# How random variable $Y$ following geometric distribution in the following situation

I have a random variable, $X$ which follows exponential distribution with parameter $\lambda$. Then, define $Y=k$ for some $a$ greater than zero such that: $ka \leq X \leq (k+1)a$. Now, my question is how can $Y$ follow geometric distribution. I am not able to understand why but book give me the answer that this $Y$ random variable will follow geometric distribution taking values $y=0,1,..$. Any help would be appreciated.

• Pick a positive number $a$. Go on, it is not hard, and there are no rules as to which number you pick, as long as it is positive. Can't think of one? Use the last four digits of your mobile phone number with a decimal point between the first two and the last two digits.
• Pick a nonnegative integer $k$. Again, just pick one number, it doesn't matter which one as long as it is nonnegative and an integer.
• Now you have $k$ and $a$ and so can compute $ka$ and $(k+1)a$. So, since $X$ is an exponential random variable with parameter $\lambda$, can you calculate the probability that $X$ takes on values between $ka$ and $(k+1)a$? Subhint: unlike the previous stuff, you will not be getting a numerical answer but rather a function of $\lambda$ which I hope will work out to be something that you can express in the form $p(1-p)^{k}$ for suitable choice of $p$; that is, you need to choose a function of $\lambda$, call it $g(\lambda)$ and say that if you set $p=g(\lambda)$, then you can write $$P\{ka \leq X < (k+1)a\} = p(1-p)^k.$$ This is the part that requires some thinking as to how to define $p$.
• Your answer above is the probability that $Y$ takes on value $k$ where $k$ is a nonnegative integer. So the general formula is $$P\{Y = k\} = p(1-p)^k, k = 0, 1, 2, \ldots$$