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

Let $X$ be a random variable on $[-1,3]$ with density $f(x) = k x^2$ (with $k \in \mathbb{R} $ to be determined) on $[-1,3]$ apart from some points s.t. $p(X=-1) = p(X=3) = \dfrac{1}{4} $ and $p(X=0) = \dfrac{1}{3}$.
What is the cumulative distribution function of $X$? How much is $p(-1 \leq X < 3) $?

To find $k$ I would just calculate $$ k = \dfrac{1}{\int_{-1}^3 x^2 \ dx} $$ and then the cumulative distribution should be $$F(x) = \int_{-1}^x f(t) \ dt $$ while $$p(-1 \leq X < 3) = 1 - p(X = 3) = \dfrac{3}{4}$$ Any suggestion would be appreciated.

share|cite|improve this question
up vote 1 down vote accepted

You can't calculate $k$ the way you suggest because the total probability due to the continuous part of the PDF isn't 1. There are three possibilities, and $$\mathrm P(X=-1) + \mathrm P(X=3) + \mathrm P(-1<X<3) = 1$$

Therefore we have

$$\mathrm P(X=-1) = \mathrm P(X=3) = \frac 1 4 \implies \mathrm P(-1<X<3) = 1 - \frac 1 4 - \frac 1 4 = \frac 1 2$$

Thus $$k = \frac {1/2}{\int_{-1}^3 x^2}$$

Then e.g. the cumulative function is zero below -1, then jumps to $\frac 1 4$ as you cross $-1$ and smoothly increases to $\frac 3 4$ at $3$, then jumps to one.

Have another go at the question!

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.