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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:

$F_Y(y)=P(Y\le y)=1-P(Y>y)= \text{?}$

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What do you mean by its decimal part? Might you have meant its integer part? Or maybe its fractional part? –  Michael Hardy Jan 18 '13 at 16:41
    
There's alway an ambiguity in the kind of notation you use. Do you mean $\lambda$ is the expected value of $X$, so that the density is $(1/\lambda)e^{-x/\lambda}$ for $x>0$, or that the rate is $\lambda$, so that the expected value is $1/\lambda$ and the density is $\lambda e^{-\lambda x}$ for $x>0$? –  Michael Hardy Jan 18 '13 at 16:44
    
to clear it up: I meant it's fractional part. and by λ I meant for it's expected value. –  adamco Jan 18 '13 at 16:46
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up vote 1 down vote accepted

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.

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And/Hence this distribution is also the distribution of $X$ conditioned by $X\leqslant1$. –  Did Jan 18 '13 at 16:58
    
@Michael Hardy: The density for $X|\{\lfloor X \rfloor = n\}$ was calculated. Then, we wish to find the density for $X - \lfloor X \rfloor | \{\lfloor X \rfloor = n\} = X - n | \{\lfloor X \rfloor = n\}$. For a random variable $X$, the density of $X - C$ is the density of $X$ shifted to the right by $C$. That is, $f_{X-C}(x) = f_{X}(x+C)$. Why, in the post, was the conditional density of $X$, given the floor part, shifted to the right $n$ units (from my viewpoint), instead of left $n$ units? –  jrand Feb 2 '13 at 2:22
    
The density of $X-C$ is the density of $X$ shifted to the LEFT by $C$, that is $f_{X-C}(x)=f_X(x+C)$. If you change $x$ to $x+C$ and $C>0$, you're shifting $C$ units to the left, not $C$ units to the right. –  Michael Hardy Feb 2 '13 at 15:55
    
@MichaelHardy: Yes, the density of $X-C$ is the density of $X$, shifted to the left by $C$ units. My previous post incorrectly stated the density shifted to the right. However, the density of the $X-\lfloor X \rfloor$ should be $\frac{\alpha e^{-\alpha x}}{1-e^{-\alpha}}$, $0<x<1$ and $0$, otherwise. For the exponential random variable, the "quantization error" doesn't depend on which "bin" the exponential random variable fell into. Personally, I'm trying to see if this is the case for any random variable (say, Gaussian $X$). In the denom., I'm getting $\Phi(\cdot)$ and the num. $exp(\cdot)$.. –  jrand Feb 3 '13 at 3:43
    
For a Gaussian random variable $X$, the quantization error does depend on the bin number. One way to see this is the sign of the density $f_X(x)$ changes (once) depending on the value of $x$. By conditioning, we cannot change the sign of the slope, only on the magnitude of the slope. By shifting, we cannot change the sign of the sign of the slope, only the position of the density. So if there exist two disjoint density regions, one increasing and one decreasing, then binning will change the magnitude, and truncate the density. The error will shift both densities to zero, but not change slope. –  jrand Feb 3 '13 at 4:53
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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 $\lambda^{-1}$.

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