Science:Math Exam Resources/Courses/MATH103/April 2013/Question 03 (b) i

MATH103 April 2013

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Question 03 (b) i
Continuous Probability. Consider $\displaystyle f(x)={\frac {x}{\sqrt {1-x}}}$. Convert $\displaystyle f(x)$ into a probability density function (pdf) $\displaystyle p(x)$ on an interval $\displaystyle (a,b)$ With $\displaystyle p(x)=cf(x)$, find the minimal a, maximal b, and the constant $\displaystyle c>0$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The function itself gives a natural restriction on how big b can be. What goes wrong with your function if b is too big?

Hint 2
The fact that a pdf is positive gives you a restriction on a.

Hint 3
All pdfs have the property that the area under the curve where the function is defined must be 1. Use this to find c.

Hint 4
To integrate the function, use a substitution.

Hint 5
The substitution you seek is $\displaystyle u=1-x$.

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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. Firstly, notice that in order for the denominator to be well defined, we require that $\displaystyle 1-x>0$ (strictly greater than 0 since the denominator of a fraction cannot be 0). This gives us that b has value 1 if it is to be maximal. Now, for a to be minimal, we notice that the denominator is always positive if a is less than 1 so that gives no restriction. However the numerator has to be positive in order to have a pdf and so we require that a is greater than 0. Lastly, the area condition on a pdf shows us that $\displaystyle 1=\int _{0}^{1}{\frac {cx}{\sqrt {1-x}}}=\lim _{q\to 1^{-}}\int _{0}^{q}{\frac {cx}{\sqrt {1-x}}}$ Let $\displaystyle u=1-x$ so that $\displaystyle du=-dx$, $\displaystyle u(0)=1$, $\displaystyle u(q)=1-q$ and substituting gives ${\begin{aligned}1&=\lim _{q\to 1^{-}}\int _{0}^{q}{\frac {cx}{\sqrt {1-x}}}\\&=\lim _{q\to 1^{-}}\int _{1}^{1-q}{\frac {-c(1-u)}{\sqrt {u}}}\\&=\lim _{q\to 1^{-}}c\int _{1-q}^{1}({\sqrt {u}}^{-1}-{\sqrt {u}})\,du\\&=\lim _{q\to 1^{-}}c(2{\sqrt {u}}-{\frac {2u^{3/2}}{3}}){\big |}_{1-q}^{1}\\&=\lim _{q\to 1^{-}}{\big (}c(2{\sqrt {1}}-{\frac {2(1)^{3/2}}{3}})-c(2{\sqrt {1-q}}-{\frac {2(1-q)^{3/2}}{3}}){\big )}\\&=c\left(2-{\frac {2}{3}}\right)\\&={\frac {4c}{3}}\end{aligned}}$ and isolating for c gives us that $c=3/4$.

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