Science:Math Exam Resources/Courses/MATH105/April 2018/Question 05 (a)

MATH105 April 2018

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Question 05 (a)

According to sociologists, the net income of the people can be modelled with a continuous random variable $X$, which has the following CDF: $$F(x)={\begin{cases}1-\left({\frac {A}{x}}\right)^{b},&{\text{if }}x\geq A,\\0,&{\text{if }}x<A,\end{cases}}$$ where $A$ and $b$ are positive constants and $b\neq 1$ ($A$ is the minimum income and $b$ is the tail index). (a) Find the PDF of the random variable $X$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall the PDF $f$ is the derivative of CDF $F$. i.e., $f=F'$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. By the Hint, it is enough to find the derivative of $F$. Since the chain rule gives $$\left(1-\left({\frac {A}{x}}\right)^{b}\right)'=-b\left({\frac {A}{x}}\right)^{b-1}\cdot \left({\frac {A}{x}}\right)'=-b\left({\frac {A}{x}}\right)^{b-1}\cdot \left(-{\frac {A}{x^{2}}}\right)={\frac {A^{b}\cdot b}{x^{b+1}}},$$ the PDF $f$ of the random variable $X$ is $$f(x)=F'(x)={\begin{cases}{\frac {A^{b}\cdot b}{x^{b+1}}},&{\text{if }}x>A,\\0,&{\text{if }}x<A.\end{cases}}$$