Does anyone know how to find this pdf? I understand that $[X]\sim\mathrm{ Ge} \left(1-e^{-\frac{1}{a}}\right)$, but I got stuck after that ($[X]$ is the integer part of $X$). I would be really grateful for some input.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Because this looks like a textbook problem, I will sketch an answer and leave some details to be completed by you.
Let the PDF of $X$, which is supported on the positive numbers $(0,\infty)$, be given by the function $f_X$. Write $Y=X-[X]$. Because this value obviously must lie between $0$ and $1$, consider an arbitrary number $y$ with $0\le y \lt 1$. The PDF of $Y$--call it $f_Y$--evaluated at $y$ is the probability that $Y \in [y, y+dy)$ for an infinitesimal $dy$. But this event describes all positive numbers whose fractional part is close to $y$; specifically,
$$X \in [y, y+dy) \cup [y+1, y+1+dy) \cup \cdots \cup [y+n, y+n+dy) \cup \cdots.$$
Because this is a countable union of disjoint events, probabilities for $X$ are given by the density $f_X$ by summing the probabilities of the events:
$$f_Y(y)dy=\Pr(Y\in[y, y+dy)) = \sum_{n=0}^\infty \Pr(X\in[y+n, y+n+dy)) = \sum_{n=0}^\infty f_X(y+n)dy.$$
The lefthand graphic plots the PDF of $X$. Colors distinguish the integer parts of $X$: blue for $[X]=0$, red for $[X]=1$, gold for $[X]=2$, and so on. The formula for $Y$ shifts each colored section leftwards against the vertical axis. The PDF for $Y$ is obtained by stacking the heights of each shifted piece, as shown in the righthand graphic.
Since $f_X(x) = e^{-x/a}$ for any positive $x$, this sum becomes
$$f_Y(y)dy = \sum_{n=0}^\infty \exp(-(y+n)/a)dy = \exp(-y/a)\left(\sum_{n=0}^\infty \exp(-a)^n\right)dy.$$
The sum is a geometric series, easily computed in closed form. Comparing the coefficients of $dy$ on each side yields a nice formula for $f_Y(y)$.
