Find $P(2Y_{(1)} < Y_{(2)})$ of a Uniformly Distributed Random Variable












1












$begingroup$



Denote $Y_{(1)} = min(Y_1,Y_2)$ and $Y_{(2)} = max(Y_1,Y_2)$.
Let $Y_1$ and $Y_2$ be independent and uniformly distributed over the interval $(0, 1)$. Find
$P(2Y_{(1)} < Y_{(2)})$.




Attempted solution:



We know that



$$
f(y_i) =
begin{cases}
1 & 0<y_i<1\
0 & else
end{cases}
$$



Therefore, we determine



$$
F(y_i) =
begin{cases}
0 & y_i < 0\
y_i & 0<y_i<1\
1 & y_i > 1\
end{cases}
$$



Using this, we can find the distribution functions for $Y_{(1)}$ and $Y_{(2)}$



$$
Y_{(1)} = 1 - (1-F(y))^2\
Y_{(2)} = F(y)^2
$$



By differentiating, we get the density functions



$$
f_{Y_{(1)}}(y) = 2(1-y)\
f_{Y_{(2)}}(y) = 2y
$$



I'm not too sure where to go from here.






probability-distributions order-statistics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Similar to the answer below, using total probability theorem, begin{align} P(2Y_{(1)}<Y_{(2)})&=P(2Y_{(1)}<Y_{(2)},Y_1<Y_2)+P(2Y_{(1)}<Y_{(2)},Y_1ge Y_2) \&=P(2Y_1<Y_2,Y_1<Y_2)+P(2Y_2<Y_1,Y_2< Y_1) \&=2times P(2Y_1<Y_2) end{align} If you can justify the steps above, the rest is straightforward.
    $endgroup$
    – StubbornAtom
    Jan 2 at 6:35
















1












$begingroup$



Denote $Y_{(1)} = min(Y_1,Y_2)$ and $Y_{(2)} = max(Y_1,Y_2)$.
Let $Y_1$ and $Y_2$ be independent and uniformly distributed over the interval $(0, 1)$. Find
$P(2Y_{(1)} < Y_{(2)})$.




Attempted solution:



We know that



$$
f(y_i) =
begin{cases}
1 & 0<y_i<1\
0 & else
end{cases}
$$



Therefore, we determine



$$
F(y_i) =
begin{cases}
0 & y_i < 0\
y_i & 0<y_i<1\
1 & y_i > 1\
end{cases}
$$



Using this, we can find the distribution functions for $Y_{(1)}$ and $Y_{(2)}$



$$
Y_{(1)} = 1 - (1-F(y))^2\
Y_{(2)} = F(y)^2
$$



By differentiating, we get the density functions



$$
f_{Y_{(1)}}(y) = 2(1-y)\
f_{Y_{(2)}}(y) = 2y
$$



I'm not too sure where to go from here.






probability-distributions order-statistics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Similar to the answer below, using total probability theorem, begin{align} P(2Y_{(1)}<Y_{(2)})&=P(2Y_{(1)}<Y_{(2)},Y_1<Y_2)+P(2Y_{(1)}<Y_{(2)},Y_1ge Y_2) \&=P(2Y_1<Y_2,Y_1<Y_2)+P(2Y_2<Y_1,Y_2< Y_1) \&=2times P(2Y_1<Y_2) end{align} If you can justify the steps above, the rest is straightforward.
    $endgroup$
    – StubbornAtom
    Jan 2 at 6:35














1












1








1


0



$begingroup$



Denote $Y_{(1)} = min(Y_1,Y_2)$ and $Y_{(2)} = max(Y_1,Y_2)$.
Let $Y_1$ and $Y_2$ be independent and uniformly distributed over the interval $(0, 1)$. Find
$P(2Y_{(1)} < Y_{(2)})$.




Attempted solution:



We know that



$$
f(y_i) =
begin{cases}
1 & 0<y_i<1\
0 & else
end{cases}
$$



Therefore, we determine



$$
F(y_i) =
begin{cases}
0 & y_i < 0\
y_i & 0<y_i<1\
1 & y_i > 1\
end{cases}
$$



Using this, we can find the distribution functions for $Y_{(1)}$ and $Y_{(2)}$



$$
Y_{(1)} = 1 - (1-F(y))^2\
Y_{(2)} = F(y)^2
$$



By differentiating, we get the density functions



$$
f_{Y_{(1)}}(y) = 2(1-y)\
f_{Y_{(2)}}(y) = 2y
$$



I'm not too sure where to go from here.






probability-distributions order-statistics







share|cite|improve this question









$endgroup$





Denote $Y_{(1)} = min(Y_1,Y_2)$ and $Y_{(2)} = max(Y_1,Y_2)$.
Let $Y_1$ and $Y_2$ be independent and uniformly distributed over the interval $(0, 1)$. Find
$P(2Y_{(1)} < Y_{(2)})$.




