Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (m)

MATH105 April 2014

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Question 01 (m)
Let $k$ be a constant. Find the value of $k$ such that $f(x)=1+k|x|$ is a probability density function on $-1\leq x\leq 1$.

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
$\int _{-\infty }^{\infty }f(x)dx=1$ when $f(x)$ is probability density function. In this problem, we are restricted to the interval [-1,1] so we interpret this by taking $f(x)=0$ for all x outside of [-1,1].

Hint 2
By symmetry, $\displaystyle \int _{-1}^{1}f(x)dx=2\int _{0}^{1}f(x)dx$ when $f(x)$ is an even function.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We must have that $\int _{-1}^{1}f(x)dx=1$ because $f(x)$ is a probability density function (which is zero outside of [-1,1]). Hence, we have: ${\begin{aligned}1&=\int _{-1}^{1}(1+k|x|)dx\\&=2\int _{0}^{1}(1+k|x|)dx\\&=2\int _{0}^{1}(1+kx)dx\\&=\left.2\left(x+{\frac {kx^{2}}{2}}\right)\right|_{0}^{1}\\&=2\left(1+{\frac {k}{2}}\right)\\&=2+k.\\\end{aligned}}$ Note that in the second equality we used that $\displaystyle \int _{-1}^{1}f(x)dx=2\int _{0}^{1}f(x)dx$ since $f(x)$ is an even function; i. e. $f(x)=f(-x)$. In the third equality, we have used that for $x>0$, $|x|=x$. Finally we get $1=2+k$. Solve it to get $k=-1$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We must have that $\int _{-1}^{1}f(x)dx=1$ since $f(x)$ is a probability density function which is zero outside of [-1,1]. Hence, we have: ${\begin{aligned}1&=\int _{-1}^{1}(1+k|x|)dx\\&=\int _{-1}^{0}(1+k|x|)dx+\int _{0}^{1}(1+k|x|)dx\end{aligned}}$ Now, we use that that $|x|={\begin{cases}x,{\text{ if }}x\geq 0\\-x{\text{ if }}x<0\end{cases}}$ to write ${\begin{aligned}1&=\int _{-1}^{0}(1-kx)dx+\int _{0}^{1}(1+kx)dx\\&=\left.\left(x-{\frac {kx^{2}}{2}}\right)\right|_{-1}^{0}+\left.\left(x+{\frac {kx^{2}}{2}}\right)\right|_{0}^{1}\\&=-\left(-1-{\frac {k}{2}}\right)+\left(1+{\frac {k}{2}}\right)\\&=2+k.\\\end{aligned}}$ Solving $1=2+k$ gives us $k=-1$.

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