# How is $\frac{(j-1)}{2}:j=3,5,7,\ldots$ found in the answer to this problem?

It goes like this: Let $X$ be the number of random number selected from $\{0,1,2,\ldots,9\}$ independently until $0$ is chosen. Find the probability mass functions of $X$ and $Y=2X+1$.

I know that the probability mass function of $X$ is $(\frac{9}{10})^{i-1}(\frac{1}{10}):i=1,2,3,4,\ldots$ because I know that $i$ is the number of picks and we keep picking a non-zero number $i-1$ times and the $i^{th}$ pick is 0 with probability $(\frac{1}{10})$.

But for the second part, why is $(\frac{9}{10})^\frac{(j-1)}{2}(\frac{1}{10})$ the answer? More importantly how is $\frac{(j-1)}{2}$ the exponent on the $(\frac{9}{10})$?

Sorry for any confusion, I know what the exponent is representing I'm just wondering how it was found.

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We have $2X+1=j$ if and only if $X=\dfrac{j-1}{2}$. Now you can recycle the result you got for $X$. For $j=3,4,\dots$ we get $$\left(\frac{9}{10}\right)^{\frac{j-3}{2}}\left(\frac{1}{10}\right).$$
Remark: In this kind of geometric random variable situation, there are unfortunately two fairly common interpretations of what we are counting. One is the interpretation you took, it is the total number of trials. Then your distribution function for $X$ is correct.
Another interpretation is that one counts the number of failures until the first success. Then $X$ takes on values $0,1,2\dots$, and $\Pr(X=i)=(9/10)^i(1/10)$. In that case, $Y$ will take on the values $1,3,\dots$, and the expression given in the OP for $\Pr(Y=j)$ is the right one.
thanks for all of your great answers, I have noticed you answer a lot of my questions. But could you elaborate on why you substituted $j$ for $Y$? – TheHopefulActuary Oct 24 '12 at 16:03
We want the distribution function of $Y$. So we want a general formula for $\Pr(Y=j)$. Now it is natural to say we want $\Pr(2X+1=j)$, which is $\Pr(2X=j-1)$, which is $\Pr(X=\frac{j-1}{2})$. I am not sure this answers the question in your comment. As to why I answer many of your questions: many years of teaching probability courses. – André Nicolas Oct 24 '12 at 16:15