Question

Problem 6, 15 Points. Let ~ ?(A), the exponential distribution. Determine and identify the probability distribution of the random variable [x], where fz] denotes the ceiling function, the smallest integer greater than or equal to z. Remark. Note that the new random variable is discrete random variable.

Answer 1

Since, X ~ Exponential( $\lambda$ ) => X > 0

Now, we observe that :

$\begin{align*} \text{When }0<X\leq 1 \ \ &\Leftrightarrow \left \lceil X \right \rceil = 1 \\ \text{When }1<X\leq 2 \ \ &\Leftrightarrow \left \lceil X \right \rceil = 2 \\ \text{When }2<X\leq 3 \ \ &\Leftrightarrow \left \lceil X \right \rceil = 3 \\ \text{When }3<X\leq 4 \ \ &\Leftrightarrow \left \lceil X \right \rceil = 4 \\ .\\ . \\ . \\ \end{align*}$

Thus, the support of $\left \lceil X \right \rceil$ is {1,2,3,...}

Now, we see that :

$\begin{align*} P(\left \lceil X \right \rceil = 1 ) &= P(0<X \leq1) =F_X(1) = 1 - e^{-\lambda} \\ P(\left \lceil X \right \rceil = 2 ) &= P(1<X \leq2 ) =P(X\leq 2) - P(X \leq 1)= F_X(2)-F_X(1)=(1 - e^{-2\lambda})-(1-e^{-\lambda}) = e^{-\lambda}-e^{-2\lambda} \\ P(\left \lceil X \right \rceil = 3 ) &= P(2<X \leq3 ) =P(X\leq 3) - P(X \leq 2)= F_X(3)-F_X(2)=(1 - e^{-3\lambda})-(1-e^{-2\lambda}) = e^{-2\lambda}-e^{-3\lambda}\\ ...\\ ...\\ ...\\ P(\left \lceil X \right \rceil = k ) &= P(k-1<X \leq k ) =P(X\leq k) - P(X \leq k-1)= F_X(k)-F_X(k-1)=(1 - e^{-k\lambda})-(1-e^{-(k-1)\lambda}) = e^{-(k-1)\lambda}-e^{-k\lambda} \\ ...\\ ... \end{align*}$

Thus, the probability mass function of $\left \lceil X \right \rceil$ is given by :

$P(\left \lceil X \right \rceil = k ) = e^{-(k-1)\lambda}-e^{-k\lambda} \ \ \ ; k =1,2,3,...$

Now, we are left with the task of identifying the distribution of $\left \lceil X \right \rceil$ . To do so, we observe that :

$\begin{align*} P(\left \lceil X \right \rceil = k ) &= e^{-(k-1)\lambda}-e^{-k\lambda} \\ &= e^{-(k-1)\lambda}-e^{-(k-1)\lambda-\lambda} \\ &= e^{-(k-1)\lambda}(1-e^{-\lambda}) \\ &= (e^{-\lambda})^{k-1}(1-e^{-\lambda}) \\ &= (1-p)^{k-1}*p \\ &\text{where } p = 1-e^{-\lambda} \end{align*}$

Now, from the last equation the distribution of $\left \lceil X \right \rceil$ can be identified as the Geometric distribtuion with paramater $p = 1-e^{-\lambda}$ .

Thus, $\left \lceil X \right \rceil \sim Geometric(p = 1-e^{-\lambda})$

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