Expected Value of a Sum

Question

Let $f(x) = |x-1|, 0 \leq x \leq 2$ be the probability density function of a random variable $X$. Find $E(X^2+X)$.

The book gives a different answer from what I got, and I have no idea why.

$E(X) = \int_{0}^2x|x-1|dx = \int_0^1 x(1-x)dx + \int_1^2 x(x-1)dx = 1$, and similarly $E(X^2)= \int_0^2x^2|x-1|dx = \int_0^1x^2(1-x)dx + \int_1^2x^2(x-1)dx = 3/2$.

So, $E(X^2+X) = E(X^2) + E(X) = 3/2 + 1 = 5/2$ is what I got, but the answer in the book is given as $13/6$. What am I doing wrong here?

The calculations check out well. Probably the book is wrong. — Sudarsan, Dec 09 '13 at 22:25
http://math.stackexchange.com/questions/83275/where-are-my-calculations-wrong-expected-value?rq=1 — Bruce B Campbell, Dec 09 '13 at 22:26

score 1 · Answer 1 · answered Dec 09 '13 at 22:30

1

$E[X+X^2] = E[(X+{1\over 2})^2-{1\over 4}] = E[(X+{1\over 2})^2] - E[{1\over 4}]$.

$E[(X+{1\over 2})^2] = \int_0^2 (x+{1\over 2})^2 |x-1| dx = {11\over 4}$, $E[{1\over 4}] = {1\over 4}$.

Hence $E[X+X^2] = {5 \over 2}$.

answered Dec 09 '13 at 22:30

copper.hat

172,524

score 1 · Answer 2 · answered Dec 09 '13 at 22:33

Firstly, $E(X) = (1/2-1/3)+(8/3-1/3)+(1/2-2) = 1/6 + 14/6 - 9/6 = 1$. So this is right.

Now, $E(X^2) = (1/3-1/4)+(4-1/4)+(1/3-8/3) \\ = 1/12 + 45/12 -28/12 = 18/12 = 3/2.$

So this is also right...

We should check that the probability density function is normalised.

So, $\int_{0}^{2} |x-1| dx = 2\cdot \int_{0}^{1} (1-x)\; dx = 2\cdot(1(1-0) - 1/2(1-0))= 1$.

So you should get $\frac{5}{2}$.

Expected Value of a Sum

2 Answers2