<s>[INST] A Problem related to the Cauchy Distribution The following problem is from the book, "Introduction to Probability" by
Hoel, Port and Stone. My answer does not match the back of the book. What did I do wrong?
Thanks,
Bob
Problem:
Let $X$ have a Cauchy density. Find the upper quartile for $X$.
Answer:
Recall the density function for the Cauchy distribution is:
\begin{align*}
f(x) &= \frac{1}{x^2+1} \\
\end{align*}
Let $a$ be the number we seek.
\begin{align*}
F(x) &= \tan^{-1}{x} \\
\int_{-\infty}^{a}  \frac{1}{x^2+1} \,\, dx &= \frac{3}{4} \\
\tan^{-1} \Big|_{-\infty}^{a} &= \frac{3}{4} \\
\tan^{-1}(a) - \frac{-\pi}{2} &= \frac{3}{4} \\
\tan^{-1}(a) + \frac{\pi}{2} &= \frac{3}{4} \\
\tan^{-1}(a) &= \frac{3}{4} - \frac{\pi}{2} \\
\end{align*}
The books's answer is $1$. What did I do wrong?
I ran the following command in R: pcauchy(1) . R returned 0.75 therefore,
I believe the book is right.
 [/INST] You forgot the factor $\frac{1}{\pi}$ in the probability density function. Correct density function is
$$
f(x) =\frac{1}{\pi(1+x^2)}
$$ so that $\int_{-\infty}^\infty f(x) dx =1$. Your problem is to find $a$ such that
$$
\frac{1}{\pi}\int_{-\infty}^a \frac{dx}{1+x^2}=\frac{\arctan a+\frac{\pi}{2}}{\pi}=\frac{3}{4}
$$ or equivalently
$$
\arctan a=\frac{\pi}{4}.
$$ This gives $a=\tan \frac{\pi}{4}=1$.
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