# uniform dist.

#### Jason

##### New member
I'm trying to review basic probability; haven't looked at it in a couple of years. Am I on the right track here?

A and B are independent random variables, uniform distribution on $[0,1]$. What is $E(min(A,B))$ ?

$\displaystyle\int_{0}^{1}\int_{0}^{a}b\,db\,da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\,db\,da$

$=\displaystyle\int_{0}^{1}\frac{a^2}{2}\,da+\int_{0}^{1}a-a^2\,da$

$=1/6+3/6-2/6=1/3$

#### CaptainBlack

##### Well-known member
I'm trying to review basic probability; haven't looked at it in a couple of years. Am I on the right track here?

A and B are independent random variables, uniform distribution on $[0,1]$. What is $E(min(A,B))$ ?

$\displaystyle\int_{0}^{1}\int_{0}^{a}b\;db\;da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\;db\;da$

$=\displaystyle\int_{0}^{1}\frac{a^2}{2}\;da+\int_{0}^{1}a-a^2\;da$

$=1/6+3/6-2/6=1/3$
That is correct, though you could use a line of explanation at the begining:

$$E(\min(A,B))=E(A|A<B)+E(B|A\ge B)$$

CB

#### Jason

##### New member
Thanks Captain. I also need to find $E((A+B)^2)$ and $E(|A-B|)$.

For the first one:
$E(A^2)+2E(A)E(B)+E(B^2)=1/3+2(1/4)+1/3)=7/6$

Could you give me a hint for the second one?

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#### Random Variable

##### Well-known member
MHB Math Helper
$\displaystyle \int_{0}^{1} \int_{0}^{1} |a-b| \ da \ db = \int_{0}^{1} \int_{b}^{1} (a-b) \ da \ db + \int_{0}^{1} \int_{0}^{b} -(a-b) \ da \ db =\frac{1}{3}$

So $\frac{1}{3}$ is the expected distance between the two points. This sort of makes since since the expected value of min{A,B} is $\frac{1}{3}$, and the expected value of max{A,B} is $\frac{2}{3}$.

#### Jason

##### New member
So I could just calculate $E(max(A,B))-E(min(A,B))$ ?

#### Random Variable

##### Well-known member
MHB Math Helper
So I could just calculate $E(max(A,B))-E(min(A,B))$ ?

$\displaystyle \int_{0}^{1} \int_{0}^{1} |a-b| \ da \ db = \int_{0}^{1} \int_{b}^{1} (a-b) \ da \ db + \int_{0}^{1} \int_{0}^{b} -(a-b) \ da \ db$

$\displaystyle = \Bigg( \int_{0}^{1} \int_{b}^{1} a \ da \ db + \int_{0}^{1} \int_{0}^{b} b \ da \ db \Bigg) - \Bigg( \int_{0}^{1} \int_{b}^{1} b \ da \ db + \int_{0}^{1} \int_{0}^{b} a \ da \ db \Bigg)$

$\displaystyle = E(\max\{A,B\}) - E(\min\{A,B\})$

#### Mr Fantastic

##### Member
I'm trying to review basic probability; haven't looked at it in a couple of years. Am I on the right track here?

A and B are independent random variables, uniform distribution on $[0,1]$. What is $E(min(A,B))$ ?

$\displaystyle\int_{0}^{1}\int_{0}^{a}b\,db\,da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\,db\,da$

$=\displaystyle\int_{0}^{1}\frac{a^2}{2}\,da+\int_{0}^{1}a-a^2\,da$

$=1/6+3/6-2/6=1/3$
It is correct. You might want to research 'Order Statistics'.