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I honestly have no idea where to start with this problem:

Suppose $X$ has the geometric distribution with PMF

$$f(x) = \frac 1 3 \left( \frac 2 3 \right)^x, x=1,2,.....$$

Determine the probability distribution of $Y=x/(x+1).$ Note that both $X$ and $Y$ are discrete random variables. To specify the probability distribution of $Y,$ specify its PMF.

Kevin Souza
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  • You should have $Y=X/(X+1)$ rather than $Y=x/(x+1). \qquad $ – Michael Hardy Oct 14 '19 at 18:53
  • I mentioned this in response to your comment on my answer, but I think you do mean to have $x = 0, 1, 2, \dots$ instead of $x = 1, 2, \dots$. Note that the probability function in your edit does not have total probability of $1$, which it needs to: https://www.wolframalpha.com/input/?i=sum+of+%281%2F3%29%282%2F3%29%5Ex+from+x%3D+1+to+infinity – Aaron Montgomery Oct 14 '19 at 19:28

1 Answers1

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To specify a variable's PMF, you need two things: (a) a list of what values it can be, and a corresponding list of probabilities of it being those values.

You already have the PMF for $X$, but let's make a table out of it to make it a bit more concrete. $$\begin{array}{r | l} x & p(x) \\ \hline 0 & 1/3 \\ 1 & \frac 1 3 \cdot \frac 2 3 = \frac 2 9 \\ 2 & \frac 1 3 \cdot (\frac 2 3)^2 = \frac 4 {27} \\ \vdots & \vdots \end{array}$$

I obtained the probabilities, denote above as $p(x)$, by plugging each value of $x$ into the given PMF.

Once you have these, you now have a complete listing of what $Y$ can be, along with all its probabilities. For instance, if $X = 0$, then $Y = \frac{X}{X+1} = \frac{0}{1} = 0$ as well; this will occur with probability $1/3$. If $X = 1$, then $Y = \frac{1}{2}$, which will occur with probability $2/9$, and so forth. Using this idea, you could make a table that looks just like the above, to get the the PMF for $Y$.

Of course, it may be more desirable to put this into a formula; you may want something compact like the expression you gave for $f(x)$. I claim that the clearest way to see what such a formula should be is to write out the first 3 or 4 lines of a table like the above, though.

Aaron Montgomery
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