’re here!

Glad it helped!

My pleasure 😊

You're most welcome!

You're very welcome!

You are welcome!

thanks!

Thank you 🙂

Thank you! Will do!

Glad it was helpful!

Hey, thanks for the kind words. I get this question often and the admittedly unsatisfying answer is no. I've found that different resources out there do a really good job at different things or at least offer different ways of viewing the same problem. When I try and learn a new topic, or review an old topic when making a video, I'll look around at lots of different resources to see which path I want to take in explaining it. That said, I think the most important part for learning (in my opinion) is to write some basic code, doesn't have to be pretty, which implements the method. That way, you can do sanity checks to see if your understanding matches to how real data would behave. Plus, you get some coding experience out of it. Sorry to not have an answer to your initial question but I hope this helps regardless!

@@ritvikmath No problem. Thanks for the long response!

Thank you very much!

I appreciate that!

No problem!

Thanks a lot 😊

You're reading my mind. I put this video out first so that in the MCMC videos (releasing next week onward), we can compare it against this. Stay tuned :)

Both. There's something called adaptive rejection sampling which creates some "proposal" distribution on the fly (See bishop book for more details). In general, rejection sampling has issues in high dimensions, Bishop's book explains it well (explained at the end of the section on adaptive rejection sampling)

So, [f(x)/Mg(x)] will be some number, say its 0.1. Then, we accept that sample with probability 0.1. That means, we generate some random number "u" from the Uniform distribution and if it is

Up

I have the same question.

Hi, the threshold is "f(s)/(Mg(s))", which is in (0,1). Since the larger result, say, f(s) greater, g(s) smaller, indicates g(s) can represent f(s) better, this sample should be accepted with greater probability. So now we can just generate u~U(0,1), and s~g(s), if u < the threshold, we accept. The process means, the more f(s)/(Mg(s)) close to 1, the higher probability u

This is a valid question and indeed something that I also had confusion over for a long time. This is a common case in Bayesian stats. For example, P(A|B) is proportional to P(B|A)P(A) / P(B) by Bayes theorem. We might care about sampling from P(A|B) which is the posterior but don't know its full form since the denominator P(B) might be difficult to compute. So, we can use Accept-Reject sampling to still sample from the posterior given only the numerator in Bayes theorem.

@@ritvikmath That makes much sense. P(B) is the normalizing constant. It is interesting questions like this come up all over the place in engineering.

The ratio you get from f(X)/Mg(X) gives you the probability of acceptance. Then you use a uniform distribution from 0 to 1 and if the value is less than the ratio, you accept. If it's bigger you reject.

You bring up a very interesting question. Usually, the distributions, p, that we want to sample from are not very nice looking (can have many peaks, noisy, etc.), so finding a distribution g that is "similar" to p can be challenging or impossible. So, instead we use a g that is "close enough" to the target and use ARS to actually sample from the target.

@@ritvikmath thx for answering my question. So ARS can smooth the potentially noisy density or find an easy alternative for p, if you have an approx/good candidate g for p, right? But what is the advantage against kernel density estimation KDE with special kernel k?

To accept the sample s, you need to sample from the uniform distribution on the interval [0, M*g(s)]. Let's say it is u. If u is less than or equal to f(s), you accept, otherwise you reject.

You can use calculus. Find the maximum value of the ratio f(x)/g(y). If you then set M in the denominator, f(x)/Mg(x), you ensure the ratio won't be greater than 1