# Probability Density Function Constants

I am trying to solve a problem from a book, Probability and Random Processes by Yates. The problem number is 3.2.5.

We are asked to find conditions on $a$ and $b$ such that the probability density function $$f_X(x) = \begin{cases} ax^2 + bx & 0 \leq x \leq 1, \\ 0 & \text{otherwise} \end{cases}$$ is a valid one. There are two conditions that need to be satisfied.

1. $f_X(x) \geq 0$ for all $x$.
2. $\int_{-\infty}^{\infty}f_X(x)\, \mathrm{d}x = 1$

Then, applying the second condition first, I get: $$a = 3 - \frac{3}{2}b$$ Applying the first condition, I get: $$x(ax+b) \geq 0$$ If I divide both sides by $x$, since $x \geq 0$, I get: $$ax + b \geq 0$$ Replacing $a$ by the known expression, $$b(1-\frac32 x)+3x \geq 0$$ And I am stuck. Can I have a hint on how to proceed?

Thanks.

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The normalization condition gives $a/3 + b/2 = 1$. I think there is a mistake in your calculation. – Ron Gordon Feb 2 '13 at 2:35
Yes, I made an error in the calculation. I should have multiplied the entire quantity on the right hand side by $3$. – jrand Feb 2 '13 at 2:40

Mainly for typing convenience, instead of solving for $a$ in terms of $b$, we solve for $b$ in terms of $a$. We have $b=\frac{6-2a}{3}$.
As you did, we look at $ax+b$, or mre precisely at $3ax+3b$, to avoid fractions. We get $$3ax+3b=3ax +6-2a.$$
The function $3ax+b$ is monotone. So we will be OK if and only if it is non-negative at both ends.
That gives $6-2a\ge 0$ and $3a+6-2a\ge 0$. Thus the condition on $a$ is $-6\le a\le 3$.
Remark: If one solves for $a$ in terms of $b$, essentially the same endpoints argument works.