Question

For the c.d.f. in Example 3.3.4, find the quantile function

F(x)=0 for x <0,

x2/3 for 0≤x≤1,

1 for x >1.

Short answer

The first quantile is 0.406

The second quantile is 0.631

The third quantile is 0.835

Step-by-step solution

Given the information

Given cdf given from continuous distribution. The cdf is given for different ranges.

State the quantile

Quantile function we defined as

F(x) = p

x = F^-1(p)

Compute the quantile function

From the given cdf for the range 0 ≤ x≤ 1, the quantile function defined as

x^2/3 = F(x) = p ….(1)

In the equation, both sides logarithm and calculate to get

\begin{aligned}\log p=\frac{2}{3}\log x\\\log x=\frac{3}{2}\log p\\x={e^{\frac{3}{2}\log p}}\end{aligned}……(2)

Therefore, we calculate the first quantile, second quantile, and third quantile.

Then for the first quantile p = 1/4 =25% in theequation (2)to get

\begin{aligned}x={e^{\frac{3}{2}\log \left( {0.25} \right)}}\\={e^{ - 0.90}}\\=0.406\end{aligned}

For the second quantile p=1/2=50% in theequation, (2)to get

\begin{aligned}x={e^{\frac{3}{2}\log \left({0.5}\right)}}\\={e^{-0.45}}\\=0.631\end{aligned}

For the third quantile p = 3/4=75% in theequation,to get

\begin{aligned}x={e^{\frac{3}{2}\log \left({0.75}\right)}}\\={e^{-0.18}}\\=0.835\end{aligned}