Science:Math Exam Resources/Courses/MATH101/April 2008/Question 01 (e)

MATH101 April 2008

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Question 01 (e)
Let $\displaystyle f(x)=kx^{2}(1-x)\quad {\text{if }}0\leq x\leq 1$ and $f(x)=0\quad {\text{if }}x<0{\text{ or }}x>1.$ For what value of the positive constant $k$ is $f(x)$ a probability density function?

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
For ƒ to be a probability density function it needs to satisfy two conditions: ƒ(x) ≥ 0 for all x, $\int _{-\infty }^{\infty }f(x)\,dx=1.$ Choose k such that both conditions are fulfilled.

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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. (a) Using the hint we see that, first of all, k needs to be a positive constant. We can calculate k explicitly by integrating: ${\begin{aligned}\int _{-\infty }^{\infty }f(x)\,dx&=\int _{-\infty }^{0}0\,dx+\int _{0}^{1}kx^{2}(1-x)\,dx+\int _{1}^{\infty }0\,dx\\&=k\int _{0}^{1}(x^{2}-x^{3})\,dx\\&=k\left.\left({\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}\right)\right|_{0}^{1}\\&=k\left({\frac {1}{3}}-{\frac {1}{4}}\right)\\&={\frac {k}{12}}.\end{aligned}}$ The only way that this integral has value 1 is when k = 12.

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