<s>[INST] Looking for a way to compute the discrete version of a continuous random distribution I just learn the way to compute the cumulative probability distribution (CDF) of a distribution in terms of some known results. For example, a random variable $Y$ is written in terms of a uniform random variable $U(0,1)$ as
$$
  Y \sim \frac{5}{U + 1}
$$
the CDF of $Y$ could be found in the following way
$$
  \begin{align*}
  \text{CDF}(y) & = F(y) = \mathbb{P}(Y\leq y) = \mathbb{P}\left(\frac{5}{U+1}\leq y\right)\\
  & = \mathbb{P}\left(U\geq \frac{5}{y}-1\right) = 1- \mathbb{P}(U<\frac{5}{y}-1)\\
  & = 2 - \frac{5}{y}
  \end{align*}
$$
so the probability distribution function (PDF) for $Y$ will be
$$
  P(y) = \dfrac{dF(y)}{dy} = \frac{5}{y^2}
$$
Compute the expectation value for $y$
$$
  \text{E}[y] = \int y\frac{5}{y^2}dy = 3.46574
$$
In the text, it is said that the uniform distribution is continuous so I apply the integral to compute the expectation. I read something online about discrete distribution. If I change the random variable as
$$
  Y \sim \left\lfloor \frac{5}{U+1} \right\rfloor
$$
how do we derive the PDF and CDF for the discrete version of $Y$? I am trying a simulation to run 1M random numbers for u as follows:
  s = 0
  for i=1 to 1000000
    s += floor(5/(rand(0, 1) + 1)
  end
  print s/1000000

this gives me an average of 2.9172 instead of 3.46574 I think the difference is originated from a great number of states that are suppressed by the floor operation. I wonder if there is any analytical way instead of a simulation to compute the result. I try to replace y in PDF with the floor(5/(u+1)) and replace the integral with a summation
$$
  \sum \left\lfloor\frac{5}{y+1}\right\rfloor \times \frac{5}{\left\lfloor\frac{5}{y+1}\right\rfloor^2} = \sum \frac{5}{\left\lfloor\frac{5}{y+1}\right\rfloor}
$$
but this gives me 1.8418, very far from 2.9172.
 [/INST] Since you calculated the original expected value correctly, you already know this, but just to be thorough for other readers, the CDF in the question is not properly specified, as $P(U < \frac{5}{y} - 1)$ is only $\frac{5}{y} - 1$ if this quantity is between 0 and 1. Otherwise, the probability is either 0 or 1. The appropriate range for $y$ where your logic holds is
$$ 0 \le \frac{5}{y} - 1 \le 1 \Longrightarrow y \le 5 \le 2y \Longrightarrow \frac{5}{2} \le y \le 5;$$
if $y < \frac{5}{2}$, we have $P(U < \frac{5}{y} - 1) = 1$ and if $y > 5$, we have $P(U < \frac{5}{y} - 1) = 0$.
Thus, we have
$$
\begin{align*}
CDF(y) &= P(Y \le y) \\
&= P(\frac{5}{U + 1} \le y) \\
&= P(U \ge \frac{5}{y} - 1) \\
&= 1 - P(U < \frac{5}{y} - 1) \\
&= \begin{cases} 0, & y < \frac{5}{2}, \\ 2 - \frac{5}{y}, & \frac{5}{2} \le y \le 5, \\ 1, & y > 5. \end{cases}
\end{align*}
$$
The PDF follows as
$$PDF(y) = \frac{d (CDF(y))}{dy} = \begin{cases} \frac{5}{y^2}, & \frac{5}{2} \le y \le 5, \\ 0, & o/w. \end{cases}$$
And the expected value follows as
$$E[Y] = \int_{-\infty}^{\infty} y \cdot PDF(y) dy = \int_{\frac{5}{2}}^{5} y \cdot \frac{5}{y^2} dy = 5\ln(5) - 5 \ln(5/2) \approx 3.466,$$
as you calculated.
For the discrete case, we can just do casework on where U falls:

*

*If $U = 0$, we have $Y = 5$.

*If $0 < U \le \frac{1}{4}$, we have $Y = 4$.

*If $\frac{1}{4} < U \le \frac{2}{3}$, we have $Y = 3$.

*If $\frac{2}{3} < U \le 1$, we have $Y = 2$.

The probability mass function for the discrete Y is then
$$P(Y = y) = \begin{cases} \frac{1}{3}, & y = 2, \\ \frac{5}{12}, & y = 3, \\ \frac{1}{4}, & y = 4. \end{cases} $$
This yields an expected value of
$$E[Y] = \sum y \cdot P(Y = y) = 2 \cdot \frac{1}{3} + 3 \cdot \frac{5}{12} + 4 \cdot \frac{1}{4} = \boxed{\frac{35}{12}},$$
which is around 2.917.
</s>