Science:Math Exam Resources/Courses/MATH103/April 2013/Question 03 (a)

MATH103 April 2013

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[hide] Question 03 (a)
Continuous Probability. Let $\displaystyle X\geq 0$ be a random variable on the interval $\displaystyle [0,a]$ with probability density function $\displaystyle p(x)$ . The mean of X is given by $\displaystyle {\overline {x}}=\int _{0}^{a}xp(x)\,dx$ . Show that $\displaystyle {\overline {x}}$ can also be computed as $\displaystyle {\overline {x}}=\int _{0}^{a}(1-F(x))\,dx$ , where $\displaystyle F(x)$ is the cumulative distribution function (cdf) of X.

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.

[show] Hint 1
Approach this question as a left hand side, right hand side proof problem. We start with the definition of mean and want to get to $\displaystyle {\overline {x}}=\int _{0}^{a}(1-F(x))\,dx$ . Starting from $\displaystyle {\overline {x}}=\int _{0}^{a}xp(x)\,dx$ , what integration techniques can we use to get a start on it?

[show] Hint 2
Try integration by parts on the mean.

[show] Hint 3
What does $F(a)$ equal?

[show] Hint 4
If you get stuck, try simplifying the other side of the equality to help.

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.

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[show] Solution
Notice that $\displaystyle F(x)$ is an antiderivative of $\displaystyle p(x)$ . If we use integration by parts, we let $\displaystyle u=x$ and $\displaystyle dv=p(x)dx$ so that $\displaystyle du=dx$ and $\displaystyle v=F(x)$ so that ${\begin{aligned}{\overline {x}}&=\int _{0}^{a}xp(x)\,dx\\&=xF(x){\big \|}_{0}^{a}-\int _{0}^{a}F(x)\,dx\\&=aF(a)-(0)F(0)-\int _{0}^{a}F(x)\,dx\\\end{aligned}}.$ Now, since the probability distribution is only defined on [0,a] then the cumulative probability satisfies F(0)=0 since nothing can happen before 0 and F(a)=1 since by x=a, there is a 100% chance that our event happened. Therefore, ${\begin{aligned}{\overline {x}}&=a-\int _{0}^{a}F(x)\,dx\\&=\int _{0}^{a}1-\int _{0}^{a}F(x)\,dx\\&=\int _{0}^{a}(1-F(x))\,dx\\\end{aligned}}$ If we missed the second last step, we could have instead simplified the other equation to see that $\displaystyle {\overline {x}}=\int _{0}^{a}(1-F(x))\,dx=a-\int _{0}^{a}F(x)$ and thus the two sides are equal. Either way this completes the proof.