Take the 2-minute tour ×
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.

I tried solving this, and I am pretty sure I am integrating this correctly, however, my solution manual shows -1 in the equation when doing this and I do not know why. The answer in the solution manual is correct.

Problem: Find the corresponding distribution function and use it to determine the probability that a random variable having the distribution function will take on a value between 0.4 and 1.6.

f(x) = x for 0 < x < 1
       2-x for 1 <= x < 2
       0 elsewhere

so for F(0.4 < x < 1.6) I did after integrating:
2(1.6) - [(1.6)^2 / 2] - [(0.4)^2 / 2] = 1.84
however the correct answer is 0.84. The solution manual has a -1 in their equation, but I do not know how they got it.

share|improve this question
Note that in this simple case, you can just calculate areas of rectangles and triangled and get the result. –  Raphael Nov 17 '10 at 15:35

2 Answers 2

up vote 3 down vote accepted

Since $f(t)$ is defined piecewise, you have to be careful with the integral. This is the source of your mistake.

If $0 < x < 1$, $$F(x) = \int_0^x f(t) dt = \int_0^x t dt = \left.\frac{1}{2}t^2\right|_0^x = \frac{1}{2}x^2,$$ which you have.

However, if $1 \leq x < 2$, then you have to break the integral that yields $F(x)$ up into pieces. This is $$F(x) = \int_0^x f(t) dt = \int_0^1 t dt + \int_1^x (2-t) dt = \left.\frac{1}{2}t^2\right|_0^1 + \left[2t - \frac{1}{2}t^2\right]_1^x $$ $$= \frac{1}{2} + 2x - \frac{1}{2}x^2 - 2 + \frac{1}{2} = 2x - \frac{1}{2}x^2 - 1.$$ Here's where the $-1$ comes in.

This is a common mistake. If it makes you feel any better, my probability students trip up over this all the time. :)

share|improve this answer
Wow, thank you very much for the well explained solution. –  Raptrex Nov 17 '10 at 6:43

First, probabilities are from $0$ to $1$, so you're certainly wrong.

Second, you calculated the cdf incorrectly for the interval $1 \leq x < 2$. It should be $$F(x) = \int_0^1 t \, \mathrm{d}t + \int_1^x 2-t \, \mathrm{d}t.$$ You forgot the first part, and integrated the second part from $0$ instead of from $1$. Since $$\int_0^1 t \, \mathrm{d}t = 1/2$$ whereas $$\int_0^1 2-t \, \mathrm{d}t = 3/2,$$ you counter $1/2$ too little and $3/2$ too much, for a total gain of $+1$.

share|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.