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

MATH103 April 2010

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[hide] 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. $A=2$ , $B=3$

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[show] 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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[show] Solution
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.