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Suppose X has the geometric pmf $f_X(x) = \frac{1}{3}(\frac{2}{3})^x, x=0, 1, 2, ...$ Determine the probability distribution of $Y = \frac{X}{(X+1)}$. Note that here both X and Y are discrete random variables. To specify the probability distribution of Y, specify its pmf.

What I have tried:

$Y = g(X)$, where $g(X) = \frac{X}{X+1}$ that is $Y = \frac{X}{X+1}$ and so we have $x = \frac{y}{1-y}$. Thus $g^{-1}(y)$. $$\begin{align}f_Y(y) &= \sum_{x \in g^{-1}(y)}f_X(x) \\ &= f_X \left(\frac{y}{1-y} \right) \\ &= \frac{1}{3}\left(\frac{2}{3} \right)^{\frac{y}{1-y}} \end{align}$$

I have identified $g^{-1}(y)$ for each $y \in Y$, now I have to sum the appropriate probabilities.

A similar question has been asked here. However they summed values in respect to X, I'm not certain about that approach as it hasn't been accepted. How do I proceed from here?

dollar bill
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  • Looks like the exactly the same question to me except that you correctly have the support including $x=0$ – Henry Mar 18 '22 at 00:27

1 Answers1

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$x \to \frac x {x+1}$ is a one-to-one function. So $Y$ takes the values $\frac n {n+1}$ with $P(Y=\frac n {n+1})=P(X=n)=\frac 1 3 (\frac 2 3)^{n}$, $n=0,1,2,...$ That is all there is to it.

If you prefer you can say $P(Y=a_n)=P(X=\frac {a_n} {1-a_n})=\frac 13 (\frac 2 3)^{a_n/(a_n-1)}, n=0,1,2,...$ where $a_n=\frac n {n+1}$.

Kavi Rama Murthy
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  • Thanks for showing me this - I pieced together the final parts by including the answer from the linked question. It's as you say, that's all there is to it. I only had to find the values of y. – dollar bill Mar 18 '22 at 16:50