Science:Math Exam Resources/Courses/MATH105/April 2011/Question 03

MATH105 April 2011

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Question 03
Is there any value of $k$ for which the function $f$ below is a probability density function? $f(x)={\begin{cases}{\frac {2k}{(k+x)(k-x)}}&{\text{for }}0\leq x\leq {\frac {k}{2}}\\0&{\text{otherwise.}}\end{cases}}$ If yes, find all such values of $k.$ If there is no such $k$, explain why.

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
A probability density function integrates to $1$.

Hint 2
Use partial fractions.

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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. Notice from the definition of $f(x)$ that if $k<0$ then $f(x)$ is always 0, which cannot be a probability density function. The same is true if $k=0$. Therefore we only need to consider positive values of $k$ as candidates. We require that $\int _{0}^{k/2}f(x)dx=1$ Thus, $\int _{0}^{k/2}{\frac {2k}{(k+x)(k-x)}}dx=1.$ To deal with the integral, we use partial fractions. We write ${\frac {2k}{(k+x)(k-x)}}={\frac {A}{k+x}}+{\frac {B}{k-x}}$. Now we multiply the equation through by $(k+x)(k-x)$ to find $2k=A(k-x)+B(k+x)$. To find $A$ and $B$ we can choose ``convenient" values for $x$. Setting $x=k$ yields $2k=0+B(k+k)$ so $B=1$. Setting $x=-k$ yields $2k=A(k-(-k))+0$ so $A=1$. Therefore, ${\begin{aligned}\int _{0}^{k/2}f(x)dx&=\int _{0}^{k/2}\left({\frac {1}{k+x}}+{\frac {1}{k-x}}\right)dx\\&=(\ln |k+x|-\ln |k-x|)|_{0}^{k/2}\\&=(\ln(3k/2)-\ln(k/2))-(\ln k-\ln k)\\&=\ln(3)+\ln(k)-\ln(2)-\ln(k)+\ln(2)\\&=\ln(3)\end{aligned}}$ Note this is independent of $k$ and it will never be $1$. Thus no such $k$ exist!

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