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

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 $\displaystyle u=-e^{-x}$

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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. 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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MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Probability density function, MER Tag Substitution, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag