# Computing pmf and cdf for a function of an exponential random variable

I'm a little stuck on this one due to the nature of the function. Here is the question:

$$\mathit{T}$$ is a $$\lambda$$ = 1 exponential random variable and $$\mathit{f(x)= \lfloor x\rfloor}$$ (largest integer not more than $$\mathit{x}$$).

Find the cdf and pmf of $$\mathit{X = f(T)}$$. What is $$\mathbb{E}$$[$$\mathit{f(T)}$$]?

I don't know how to work with the "floor" portion of the question. I'm sure once I got past that I could do the rest of the question.

• I suppose I'm really stuck on how to find X=f(T), because of the "floor" function on x – user610107 Jan 31 at 1:55

Let $$Y = \lfloor X \rfloor$$, where $$X \sim \operatorname{Exponential}(\lambda)$$ is exponentially distributed with rate $$\lambda$$. Then clearly $$Y \in \{0, 1, 2, \ldots\}$$, since $$X \ge 0$$ means $$\lfloor X \rfloor$$ takes on nonnegative integer values.

Specifically, $$Y = 0$$ if $$0 \le X < 1$$, so $$\Pr[Y = 0] = \Pr[0 \le X < 1] = F_X(1) - F_X(0) = 1 - e^{-\lambda}.$$ Similarly, $$Y = 1$$ if $$1 \le X < 2$$, so $$\Pr[Y = 1] = \Pr[1 \le X < 2] = F_X(2) - F_X(1) = (1 - e^{-2\lambda}) - (1 - e^{-\lambda}) = e^{-\lambda} - e^{-2\lambda}.$$ And it is easy to see that in general, for a nonnegative integer $$y$$, $$\Pr[Y = y] = \Pr[y \le X < y+1] = F_X(y+1) - F_X(y) = e^{-\lambda y} - e^{-\lambda(y+1)}.$$

The only part remaining is to simplify this expression and determine the kind of distribution $$Y$$ actually is.