1. ## exponential distribution question

Let $\displaystyle Y$ be an exponentially distributed random variable with mean $\displaystyle \beta$. Define a random variable $\displaystyle X$ in the following way: $\displaystyle X=k$ if $\displaystyle k-1 \leq Y \leq k \ \ \mbox{for} \ \ k =1,2,...,n$

a) Find $\displaystyle P(X=k) \ \ \forall \ \ k=1,2,...,n$

b) Show that your answer to part a) can be written as:

$\displaystyle P(X=k) =(e^{-\frac{1}{\beta}})^{k-1}(1-e^{-\frac{1}{\beta}}) \ \ \forall \ \ k =1,2,...,n$

I would think for a) it would be $\displaystyle \int_{k-1}^{k} \frac{1}{\beta} e^{-\frac{x}{\beta}} dy = -e^{-\frac{x}{\beta}} \bigg{|}^{k}_{k-1} = -e^{-\frac{k}{\beta}}+e^{-\frac{k-1}{\beta}}$, but this doesn't seem right since k is not continuous.

and for b) I would think that you have to manipulate the function you got in a) to get what is shown, but am clueless on how to do so. Any help would be greatly appreciated.

Besides, this will be confirmed by b). You need to expand $\displaystyle (e^{-\frac{1}{\beta}})^{k-1}(1-e^{-\frac{1}{\beta}})$ (into a sum of two terms) and use the properties $\displaystyle e^ae^b=e^{a+b}$ and $\displaystyle \left(e^a\right)^b=e^{ab}$, in order to write it like your result in a).
(And if you can do that, then you can use the same computations in the reverse order to go from the result a) to the expected formula, if you prefer)

3. Originally Posted by lllll
Let $\displaystyle Y$ be an exponentially distributed random variable with mean $\displaystyle \beta$. Define a random variable $\displaystyle X$ in the following way: $\displaystyle X=k$ if $\displaystyle k-1 \leq Y \leq k \ \ \mbox{for} \ \ k =1,2,...,n$

a) Find $\displaystyle P(X=k) \ \ \forall \ \ k=1,2,...,n$

b) Show that your answer to part a) can be written as:

$\displaystyle P(X=k) =(e^{-\frac{1}{\beta}})^{k-1}(1-e^{-\frac{1}{\beta}}) \ \ \forall \ \ k =1,2,...,n$

I would think for a) it would be $\displaystyle \int_{k-1}^{k} \frac{1}{\beta} e^{-\frac{x}{\beta}} dy = -e^{-\frac{x}{\beta}} \bigg{|}^{k}_{k-1} = -e^{-\frac{k}{\beta}}+e^{-\frac{k-1}{\beta}}$, but this doesn't seem right since k is not continuous.

and for b) I would think that you have to manipulate the function you got in a) to get what is shown, but am clueless on how to do so. Any help would be greatly appreciated.
Your answer to a) is correct. Why does the fact that k is discrete bother you?

b) Note that $\displaystyle e^{-(k-1)/\beta} - e^{-k/\beta} = e^{-(k-1)/\beta} - e^{-k/\beta} \, e^{1/\beta} \, e^{-1/\beta}$

$\displaystyle = e^{-(k-1)/\beta} - e^{-(k-1)/\beta} \, e^{-1/\beta} = e^{-(k-1)/\beta} \left( 1 - e^{-1/\beta}\right)$.