Could you make a ""The expectation and variance for the log-normal random variable"" video? Your normal approach is so easy to understand. I only have lognormal that has not been proved yet. Other methods on the Internet are not as concise and easy to understand as you said.
I’m able to follow you with everything except for around 8:40 when you say something to the effect of “this substitution of alpha for 1/2 allows us to take the negative derivative of the integrand”. Why exactly are we able to take the negative derivative of y^2e^-alphay^2?
Thank you very much for this video, it has been very useful! Just a comment, is it possible that the substitution (at 9:20) is written backwards on the slide? I think it should be Y = z / sq (2 * alpha) instead of z = y / sq (2 * alpha) ... Anyway, that's just a detail, thanks again for the video!
Great video, thanks! And for whoever's watching it and wondering - at 09:19 there's a small mistake: the substitution is z = y * sqrt(2 \alpha), and not the other way around as shown. :)
Danke
Could you make a ""The expectation and variance for the log-normal random variable"" video? Your normal approach is so easy to understand. I only have lognormal that has not been proved yet. Other methods on the Internet are not as concise and easy to understand as you said.
I’m able to follow you with everything except for around 8:40 when you say something to the effect of “this substitution of alpha for 1/2 allows us to take the negative derivative of the integrand”. Why exactly are we able to take the negative derivative of y^2e^-alphay^2?
Thank you very much ....
Awesome explanation
thank you so much you saved my life
Thank you very much for this video, it has been very useful!
Just a comment, is it possible that the substitution (at 9:20) is written backwards on the slide? I think it should be Y = z / sq (2 * alpha) instead of z = y / sq (2 * alpha) ... Anyway, that's just a detail, thanks again for the video!
Thanks for your comment. Yes, you are correct. Thanks for pointing this out. I am glad that you found the video useful.
this was so helpful thank you - christoph
Great video... Thank you for your help
Amazing!
Thank you !!
❤❤❤
thank you so much
Great video, thanks! And for whoever's watching it and wondering - at 09:19 there's a small mistake: the substitution is z = y * sqrt(2 \alpha), and not the other way around as shown. :)
非常好 good🎉我愛中國
why is dy/dx = 1/sigma ?
Taking derivative with respect to x of both sides of previous equation in 4:47 (y=x-mu/sigma) might help.
If you dont want to explain sth dont do it.
Very poor to just state results without expaining where it came from.
He explained the best