Expected value of two uniformly distributed random variables

[DottZakapa]

Homework Statement

$X_1$ and$X_2$ are uniformly distributed with parameters $(0,1)$
then:
$E[min {X_1,X_2}]=$

Relevant Equations

Probability

$X_1$ and$X_2$ are uniformly distributed random variables with parameters $(0,1)$
then:
$E \left[ min \left\{ X_1 , X_2 \right\} \right] =$

what should I do with that min?

 

[etotheipi]

If $Z = \text{min}(X_1, X_2)$, then $Z > z$ iff both $X_1$ and $X_2$ are greater than $z$. It follows that$$P(Z > z) = P(X_1 > z)P(X_2 > z)$$You can use this to find the cumulative probability function $F_Z(z) = P(Z<z)$ in terms of $F_{X_1}(z)$ and $F_{X_2}(z)$, the last two of which are known since we're given that $X_1$ and $X_2$ are uniformly distributed in the interval $[0, 1]$.

 

[DottZakapa]

It should be something like this?
$f_{x_{1}} \left(z\right) = \frac 1 {b-a}$
$f_{x_{2}} \left(z\right) = \frac 1 {b-a}$
$F_{x_{1}} \left(z\right) = \frac {z-a} {b-a}$
$F_{x_{1}} \left(z\right) = \frac {z-a} {b-a}$

but, being $a=0$ and $b=1$

$F_{x_{1}} \left(z\right) = z$
$F_{x_{1}} \left(z\right) = z$
so
$F_{x_{1}} \left(z\right)*F_{x_{1}} \left(z\right)=z^2$

being by definition

$E \left[x\right] = \int_a^b \left(1-F_x\left(z\right)\right) \, dz$

so i'll have to integrate

$\int_0^1 \left(1-z^2 \right ) \, dz$ is it correct?

 

[etotheipi]

Your $F_{X_1}(z) = z$ and $F_{X_2}(z) = z$ are correct, but the rest is not. Start with$$\begin{align*}P(Z > z) &= P(X_1 > z)P(X_2 > z) \\ \\ 1-F_Z(z) &= (1-F_{X_1}(z)) (1-F_{X_2}(z)) = 1 - F_{X_1}(z) - F_{X_2}(z) + F_{X_1}(z)F_{X_2}(z) \\ \\ F_Z(z) &= F_{X_1}(z) + F_{X_2}(z) - F_{X_1}(z)F_{X_2}(z) = 2z - z^2 \end{align*}$$and this last expression you might notice is a statement of the inclusion-exclusion principle.

Now you have the cumulative function $F_Z(z)$, you can find $\mathbb{E}(Z)$ either by integrating $1 - F_Z(z)$, which is the way you suggested, or by finding $f_Z(z) = \frac{dF_Z(z)}{dz}$ and integrating $z f_Z(z)$.

 

[DottZakapa]

ok so the thinking is :
being $X_1>z$ and $X_2>z$
for each of them I consider the Survival Function $\overline{F_{x}}$, that is
$\overline{F_{x_1}}=1-F_{x_1}\left(z\right)= 1-z$ same for the other $\overline{F_{x_2}}=1-F_{x_2}\left(z\right)= 1-z$
and
$\overline{F_{x_{1}} \left(z\right)}*\overline{F_{x_{1}} \left(z\right)}=\left(1-z\right)*\left(1-z\right)= \left(1-z\right)^2= z^2-2z+1$

and integrate as
$\int_0^1 \left(z^2-2z+1 \right ) \, dz$

 

[etotheipi]

I haven't come across the term 'survival function', but from how you use it I'll just assume it's 1 minus the cumulative probability. Then yes,$$\overline{F_{Z}}(z) = \overline{F_{X_1}}(z) \overline{F_{X_2}}(z) = (1-z)^2$$which you can just integrate up like$$\mathbb{E}(Z) = \int_0^1 \overline{F_{Z}}(z) dz$$