Geometric distribution of a random variable

Question

Hello, I am struggling with how to do this question. Could somebody please help me? Thank you!

Answer 1

For some positive integer $r$, we have that

$P(X>r)$

$=P(\lceil\frac{lnU}{ln(1-p)}\rceil>r)$

$=P(\frac{lnU}{ln(1-p)}>r)$

$=P(log_{1-p}^{}U>r)$

$=P(U<(1-p)^{r})=(1-p)^{r}$

so $X$ is the geometric distribution.

Answer 2

$X=n$ iff $n \leq \frac {\ln\, U} {\ln(1-p)} <n+1$ iff $(n+1) \ln(1-p) \leq \ln\, U <n\ln(1-p)$ iff $(1-p)^{n+1} \leq U < (1-p)^{n}$ so $P(X=n)=(1-p)^{n} -(1-p)^{n+1} =(1-p)^{n}(1-(1-p))=p(1-p)^{n}$