# Solved – Finding a probability mass function for $[x]$ is defined as the largest integer $n$ such that $n \leq x$

exponential distributionprobability

If $X$ is an exponential random variable with parameter $\lambda$ and we have that $[x]$ is defined as the largest integer $n$ such that $n \leq x$.

If $Y$ is defined as $$Y = \left[ X\over a \right]$$ where $a>0$ then how would you find the probability mass function of $Y$?

$n \leq x$ won't suffice as it does not translate that n is the larger integer that fulfill the property.

you have to work with $n \leq x < n+1$.

$$P(Y=n) = P(n \leq \frac{x}{a} \leq n+1)$$ $$P(Y=n) = P(na \leq x \leq (n+1)a)$$ $$P(Y=n) = F_{x}((n+1)a) - F_{x}(na)$$

With the expression of the cumulative distribution function $F_{x} = 1 - e^{-\lambda x}$

$$P(Y=n) = 1 - e^{-\lambda (n+1)a} - 1 + e^{-\lambda na}$$

$$P(Y=n) = - e^{-\lambda (n+1)a} + e^{-\lambda na}$$

$$P(Y=n) = 2e^{-\lambda (n+\frac{1}{2})a}sh(\frac{\lambda a}{2})$$

You can observe the effects of n with this second expression.