Geometric distribution and expectation of polynomials of random variable

Asked Feb 12 '18 at 20:12

Active Feb 12 '18 at 20:12

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Let $X$ be a random variable geometrically distributed. I have known that geometric distribution is memoryless which means

$P(X>i+j|X>i)=P(X>j)$.

Now I want to compute the value of $E(X(X-1)(X-2)...(X-r+1))=\frac{r!p^r}{(1-p)^r}$.

I see the answer from this post below. But I really don't understand what does $U$ in this answer mean. Is there another way to interpret the solution? Expectation and variance of the geometric distribution

asked Feb 12 '18 at 20:12

And asking about the meaning of U by posting a comment over there is not an option because? – Did Feb 12 '18 at 21:05
Sorry for that. But I was asking for another interpretation of the solution. – You_Don't_Know_Who Feb 12 '18 at 23:02

0 Answers0