Science:Math Exam Resources/Courses/MATH105/April 2013/Question 01 (l)

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MATH105 April 2013

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Question 01 (l)
Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Let $k$ be a constant. Find the value of $k$ such that $\displaystyle f(x)=ke^{-x}e^{(-e^{-x})}$ is a probability density function on $(-\infty ,1]$.

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
The integral of a probability density function over the given interval must be 1.

Hint 2
Next, use a substitution to help evaluate the corresponding integral.

Hint 3
Use the substitution $\displaystyle u=-e^{-x}$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

We proceed in the hint and try to solve for k where $\displaystyle 1=\int _{-\infty }^{1}ke^{-x}e^{(-e^{-x})}$ First, we change the improper integral to $\displaystyle 1=\lim _{b\to -\infty }\int _{b}^{1}ke^{-x}e^{(-e^{-x})}$ Let $\displaystyle u=-e^{-x}$ so that $\displaystyle du=e^{-x}dx$ and the endpoints change to $\displaystyle u(1)=-e^{-1}\qquad u(b)=-e^{-b}$ This gives ${\begin{aligned}1&=\lim _{b\to -\infty }\int _{b}^{1}ke^{-x}e^{(-e^{-x})}\\&=\lim _{b\to -\infty }\int _{-e^{-b}}^{-e^{-1}}ke^{u}\,du\\&=\lim _{b\to -\infty }ke^{u}{\Bigr |}_{-e^{-b}}^{-e^{-1}}\\&=\lim _{b\to -\infty }k(e^{-e^{-1}}-e^{-e^{-b}})\end{aligned}}$ Since $\lim _{b\to -\infty }-b=+\infty$ we have that $\lim _{b\to -\infty }e^{-b}=+\infty$. Using this yields that $\lim _{b\to -\infty }-e^{-b}=-\infty$ and hence $\lim _{b\to -\infty }e^{-e^{-b}}=0$. Thus the second term in the long equation above vanishes and we finally conclude that $\displaystyle 1=ke^{-e^{-1}}$ Solving gives $\displaystyle e^{e^{-1}}=k$

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