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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.

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

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