Probability density function of an exponential minus its integer part: I have an exponential distribution $X\sim exp(\lambda)$, and another distribution $T=$ $X - \lfloor{X}\rfloor$ and I need to find the pdf the random variable T.
I know the pdf of $\lfloor{X}\rfloor$, which is $P(\lfloor{X}\rfloor = k) =(1-e^{-\lambda})(e^{-\lambda k})$.
Here's what I did up until now:
$F_T = P(T\leq t) = P(X-\lfloor{X}\rfloor \leq t) = \sum\limits_{k=0}^\infty P( X \leq t +k | \lfloor{X}\rfloor =k) = $
$ \sum\limits_{k=0}^\infty\frac{P( X \leq t +k , \lfloor{X}\rfloor =k)}{P(\lfloor{X}\rfloor =k)}$
But I am pretty sure $X$ and $\lfloor{X}\rfloor$ aren't independent which means I don't really know how to proceed.
Thanks in advance.

Ok I found a solution thanks to a comment, here it is:
$$ P(T\leq t) = \sum\limits_{n=0}^\infty P(n \leq X \leq n + t) = \sum\limits_{n=0}^\infty {F_X(n+t)-F_X(n)}$$
hence
$$P(T\leq t) =
\sum\limits_{n=0}^\infty [1-e^{-\lambda(n+t)} - (1-e^{\lambda n})]=  \frac{1}{1-e^{-\lambda}}- \frac{e^{-\lambda t}}{1-e^{-\lambda}} = \frac{1-e^{-\lambda t}}{1-e^{-\lambda}}$$
Thus, on $(0,1)$, the density is:
$$ f_T(t)= \frac{\lambda e^{-\lambda t}}{1-e^{-\lambda}}$$
