Negative binomial distribution

#1
Hello,

I am a student at the university of Hasselt (Belgium) in management, and I had to follow a statistical course. It is for me very difficult since I simply don't have any background in mathematics and statistics.

I am struggling with an exercise that might be easy but I really don't know where to start and so what to do. I know how to answer question b) but I don't know what are the mean and variance of Y. And I just don't know how to do a), which , I suppose, will help me to do b)

So if some could help me, I would be grateful!
 

rogojel

TS Contributor
#2
hi,
first of all I would say that this is an amazing exercise for someone studying management - anyone expecting future managers to be fluent in higher mathematics has not seen many people in this role :p. Nevertheless, a few hints:

I do not think you need point a) to solve b). a) is talking about the distribution of X, you only need the mean and the variance of Y to solve b).

Concerning point a) you just need to apply the total probability formula to calculate the distribution of x - which reduces to calculating a nice integral, once you substituted the formulas for the conditional probalty of x|y and y. Good luck !

regards
rogojel
 
#3
Hi! Thanks for your answer. Yes I must admit at first I hated the course but now that I get 90% of it, I just like it haha.

So for a), you mean that P(X) = ∑ P(X|Y)P(Y)
And I already have P(X|Y), right. P(Y) is supposed to have a gamma distribution, but don't I need to have the value for alpha and beta ? for both a) and b) ? Sorry if it's obvious I don't get it :mad:
 

rogojel

TS Contributor
#4
hi,
yes, except that you will have an integral instead of a sum. The way the problem is formulated, unless the alpha and beta were specified earlier, you just have to keep them as constants. Question a) only asks you to prove that the distribution of X is negative binomial, so having an alpha and a beta in the formula is no problem.

I guess the same will have to do for b) as well.

regards
rogojel
 
#7
Erf, even with that i can't do it, i must be terrible at statistics / mathematics xD

The whole part without "y" can go outside the integral right ? But what do I do with the "x!" within the integral ?

Does Γ(alpha) = (1 - alpha)! ?

Also I usually have limits, like Y lies within a and b, here is it infinity ? Or is an integral without doing this substition at the end ?
 

JesperHP

TS Contributor
#8
1) take constant outside integral (yes...The whole part without "y")
2) Then use that [math]\int_0^\infty y^{\alpha-1}\exp(-\beta y) dy = \frac{\Gamma(\alpha)}{\beta^\alpha} [/math] ... this is probably proven in curriculum... or you can derive it yourself from definition of gammafunction and a simple substitution...
3) you then get something with some Gammafunctions and x! and [math](1+\beta)^\alpha[/math] and some other stuff.... it does not look like the negative binomial but will apparently be when reparameterized



and by the way you are not necessarily bad at math.... the integral is in a sense not really "solved" since the gammafunction just is an integral so if you are staring at the integral thinking OMG it just shows you have never learned to manipulate and utilize the gammafunction...but hey youre learning now :)

PS... [math]\Gamma(\alpha) = (\alpha-1)![/math]

also if you have any reference for an example of actually applying this model I would be very interested...??
 
Last edited: