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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.

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    yes that makes perfect sense now. Thank you!2012-10-24