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Let X be a Geom($\frac{1}{2}$) random variable, and define Y=$X^{-1}$

What is the p.m.f. of Y ?

attempt:

pmf of a Geom RV in general form is $p(1-p)^{k-1}$

There is this similar question, not for the same type of RV, but it doesn't actually explain, the accepted answer states "You can finish it from there"

I've also watched this video, but in the example the range of values is explicitly defined.

I'm not sure how to relate the p.m.f of X to p.m.f of Y for this specific case of mine

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    $\begingroup$ Here, the range of values is not explicitly defined, but you can deduce that it is $\{1,2,3,4,\dots\}$, because that is what is support of a geometric random variable is. Then apply the same method in the video, the only difference being that the underlying set is infinite. $\endgroup$ – Mike Earnest Apr 18 at 0:12
  • $\begingroup$ to the person down voted; why the down vote? you can't just down vote and not explain what is wrong. There are already answers on the question. $\endgroup$ – Kal Apr 20 at 23:14
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$X$ takes positive integer valyes and $Y$ takes the values $1,\frac 1 2,\frac 1 3,...$. We have $P(Y=\frac 1 n)=P(X=n)=p(1-p)^{n-1}$.

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  • $\begingroup$ where are you getting those values? This answer does not not answer my question. Nor does it provide sufficient explanation. No source. why is this up voted? $\endgroup$ – Kal Apr 18 at 0:01
  • $\begingroup$ If $x \in \mathbb N$ the $\frac 1 x \in \{1,\frac 1 2,\frac 1 3,...\}$ right? $\endgroup$ – Kavi Rama Murthy Apr 18 at 0:02
  • $\begingroup$ Your answer states it takes values 1, 1/2,1/2 ?? $\endgroup$ – Kal Apr 18 at 0:05
  • $\begingroup$ Sorry, that was a typo. $\endgroup$ – Kavi Rama Murthy Apr 18 at 0:06
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Since $X$ is concentrated on the positive integers, $Y=1/X$ is concentrated on the set $S=\{1/k\colon\, k\geq 1\}$ i.e. $P(Y\in S)=1$. Hence it sufficies to compute the probability mass function of $Y$ for points in $S$. But $$ P(Y=k^{-1})=P(X=k)=2^{-k} $$ for $k=1,2,3, \dotsc$ as desired.

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  • $\begingroup$ Could you elaborate on the answer please? and why is it so different than the answer Kavi Rama Murthy has given? $\endgroup$ – Kal Apr 20 at 17:36
  • $\begingroup$ It is exactly the same answer. Just take p=1/2 $\endgroup$ – Foobaz John Apr 20 at 18:38
  • $\begingroup$ that'll give (1/2)*(1/2)^(n-1) ?? asking again: could you show the steps in your answer? $\endgroup$ – Kal Apr 20 at 18:44
  • $\begingroup$ Which simplified gives $2^{-n}$. The same as my answer no? $\endgroup$ – Foobaz John Apr 20 at 23:49

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