probability: finding Probability density function Let $X\sim\exp(λ)$ and $Y$ equals its decimal part.
How would you find the probability density function of $Y$?
I started by looking for $F_Y(y)$ but got stuch in this level:
$F_Y(y)=P(Y\le y)=1-P(Y>y)= \text{?}$
 A: Let $f$ be a bounded continuous fonction. Splitting up $\int_0^\infty=\sum_{n=0}^\infty \int_n^{n+1}$, we have:
$$
E[f(Y)] = \sum_{n=0}^\infty \int_0^1 f(u) \lambda e^{-\lambda(n+u)}du = \int_0^1 f(u) \frac{\lambda e^{-\lambda u}}{1-e^{-\lambda}}du.
$$
From here, you easily recover that the density of $Y$ with respect to Lebesgue's measure on $[0,1]$ is
$$
u \mapsto \frac{\lambda e^{-\lambda u}}{1-e^{-\lambda}}
$$
If you want $\lambda$ to be the expected value of $X$, replace every $\lambda$ by $\lambda^{-1}$.
A: I'm going to assume "decimal part" means "fractional part", so that it is $X-\lfloor X\rfloor$, where $\lfloor X\rfloor$ is the greatest integer less than or equal to $X$.
I will write the density as $\alpha e^{-\alpha x}$ for $x>0$ (and $0$ for $x<0$), and let you worry about whether $\alpha=\lambda$ or $\alpha=1/\lambda$.
The conditional density of $X$ given that $n\le X<n+1$, where $n\ge0$ is an integer, is
$$
x\mapsto \frac{\alpha e^{-\alpha x}}{\int_n^{n+1} e^{-\alpha w}\left(\alpha\,dw\right)}
= \frac{\alpha e^{-\alpha x}}{\left(e^{-\alpha n} - e^{-\alpha (n+1)}\right)}.
$$
The fractional part is just this shifted $n$ units leftward, so the conditional density of the fractional part, given that the integer part is $n$, is
$$
\frac{\alpha e^{-\alpha(x+n)}}{\left(e^{-\alpha n} - e^{-\alpha (n+1)}\right)} = \frac{\alpha e^{-\alpha x}}{\left(1-e^{-\alpha}\right)} \qquad \text{for }0<x<1,\text{ and }0\text{ otherwise}.
$$
Lo and behold: This does not depend on $n$.  The conditional distribution of the fractional part, given the integer part, does not depend on the integer part.  Therefore the fractional part and the integer part are independent.  And above, we have the density of the fractional part.
