crybloodwing
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- Aug 22, 2017
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Let X1, X2, X3 and X4 be negative binomial random variables with n=1 and p=p0. Find the moment generating function of W where W=X1+X2+X3+X4. Determine the Expectation and Variance of W.
I know that the negative binomial mgf is (p/(1-(1-p)et))r. I also know that the mgf of W would be MX1* MX2* MX3 * MX4.
I believe that I would just have (p/(1-(1-p)et))X1.....and so on for each X.
I would end up with (p/(1-(1-p)et))X1+X2+X3+X4
For the Expectation, it is (r(1-p))/p and all the expectations are added together. So I would end up with (X1(1-p))/p + (X2(1-p))/p + (X3(1-p))/p + (X4(1-p))/p. And then I could simplify that into ((X1+X2+X3+X4)(1-p))/p
For variance, it would be the same but the denominator would be p2.
Is this right? If so, where does the n=1 and p=p0 come into play? I have never seen n used in a negative binomial before and I know it stands for sample size.
I know that the negative binomial mgf is (p/(1-(1-p)et))r. I also know that the mgf of W would be MX1* MX2* MX3 * MX4.
I believe that I would just have (p/(1-(1-p)et))X1.....and so on for each X.
I would end up with (p/(1-(1-p)et))X1+X2+X3+X4
For the Expectation, it is (r(1-p))/p and all the expectations are added together. So I would end up with (X1(1-p))/p + (X2(1-p))/p + (X3(1-p))/p + (X4(1-p))/p. And then I could simplify that into ((X1+X2+X3+X4)(1-p))/p
For variance, it would be the same but the denominator would be p2.
Is this right? If so, where does the n=1 and p=p0 come into play? I have never seen n used in a negative binomial before and I know it stands for sample size.