# Poisson probability and expectation

I have a Poisson process with parameter $\lambda=1$ and arrival times $X_1<X_2<X_3<\cdots$

I round down the arrival times to the nearest hundredth and call it a new process such that $R_1<R_2<R_3<\cdots$ (e.g. if $X_i=4.239,$ then $R_i=4.23$).

Let $T$ denote the steps until the first integer appears in $R_i$.

So for example, if $X_1 = 1.021<X_2=2.438<X_3=4.007$, then $R_1 = 1.02<R_2=2.43<R_3=4.00$ and $T = 3$ since the first integer appears at the third step "$R_3$" and $R_T=4.00$.

How do I calculate $P(R_T=5)$ and also $E[X_T]?$

The way I am thinking about solving this is by using geometric distribution with probability of success = 1/100 (because we will have an integer only if the first two digits after the decimal point are zeros).

But I am not sure if this is the right approach to solve it, and I also wonder how to make use of the fact that we have a Poisson process with parameter $\lambda=1$?

Any help would be appreciated. Thanks.

To go more exact, as the inter-arrival times of a Poisson process would be exponentially distributed with the same parameter; $X_n$ is the sum of $n$ independent exponentials, each with parameter $1$. This means $X_i$ are Gamma distributed with parameters $(2i, i)$. You need to find the probability that $X_i$ is within $[k-0.005, k + 0.004]$ for some integer $k$, so you need its cumulative probability distribution, which I think can be expressed using the incomplete Gamma function.
However, given the small size of the interval, perhaps we can approximate by the pdf at $k$ multiplied by $0.009$. With this approximation, we have for natural numbers $n, k$:
$$P[R_n = k] \approx 0.009\dfrac{n^{-2 n} e^{\tfrac{-k}{n}} k^{2 n-1}}{(2n-1)!}$$
You can tabulate this for a few initial values of $n, k$ to see if it matches with a simpler approximation such as geometric. Please double check all of the above before using though...