In relation to the summation, p is a constant, then p can be moved outside the sign of summation. But with respect to the derivative, p is the variable, and can not be moved out.
This is due to the fact that you can differentiate a power series. You cannot do that for any infinite sum, even if it converges without knowing that the infinite sum of the derivative converges uniformly. Check out "interchanging the order of differentiation and summation" on maths stackexchange
Hi, So that depends on how the geometric random variable is defined. Some define it to be the trial number on which the first success occurs. In this case, the mean is 1/p. However, some define it to be the number of failures that occur before the first success. In this case, the mean will be 1 less than 1/p, or 1/p - 1, or (1-p)/p. Be aware that several of the classic/well-known random variables are defined in different ways. The geometric is not the only one. The negative binomial is another. Some define the negative binomial to model the number of trials until the rth success. However, some define it to be the number of additional trials needed beyond r to achieve the nth success. These different definitions can result in shifting the mean.
@@mojones1 Thanks a lot Matthew. I was taking the probability mass function as p(1-p)^k and not p(1-p)^k-1. As you said it depends on how the geometric random variable is defined .
@@pavanadapa5832 The reason why we multiplied with that is because (1-p)^k in the previous result can be written as (1-p)^k-1 * (1-p)^1. Essentially we just treated (1-p) as the base and add the exponent. So by doing so, you can multiply p(1-p)/p since p cancel each other and makes no difference. But its actually multiplied by the expression since it eventually helps in deriving the subsequent steps of the proof. For example, with a numerical example: lets say 3^k+1 can be written as 3^k + 3. So that value 3 can be written with any coefficent that would not affect it and let it be p, So that would be p(3/p) and thus is still 3. Thats how then we multiply with p(1-p/p) in the proof. Hope that helps
Hey man, even though you made this video 8 years ago I'd like to let you know you completely saved my sanity today. THANK YOU!
best one to explain the variance of geormetric dist
觉得这个证明是最方便的一个了，学习到了，谢谢！
shouldn't we set (1-p) as a seperate variable to derive it?
Excellent video, thank you...
Excuse me. I want to ask you a question.
What kind of conditions that we "can't" move the d/dp outside?
In relation to the summation, p is a constant, then p can be moved outside the sign of summation. But with respect to the derivative, p is the variable, and can not be moved out.
This is due to the fact that you can differentiate a power series. You cannot do that for any infinite sum, even if it converges without knowing that the infinite sum of the derivative converges uniformly. Check out "interchanging the order of differentiation and summation" on maths stackexchange
so NB, too 抽象，I love this way lol. How did u conduct the formula in this way? Genius!
Wonderful....
Thanks 😊
excellent
Isn't the expectation value of a geometric 1-p/p and not 1/p?
Hi, So that depends on how the geometric random variable is defined. Some define it to be the trial number on which the first success occurs. In this case, the mean is 1/p. However, some define it to be the number of failures that occur before the first success. In this case, the mean will be 1 less than 1/p, or 1/p - 1, or (1-p)/p.
Be aware that several of the classic/well-known random variables are defined in different ways. The geometric is not the only one. The negative binomial is another. Some define the negative binomial to model the number of trials until the rth success. However, some define it to be the number of additional trials needed beyond r to achieve the nth success. These different definitions can result in shifting the mean.
@@mojones1 Thanks a lot Matthew. I was taking the probability mass function as p(1-p)^k and not p(1-p)^k-1.
As you said it depends on how the geometric random variable is defined .
At 3:39 why did you multiply by p*(1-p/p) thats the only part of the proof that i'm confused on
Ohh never mind, i now get it why you multiplied by it. Your vid was so helpful , thanks very much
@@phuthangmaponopono Can you explain to me why we multiply by p*(1-p/p)
@@pavanadapa5832 The reason why we multiplied with that is because (1-p)^k in the previous result can be written as (1-p)^k-1 * (1-p)^1. Essentially we just treated (1-p) as the base and add the exponent. So by doing so, you can multiply p(1-p)/p since p cancel each other and makes no difference. But its actually multiplied by the expression since it eventually helps in deriving the subsequent steps of the proof. For example, with a numerical example: lets say 3^k+1 can be written as 3^k + 3. So that value 3 can be written with any coefficent that would not affect it and let it be p, So that would be p(3/p) and thus is still 3. Thats how then we multiply with p(1-p/p) in the proof.
Hope that helps
why but why is there a negative sign before derivative there? Desperate to know!
Using the chain rule to derivate p. The p is actually a function of f(p)=1-p. The sign of p is -1.
yeah I noticed that later.