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X has exponential distribution with probability density function $f(x)=\lambda e^{-\lambda x}\;$ for $x > 0$. We define $Y:=[X]$ (i.e Y is floor function of X). Compute $P(X=k)$ for $ k=0,1,2, ...$ Find distribution of Y ( including the parameters).

First step: We need to compute $P(X=k)$ for $ k=0,1,2, ...$

we know that distribution function of X is:

$F(x)=P[X \le x]=\int_{-\infty}^{x} 1-e^{-\lambda x} dx$

But then shouldn't always be the probability of $k=0,1,2 ...$ zero? Because we don't compute any area under the function.

But I think my approach is in some way wrong and I am not really sure, if I understand the example.

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    Try integrating over each unit. Use the definition of floor function. – 2017-01-19
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    Like this: $\int_{1}^{2}1-e^{-\lambda x} dx$ for x=1 ? – 2017-01-19
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    "Because we don't compute any area under the function": re-consider this assumption. $[x]=k\iff k\le x$x=k$.) – 2017-01-19
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    @Muffy Pretty much. Check Yves's comment. – 2017-01-19

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$P(X=k) = 0 $ for any $k \in \mathbb{R}$ -- $X$ is continuous valued, so it doesn't take any individual value with positive probability.

From the definition of $Y$, we have $P(Y=k) = P([X] = k) = P(k \leq X < k+1) = \int_k^{k+1} f_X(x) dx$ where $f_X(x) = \lambda e^{-\lambda x}$ for $x>0$ and $0$ otherwise is the density of $X$.

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    Okay. I understand it. But then I have also computed the $Y:=[X]$ because X is already floor function right? – 2017-01-19
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    The second paragraph computes the distribution of $Y$, which takes on values $0,1,2,\ldots$. – 2017-01-19
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    I see now. thank you. Your answer is really helpful. – 2017-01-19