Probabilities, and functions of random variables

Question

Let $X$ have an exponential distribution with parameter $\lambda > 0$, and let $Y := \left \lceil X \right \rceil$. Show that $\mathbb{P}(Y=k) = (1-p)^{k-1}p$ for each $k \in \mathbb{N}$, where $p=\mathbb{P}(X<1)$. Hint: show that $\mathbb{P}(Y>k)=(1-p)^{k}$.

If I am correct, the density function of X is $\lambda e^{-\lambda x}$, and

$\mathbb{P}(X<1)=1-\mathbb{P}(X\geq 1)=1-\int_{1}^{\infty} \lambda e^{-\lambda x} \;\mathrm{d}x = 1-e^{-\lambda}$.

So $p=1-e^{-\lambda}$.

This would mean the first step would be to show that

$\mathbb{P}(Y>k)=(1-p)^{k} = e^{-\lambda k}$

But whenever I try to do this, I run into the ceiling function of which an inverse cannot be found, and Y depends on X. Im just looking for a push in the right direction!

Answer 1

If $Y = \lceil X \rceil$, then $$\begin{align*} \Pr[Y = k] &= \Pr[k-1 < X \le k] \\ &= \Pr[X \le k] - \Pr[X \le k-1] \\ &= e^{-\lambda(k-1)} - e^{-\lambda k} \\ &= e^{-\lambda (k-1)}(1 - e^{-\lambda}) \\ &= (1-(1-e^{-\lambda}))^{k-1}(1-e^{-\lambda}) \\ &= (1-p)^{k-1} p.\end{align*}$$

Answer 2

For $x\geq0$ we have $P\left\{ X>x\right\} =e^{-\lambda x}$ and $X>k\iff Y>k$ for $k\in\mathbb{N}$.

So $P\left\{ Y>k\right\} =e^{-\lambda k}$.

Here $1-p=e^{-\lambda}$ so we could also write $P\left\{ Y>k\right\} =\left(1-p\right)^{k}$.

$P\left\{ Y=k\right\} =P\left\{ Y>k-1\right\} -P\left\{ Y>k\right\} =\cdots$

Answer 3

For positive integer $k$, $Y=k$ if and only if $k-1<X\le k$. You can find this probability by integrating your pdf.