Science:Math Exam Resources/Courses/MATH103/April 2010/Question 01 (f)

MATH103 April 2010

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Question 01 (f)

Multiple Choice Question: Exactly ONE of the answers provided is correct. There is no partial credit in this questions. A continuous random variable $X$ has a probability density function of the form $f(x)=Ax+Bx^{2},\quad {\text{for}}\quad -1\leq x\leq 1,$ where $A$ and $B$ are constants. The values of $A$ and $B$ should be i. $A=1$, $B=1$ ii. $A=1$, $B=3/2$ iii. $A=0$, $B=3/2$ iv. $A=0$, $B=2/3$ v. $A=2$, $B=3$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that a probability density function f(x) has to satisfy two conditions: $\int _{-\infty }^{\infty }f(x)\,dx=1,$ $f(x)\geq 0$, for all 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. Using the hint, we first integrate f(x) and make sure it integrates to 1. Keep in mind that $f(x)=0$ if $x\not \in [-1,1]$: ${\begin{aligned}1&=\int _{-\infty }^{\infty }f(x)\,dx\\&=\int _{-1}^{1}f(x)\,dx\\&=\int _{-1}^{1}(Ax+Bx^{2})\,dx\\&=\left.A{\frac {x^{2}}{2}}+B{\frac {x^{3}}{3}}\right|_{-1}^{1}\\&=B{\frac {2}{3}}.\end{aligned}}$ Note that it is no surprise that A cancels here, because the term Ax is odd an thus cancels when integrating from -1 to 1. The calculation above tells us that B = 3/2. The constant A needs to ensure that $f(x)\geq 0$, for all x. For very small (in absolute value) negative numbers x, the linear term Ax dominates the quadratic term 3/2 x2, so we suspect that A = 0 is the only option to keep f(x) positive. But let's do the math: To show that f(x) is positive it suffices to show that f(x) is positive at the global minimum of $f(x)$. By the extreme value theorem, this global minimum will occur at either a critical point, or at the endpoints of the interval. We can find a critical point (convince yourself it is a local minimum) by setting f'(x) to zero: $\displaystyle f'(x)=A+2(3/2)x=A+3x=0.$ Hence x = -A/3 is the local minimum of f(x). At this point the value of f is $\displaystyle f(-A/3)=A(-A/3)+3/2(-A/3)^{2}=-A^{2}/3+A^{2}/6=-A^{2}/6\leq 0$ for any A. Since we need all points, including this critical point, to satisfy $f(x)\geq {}0$ then we demand A=0. With this choice of A, notice that $f(-1)=f(1)=3/2>0$ so we have that the local minimum is the global minimum and therefore that $f(x)\geq {0}$ for all x. Final answer: iii.

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