概率密度函数下区间概率计算结果的验证请求

Hey Diego! Great job working through this probability problem—let's break down your solution step by step to confirm it's correct:

Your Problem Recap #

What is the probability that a randomly chosen $x$ in $[0,+\infty)$ is less than 1, if the probability density function on $[0,+\infty)$ is given by:
$ρ(x)=\frac{r}{1+x^2}$ for some $r$.

Step 1: Normalizing the PDF to find $r$ #

You correctly started by verifying the PDF's core requirements:

• First, you noted $\rho(x) \geq 0$ (since we'll end up with a positive $r$, this holds for all $x \in [0,+\infty)$)
• Then you computed the normalization integral:
  $$
  \int_0^{+\infty} \frac{r}{1+x^2}dx = r\int_0^{+\infty} \frac{1}{1+x^2}dx = r \cdot \frac{\pi}{2}
  $$
  Setting this equal to 1 gives $r = \frac{2}{\pi}$—this is 100% correct! The integral of $\frac{1}{1+x^2}$ over $[0,+\infty)$ is indeed $\frac{\pi}{2}$ (since $\arctan(x)$ approaches $\frac{\pi}{2}$ as $x \to +\infty$ and equals 0 at $x=0$).

Step 2: Calculating the Probability for $x < 1$ #

Next, you integrated the normalized PDF over $[0,1]$:
$$
\int_0^1 \frac{2}{\pi} \cdot \frac{1}{1+x^2}dx = \frac{2}{\pi} \left( \arctan(1) - \arctan(0) \right)
$$
Since $\arctan(1) = \frac{\pi}{4}$ and $\arctan(0) = 0$, this simplifies to:
$$
\frac{2}{\pi} \cdot \frac{\pi}{4} = \frac{1}{2}
$$
This result is also correct—every part of your calculation checks out perfectly.

So to sum up: every step of your solution is right, and the final probability of $\frac{1}{2}$ is accurate. Nice work!

备注：内容来源于stack exchange，提问作者Diego Márquez