Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.