I want to solve this problem of probability respect the Exponential Distribution:

Show that $x \longmapsto e^{-x}$ for $x \geq 0$ and $x \longmapsto 0$ for $x<0$ is a density. Calculate the probability that a corresponding random variable $X$ belongs to the interval $[2,3]$.

My attempt is the following: $$\int_{0}^{\infty} e^{-x} dx = e^{-x}|_{0}^{\infty}=1$$ Then it is a density. Now consider $$P(\{2\leq X(w) \leq 3\})= e^{-2}-e^{-3}$$ then the median is $log \ 2$

For example if we consider $x \longmapsto ae^{-ax}$ for $x \geq 0$ and $x \longmapsto 0$ for $x<0$ then $$P(\{2\leq X(w) \leq 3\})= e^{-2a}-e^{-3a}$$ and the median is $a^{-1} log \ 2$

Is it correct this answer?

Now, for $n=1,2,3,\ldots$. Let $Y_n$ denote $n^{-1}$ multiplied by the largest integer that is no larger than nX. How can I indentify the distribution function of the random variable $Y_n$?

Someone can help me to solve this, please. Thanks for your time and help everyone.


HINT $$ \Bbb P\left(Y_n =k\right)=P\left(\frac{ \lfloor nX \rfloor}{n}=k\right)=P\left(nk\leq nX <n(k+1)\right)=\int_{k}^{k+1}f_X(x)\,\mathrm dx $$

  • $\begingroup$ if I understand you I have first compute the $P[Y_n \leq y]$ but we have to note that $$ Y_n \leq y \Rightarrow \frac{ \lfloor nX \rfloor}{n} \leq y \Rightarrow \lfloor nX \rfloor \leq ny $$ then $$ nX \leq \lceil ny \rceil \Rightarrow X \leq \frac{\lceil ny \rceil}{n}$$. With this $$ P[X \leq \frac{\lceil ny \rceil}{n}] = \int_{0}^{\frac{\lceil ny \rceil}{n}} e^{-x} dx = ¿\sum_{x=0}^{\infty} e^{-x}?$$ Is this correct? $\endgroup$ – Knight Oct 4 '17 at 17:54
  • $\begingroup$ $Y_n$ is a discrete variable! Let $k$ an integer, you have to evaluate $\Bbb P\left(Y_n =k\right)$. Observe that $\lfloor Z\rfloor=k$ iff $k\le Z<k+1$ $\endgroup$ – alexjo Oct 4 '17 at 18:38
  • $\begingroup$ Yea I know that $Y_n$ takes values in $0,\frac{1}{n},\frac{2}{n},\ldots $ but then $$ \int_{k}^{k+1} f_{X} (x) dx =\int_{k}^{k+1} e^{-x} dx = e^{-(k+1)}-e^{-k}? $$ $\endgroup$ – Knight Oct 4 '17 at 18:52
  • $\begingroup$ Yes $$\Bbb P\left(Y_n =k\right)=\int_{k}^{k+1}f_X(x)\,\mathrm dx=F_X(k+1)-F_X(k)=\mathrm e^{-(k+1)}-\mathrm e^{-k}$$ $\endgroup$ – alexjo Oct 4 '17 at 19:04
  • $\begingroup$ Sorry but ask again but for example we saw that $X$ r.v. takes values in $n=0,1,\ldots$ and $Yn$ r.v. takes values in $n=0,\frac{1}{n},\frac{2}{n},\ldots$ cause it is discrete, but If you note it is following asimilar behavior to the geometric distribution, How can I prove that this $Yn$ has the same type as the geometric distribution ? $\endgroup$ – Knight Oct 6 '17 at 0:14

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