Expected value of two uniformly distributed random variables

  • Thread starter DottZakapa
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  • #1
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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?
 

Answers and Replies

  • #2
etotheipi
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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]##.
 
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  • #3
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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?
 
  • #4
etotheipi
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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)##.
 
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  • #5
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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 ##
 
  • #6
etotheipi
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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$$\mathbb{E}(Z) = \int_0^1 \overline{F_{Z}}(z) dz$$
 
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