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I have that the conditional probability density of $Y|X$ is as such

$f_{Y | X} \propto x^{y - 1}(1-x)^{n-y-1}\alpha^{n-y}\beta^{y}$

where $\alpha, \beta$ are constants in $(0, 1)$, $x$ is a random variable in $(0, 1)$, and $y = 0, 1, 2, ..., n$.

I want to determine what distribution this conditional probability density belongs to. My immediate thought was the geometric distribution, but this seems impossible. If I shed the terms that aren't a function of $y$ then I get

\begin{align*} f_{Y | X} &\propto x^{y}(1-x)^{-y}\alpha^{-y}\beta^{y} \\ &= (x\beta)^{y}((1-x)\alpha)^{-y} \\ &= \left(\frac{x\beta}{(1-x)\alpha}\right)^{y} \end{align*}

but I can't recall any distributions that match that either.

I would be very grateful for any assistance.

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    $\begingroup$ Isn't that a geometric distribution with parameter $p=(x\beta)/((1-x)\alpha)$? $\endgroup$
    – Thomas Lumley
    Commented Apr 22 at 5:30
  • $\begingroup$ If it is a density, i.e. from a continuous distribution, then perhaps an exponential distribution with parameter $\lambda= -\log_e \left(\dfrac{x\beta}{(1-x)\alpha}\right)$ $\endgroup$
    – Henry
    Commented Apr 22 at 9:46
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    $\begingroup$ But you say $y = 0, 1, 2, ..., n$, so not a continuous distribution, i.e. you do not have a density. If $y$ cannot exceed $n$, then you have a probability mass function from a truncated geometric distribution. $\endgroup$
    – Henry
    Commented Apr 22 at 9:50
  • $\begingroup$ That algebraic rearrangement is a good start. Perhaps writing $q=x\beta/((1-x)\alpha),$ which expresses the conditional probability mass function of $y$ more simply as $q^y$ (for integers $0\le y\le n$), will help you see the answer. Among the many possible characterizations it can be viewed as a form of a truncated Geometric distribution. $\endgroup$
    – whuber ♦
    Commented Apr 22 at 14:06

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