+Michael Zeng it is recursive... the way he simplifies Sum(ke^(-lambda))/k! to equal lambda is the same to simplify Sum(je^(-lambda))/j! to equal lambda. He should have used E(k) and E(j) that is where the confusion happened . But anyways both E(k) and E(j) simplify to lambda.
Very easy to understand
Thank u for making my concepts clear...
Love from India 🇮🇳
Doing the work most professors can’t/couldn’t do
The website is not working. It is showing a 404 error.
replace .html with .pdf
thanks for making the concepts so clear... you should consider making more videos.. much love
i have a quiz tmrw....... its 1:30am now......your videos are awesome
i m in the exactly same situation....and it's 1:30
Here 12.45a
3: 14 am 😅
Still reviewing your video. I have my exam at 7:00pm
Please tell me about pdf of Poisson Distribution
also,in 17:01, what kind of expansion is that again? I dont understand it.
+Michael Zeng Taylor Series, the one he showed before too while proving sum of PMF = 1.
why it is equal to E[x] in 13:45 with the extra j????
+Michael Zeng it is recursive... the way he simplifies Sum(ke^(-lambda))/k! to equal lambda is the same to simplify Sum(je^(-lambda))/j! to equal lambda. He should have used E(k) and E(j) that is where the confusion happened . But anyways both E(k) and E(j) simplify to lambda.
well done
very clearly!
amazing
great!