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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

Expert verified

The first quantile is 0.406

The second quantile is 0.631

The third quantile is 0.835

Step by step solution

01

Given the information

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

02

State the quantile

Quantile function we defined as

F(x) = p

x = F-1(p)

03

Compute the quantile function

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

x2/3 = 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}

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