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Let $X$ be a random variable with probability density function \begin{equation} f(x)=\frac{1}{2} e^{-|x|},\;\; -\infty<x<\infty. \end{equation} What is expected value of $\lfloor X \rfloor$, i.e. \begin{equation} E( \lfloor X \rfloor), \end{equation} where $\lfloor X \rfloor$ denotes the greatest integer of $X$.

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  • $\begingroup$ @Chinny84 greatest integer command is not working properly here. So, I just denote by some other notation but I have defined this notation. $\endgroup$ Jan 27, 2017 at 17:37
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    $\begingroup$ how far have you got? could you work out the expected value without the greatest integer part, using integration? $\endgroup$
    – Cato
    Jan 27, 2017 at 17:44
  • $\begingroup$ find prob of greatest integer being n, then do infinite summation of nP(N = n) to get average - the summation is different for n<0 and n>=0 $\endgroup$
    – Cato
    Jan 27, 2017 at 17:56
  • $\begingroup$ @Cato could you please elaborate it? $\endgroup$ Jan 27, 2017 at 18:02

2 Answers 2

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Let's define, for ease of notation, $Z=\lfloor X \rfloor$.

To get the expectation, let's first find the distribution law of $Z$.

Let $n\in\mathbb{N}_0$. Then $P(Z=n)=P(n\leq X < n+1)=\int_{n}^{n+1}\frac{1}{2}e^{-x}dx=\frac{e-1}{2e^{n+1}}$.

Because of the symmetry of the function $\frac{1}{2}e^{-|x|}$ around $0$, we have that $P(Z=-n-1)=P(-n-1\leq X<-n)=P(n\leq X < n+1)=P(Z=n)$.

Now we can simply calculate the expectation: \begin{align} E(Z) &= \sum\limits_{k=-\infty}^{+\infty}kP(Z=k)\\ &=\sum\limits_{n=0}^{+\infty}nP(Z=n)+\sum\limits_{k=-\infty}^{-1}kP(Z=k)\\ &=\sum\limits_{n=0}^{+\infty}nP(Z=n)+\sum\limits_{n=0}^{+\infty}(-n-1)P(Z=-n-1)\\ &= \sum\limits_{n=0}^{+\infty}nP(Z=n)+\sum\limits_{n=0}^{+\infty}(-n-1)P(Z=n)\\ &= -\sum\limits_{n=0}^{+\infty}\frac{e-1}{2e^{n+1}} = \frac{1-e}{2e}\cdot\frac{1}{1-\frac{1}{e}}\\ &=-\frac{1}{2} \end{align}

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In fact the answer is $- 1/2$ for any density function which is symmetric about zero, and has finite expectation. The point is that $\lfloor x \rfloor + \lfloor -x \rfloor = -1$ for every non-integer $x$. So $E \lfloor X \rfloor + E \lfloor -X \rfloor = -1$ given finite expectation, and then (given symmetry) $E \lfloor -X \rfloor = E \lfloor X \rfloor $.

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