Does anyone know what the following distribution is, please? I can tell it is discrete and $\theta$ is somehow a probability. Thank you!
$$\mathbb P_X(x; \theta) = (x-1) \theta^2 (1-\theta)^{x-2}, \ \ x=2,3,\ldots; \ \ 0 < \theta < 1.$$
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$\begingroup$ What distributions do you think it looks similar to? $\endgroup$– jskMay 3, 2014 at 7:04
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$\begingroup$ @jsk I guess it is some counting model maybe with shift. It is not Binomial since it has only one parameter. It is not poisson, either since it has no exponential. It is not geometric due to the $(x-1)$. Nor is negative binomial or hypergeometric. That is pretty much all discrete cases I can think of. $\endgroup$– LaTeXFanMay 3, 2014 at 7:19
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$\begingroup$ I need to find expectation of some function of $X$ such as $\frac{1}{X+7}$. On a second thought, I guess it does not really matter what distribution it is since I do not think it will be helpful for solving the expectation. $\endgroup$– LaTeXFanMay 3, 2014 at 7:33
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$\begingroup$ 20826 -- I'm afraid you ruled out the distribution it actually is in your comments. $\endgroup$– Glen_bMay 3, 2014 at 7:39
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$\begingroup$ @Glen_b Yes, I see that. But I also realize that it does not help much to find out the distribution. How to find the expectation of $\frac{1}{x+7}$ and $\frac{1}{(x+7)^2}$ easily without using the definition of expectation? $\endgroup$– LaTeXFanMay 3, 2014 at 7:44
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1 Answer
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It's a negative binomial (the number of trials form) with $p=1-\theta$ and $r=2$
http://en.wikipedia.org/wiki/Negative_binomial#Alternative_formulations
$\Pr(X = x) = {x-1 \choose x-r} (1-p)^r p^{x-r} \quad\text{for }x = r, r+1, r+2, \dots, $
