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Random variable $X$ has exponential distribution with parameter $\lambda>0$. Let $T=\lfloor X\rfloor $, $R=\{X\}$ where $\lfloor x\rfloor$ is floor from number $x\in\mathbb{R}$ and $\{x\}$ is it's fraction part. What is the distribution and expected value of $T$ and $R$?

I tried doing that the following way:

$$\mathbb{P}(T=k)=\mathbb{P}(X\in[k,k+1))=\int_k^{k+1}\lambda e^{-\lambda x}=-e^{-\lambda x}\Big|_k^{k+1}=\frac{e^{\lambda}-1}{e^{\lambda +k}}$$ so $$\mathbb{E}T=\sum_{k=0}^{\infty}\frac{k(e^{\lambda}-1)}{e^{\lambda +k}}=\frac{e^{\lambda}-1}{e^{\lambda}}\sum_{k=0}^{\infty}\frac{k}{e^{\lambda k}}=\frac{e^{\lambda}-1}{e^{\lambda}}\frac{e^{\lambda}}{(e^{\lambda}-1)^2}$$

I took the last equality from Wolfram, yet I'm unsure of it's origins. Where does it come from? Secondly, I'm clueless about $R$ distribution. Could you help?

Jenaro Fernández
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Jules
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    The most interesting part of the exercise is what comes next: Show that $[X]$ and ${X}$ are in fact independent. And for bonus points: Are there other (say, with a density) distributions on $(0,+\infty)$ which satisfy this independence property? – Did Sep 02 '14 at 13:15
  • https://math.stackexchange.com/q/805796/321264 – StubbornAtom May 27 '20 at 07:23

2 Answers2

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Since the pdf of $X$ is $\lambda e^{-\lambda x}\mathbb{1}_{x\geq 0}$ we have: $$\forall m\in\mathbb{N},\quad\mathbb{P}[\lfloor X\rfloor\leq m]=\int_{0}^{m+1}\lambda e^{-\lambda x}\,dx = 1-e^{-\lambda(m+1)}$$ hence the pdf of $\lfloor X \rfloor$ is $(e^{\lambda}-1)e^{-\lambda(m+1)}\mathbb{1}_{m\in\mathbb{N}}$ and: $$\mathbb{E}[\lfloor X\rfloor]=(e^{\lambda}-1)\sum_{m=0}^{+\infty} m e^{-\lambda(m+1)}=\frac{1}{e^\lambda-1}$$ since for any $-1<y<1$: $$\sum_{m=0}^{+\infty} m y^m = y\cdot\frac{d}{dy}\sum_{m=1}^{+\infty}y^m=\frac{y}{(1-y)^2}.$$ For the second part, we have: $$\forall t\in(0,1),\qquad \mathbb{P}[\{X\}\leq t]=\sum_{n\in\mathbb{N}}\int_{n}^{n+t}\lambda e^{-\lambda x}dx = (1-e^{-\lambda t})\sum_{n\in\mathbb{N}}e^{-\lambda n}=\frac{1-e^{-\lambda t}}{1-e^{-\lambda}}$$ hence the pdf of $\{X\}$ is $\frac{\lambda}{1-e^{-\lambda}}e^{-\lambda x}\mathbb{1}_{x\in(0,1)}$. By the linearity of $\mathbb{E}$, $$\mathbb{E}[\{X\}]=\mathbb{E}[X]-\mathbb{E}[\lfloor X\rfloor] = \frac{1}{\lambda}-\frac{1}{e^{\lambda}-1}.$$

Jack D'Aurizio
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  • $\lfloor{X\rfloor}$ is discrete, it can't have a pdf. – StubbornAtom Feb 24 '18 at 14:35
  • @StubbornAtom: a pDf can be a distribution, like $\delta(x)$ for the random variable which is always zero. – Jack D'Aurizio Feb 24 '18 at 14:41
  • If by pdf you mean density function, then I am supposed to differentiate the CDF to get the pdf, which is not what you do. Your answer is correct, but as $\lfloor{X\rfloor}$ is Geometric, its support in this case should be $\mathbb{N}\cup{0}$. – StubbornAtom Feb 24 '18 at 14:54
  • @StubbornAtom: according to the notation I am used to, $0$ is already an element of $\mathbb{N}$. – Jack D'Aurizio Feb 24 '18 at 14:56
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To calculate the expected value, you can also notice that the floor function follows a geometric distribution (number of failures before the first success) with parameter p=1-exp{-lambda}.

Adrien Peltzer
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