Search results
Results from the WOW.Com Content Network
1.E(X) = n p, or 2.E(Y) = n(1 − p) p 1. E (X) = n p, or 2. E (Y) = n (1 − p) p. Proof for 1.: Proof for the calculation of mean in negative binomial distribution. Proof for 2: Although I can't find a concrete proof on stackexchange, this is the expected value used in the wikipedia article for negative binomials, and I have also seen this ...
The reason is that if we recall the PMF of a negative binomial distribution, Pr [X = x] = (r + x − 1 x)pr(1 − p)x, the relationship between the factors pr and (1 − p)x are such that the bases must add to 1. So long as 0 <(1 − p)eu <1, we can think of this as a Bernoulli probability of a single trial; i.e., let 1 − p ∗ = (1 − p)eu ...
Due to the differences in notation for the formula of the CDF of negative binomial distribution from Wikipedia, ScienceDirect and Vose Software, I decide to rewrite it in the way that I can easily
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
A final word: perhaps the most elegant computation is to exploit the fact that the negative binomial distribution is a generalization (i.e., a sum of IID) geometric random variables. But the purpose of this answer is to show how the computation can be done purely as an algebraic manipulation with very few prerequisites.
I am using the definition of the negative binomial distribution from here. This is the same definition that Matlab uses. For convenience, P(k) = (r + k − 1 k)pr(1 − p)k, where p is the probability of success. P(k) is the probability of k failures before r successes. The probability generating function is supposed to be, g(x) = (p 1 − (1 ...
the moment generating function of the negative binomial distribution. 0.
I was asked to derive the mean and variance for the negative binomial using the moment generating function of the negative binomial. However i am not sure how to go about using the formula to go out and actually solve for the mean and variance.
My textbook did the derivation for the binomial distribution, but omitted the derivations for the Negative Binomial Distribution. I know it is supposed to be similar to the Geometric, but it is not only limited to one success/failure. (i.e the way I understand it is that the negative binomial is the sum of independent geometric random variables).
1. There are two distributions called Geometric. 1. The distribution of Bernoulli trials until a failure. ( This is sometimes called the Shifted Geometric Distribution. Assuming: Z = X + Y and P(X = k) = P(Y = k) = (1 − p)pk − 1, k ∈ {1, 2...} then. P(Z = z) = z − 1 ∑ k = 1P(X = k)P(Y = z − k) Note the summation bounds = z − 1 ∑ ...