Calculate the 30th and 60th percentiles of X.

Question

The following function is the CDF of a mixed distribution $$F(x)=\begin{cases} 0,& \text{when }x<0\\ \frac{x+1}{8},&\text{when } 0≤x<1\\ \frac{x+3}{8},&\text{when } 1≤x<3\\ 1,&\text{when } x≥3. \end{cases}$$ Calculate $X_{0.3}$ and $X_{0.6}$, the 30th and 60th percentiles of $X$, if $X$ is a r.v. with $F(x)$ given above.

So far, I only calculated the 60th percentile as follows: \begin{align} X_{0.6} &= F^{-1}(3/5)\\ \frac{x + 3}{8} &= \frac35\\ x+3 &= \frac{24}5,\end{align} so $x = 9/5$.

I know I have to solve $X_{0.30} = F^{-1}(3/10)$, but I do not know which piece of the definition of $F$ to use because no piece produces values that would include $0.3$ in their range.

Can anyone help me calculate this percentile? Thank you

Answer 1

It spends $25\%$ of the time below $1$; and $50\%$ of the time above $1$, so it spends $25\%$ of the time with the exact value $1$.

Just as the median of a list of numbers is the middle one, although several other numbers may have that value, the $30^{th}$ percentile is exactly $1$, as $25\%$ of the numbers are below it and at least $5\%$ of the numbers are equal to it.

Answer 2

Percentiles are really nice when $F$ is invertible and the distribution is continuous. In this example, the distribution is not continuous; it is mixed. That means you have to check point masses. \begin{align*} F(1-)&=\lim_{x\to 1^-}F(x)=0.25\\ F(1)&=0.5 \end{align*} This tells us that $\mathbb{P}(X=1)=0.25$. That means that the 30$^{\text{th}}$ percentile is at $x=1$, or \begin{equation*} \pi_{0.3}=1 \end{equation*} Note that you could have other percentile values also being $1$, say $\pi_{0.35}=1$ too!