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Just had a bash at this question for my Intro to Maths Stats module...I got to the end with a probability density function rather than a probability mass function, namely $f_Y(y) = \lambda a e^{-\lambda a y}.$ Obviously I'm missing some subtleties with the floor function that makes the new r.v. into a probability mass function instead. Anyways, here it is...

Suppose $X$ is an $exponential(\lambda)$ r.v. given by 0 for $x$ < 0 and $\lambda e^{-\lambda x}$ for $x \geq 0$. Recall the function $\lfloor x\rfloor$ is defined as the largest integer $n \leq x.$

Let $Y$ be defined by $ Y = \lfloor \frac{X}{a} \rfloor$, where $a$ > 0. Find the probability mass function of $Y$ and hence deduce that $Y$ is a geometric r.v., stating its parameter.

Thanks in advance (should be a quick one!) Sam

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Hint:* $Y$ is clearly discrete taking values in the set of non-negative integers, due to the the flooring. Then, for any integer $n\geq 0$ we have $$P(Y = n) = P\left(X\in [an, a(n+1))\right) $$ $$= \int_{an}^{a(n+1)} \lambda\mathrm e^{-\lambda x}dx= (1-p)^np$$ where $p = 1 - e^{-\lambda a} \in (0,1),$ as $\lambda>0$ and $a>0$.

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  • $\begingroup$ This extra clarification complements Batman's work. I now have a great model solution. Thanks to you both! $\endgroup$ – user142671 May 22 '14 at 21:24
  • $\begingroup$ Ok. Please accept one of our answers (left hand side upper corner) so that this question appears as answered in the site's statistics. Good luck with your studies! $\endgroup$ – ir7 May 22 '14 at 21:34
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$Y$ takes on value $k \in \{0,1,\ldots,\}$ if $k = \lfloor \frac{X}{a} \rfloor$. Thus, $P(Y=k) = P( \lfloor \frac{X}{a} \rfloor = k) = P(k \leq \frac{X}{a} < k+1) = P(ak \leq X < a(k+1)) = \int_{ak}^{a(k+1)} f(x) dx$ where $f(x)$ is the density of $X$.

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  • $\begingroup$ Oh dear, quite obvious really then. Must be too late in the day... $\endgroup$ – user142671 May 22 '14 at 21:22

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