hi guys thank you very much for the tutorial videos.God bless you all.wish you cd also upload lessons on transformation of variables techniques as well as questions involving beta distributions and gamma as well.Thanks though
This is good stuff but you could have easily done a change of base and call x-1 y, then multiple and divide inside series by (1-p)e^t. The [(1-p)e^t]^-1 will combine with [(1-p)e^t]^x to easily get you the moment. This looks simpler to me, your style is fine too.
Why is e^t * (1-p) < 1 ? I understand 1-p is always = 1, because 't' is non negative (correct me if this understanding is wrong) Then, how cum the above product be less than 1? This is a crucial assumption used in derivation for this lesson. Please explain. Need this URGENTLY!!!
The MGF of a negative binomial is only defined for the values t < ln(1/(1-p)). You can think of {t: t < ln(1/(1-p)) } as the domain of the function M(t).
You didn't include the Mode or describe the Memorylessness property as described in part 1 of Geometric Distributions, just fyi.
hi guys thank you very much for the tutorial videos.God bless you all.wish you cd also upload lessons on transformation of variables techniques as well as questions involving beta distributions and gamma as well.Thanks though
I hope many people studying SOA will watch this lecture!
This is good stuff but you could have easily done a change of base and call x-1 y, then multiple and divide inside series by (1-p)e^t. The [(1-p)e^t]^-1 will combine with [(1-p)e^t]^x to easily get you the moment. This looks simpler to me, your style is fine too.
Should r be |r|
@6:53, shouldn't r be |r|
how can we take out variance without double differentiating moment generating function or by another method
Why is e^t * (1-p) < 1 ? I understand 1-p is always = 1, because 't' is non negative (correct me if this understanding is wrong)
Then, how cum the above product be less than 1? This is a crucial assumption used in derivation for this lesson. Please explain. Need this URGENTLY!!!
The MGF of a negative binomial is only defined for the values t < ln(1/(1-p)). You can think of {t: t < ln(1/(1-p)) } as the domain of the function M(t).
Thanks for the explanation !
The expectation? is it E(X) = 1/p or E(X) = (1-p)/p or these two are the same. Please explain I am confused
I found out myself thanks
how to find Factorial MGF after finding MGF?
you are a ****ing genius...
Its really helpfull
thanks
Wonderful. Thanks.:D
Why is in minute 6: -1= 1-e^t(1-p)?
MyJansch, not sure I understand your question. Are you referring to the denominator or numerator?
In the numerator, we have to substract the -1, but instead of substracting minus 1 in the numerator you substracted 1-e^t(1-p)
Thank you, sir.
Hi, can any one explain to me why the Sigma of -1 is -1 instead of infinity?
ok, i get it, the -1 is not inside of the sigma.
Variance is E[(x-)^2]
Please upload more video