Attempted solution:



We know that



$$
f(y_i) =
begin{cases}
1 & 0<y_i<1\
0 & else
end{cases}
$$



Therefore, we determine



$$
F(y_i) =
begin{cases}
0 & y_i < 0\
y_i & 0<y_i<1\
1 & y_i > 1\
end{cases}
$$



Using this, we can find the distribution functions for $Y_{(1)}$ and $Y_{(2)}$



$$
Y_{(1)} = 1 - (1-F(y))^2\
Y_{(2)} = F(y)^2
$$



By differentiating, we get the density functions



$$
f_{Y_{(1)}}(y) = 2(1-y)\
f_{Y_{(2)}}(y) = 2y
$$



I'm not too sure where to go from here.






probability-distributions order-statistics




probability-distributions order-statistics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 0:01









Bryden CBryden C

31918




31918












  • $begingroup$
    Similar to the answer below, using total probability theorem, begin{align} P(2Y_{(1)}<Y_{(2)})&=P(2Y_{(1)}<Y_{(2)},Y_1<Y_2)+P(2Y_{(1)}<Y_{(2)},Y_1ge Y_2) \&=P(2Y_1<Y_2,Y_1<Y_2)+P(2Y_2<Y_1,Y_2< Y_1) \&=2times P(2Y_1<Y_2) end{align} If you can justify the steps above, the rest is straightforward.
    $endgroup$
    – StubbornAtom
    Jan 2 at 6:35


















  • $begingroup$
    Similar to the answer below, using total probability theorem, begin{align} P(2Y_{(1)}<Y_{(2)})&=P(2Y_{(1)}<Y_{(2)},Y_1<Y_2)+P(2Y_{(1)}<Y_{(2)},Y_1ge Y_2) \&=P(2Y_1<Y_2,Y_1<Y_2)+P(2Y_2<Y_1,Y_2< Y_1) \&=2times P(2Y_1<Y_2) end{align} If you can justify the steps above, the rest is straightforward.
    $endgroup$
    – StubbornAtom
    Jan 2 at 6:35
















$begingroup$
Similar to the answer below, using total probability theorem, begin{align} P(2Y_{(1)}<Y_{(2)})&=P(2Y_{(1)}<Y_{(2)},Y_1<Y_2)+P(2Y_{(1)}<Y_{(2)},Y_1ge Y_2) \&=P(2Y_1<Y_2,Y_1<Y_2)+P(2Y_2<Y_1,Y_2< Y_1) \&=2times P(2Y_1<Y_2) end{align} If you can justify the steps above, the rest is straightforward.
$endgroup$
– StubbornAtom
Jan 2 at 6:35




$begingroup$
Similar to the answer below, using total probability theorem, begin{align} P(2Y_{(1)}<Y_{(2)})&=P(2Y_{(1)}<Y_{(2)},Y_1<Y_2)+P(2Y_{(1)}<Y_{(2)},Y_1ge Y_2) \&=P(2Y_1<Y_2,Y_1<Y_2)+P(2Y_2<Y_1,Y_2< Y_1) \&=2times P(2Y_1<Y_2) end{align} If you can justify the steps above, the rest is straightforward.
$endgroup$
– StubbornAtom
Jan 2 at 6:35










1 Answer
1






active

oldest

votes


















1












$begingroup$

Since $avee b=(a+b)/2+|a-b|/2$ and $awedge b=(a+b)/2-|a-b|/2$,



begin{align}
mathsf{P}(2Y_{(1)}<Y_{(2)})&=mathsf{P}(Y_1+Y_2<3|Y_1-Y_2|) \
&=mathsf{P}(Y_1+Y_2<3(Y_1-Y_2))+mathsf{P}(Y_1+Y_2<3(Y_2-Y_1)) \
&=mathsf{P}(2Y_2<Y_1)+mathsf{P}(2Y_1<Y_2)=1/2.
end{align}




$$
mathsf{P}(2Y_2<Y_1)=int_0^1mathsf{P}(Y_2<y/2)dy=int_0^1(y/2)dy=1/4.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Alright, that makes sense. I'm still not too sure how to continue, however.
    $endgroup$
    – Bryden C
    Jan 2 at 2:57











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1 Answer
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1 Answer
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1












$begingroup$

Since $avee b=(a+b)/2+|a-b|/2$ and $awedge b=(a+b)/2-|a-b|/2$,



begin{align}
mathsf{P}(2Y_{(1)}<Y_{(2)})&=mathsf{P}(Y_1+Y_2<3|Y_1-Y_2|) \
&=mathsf{P}(Y_1+Y_2<3(Y_1-Y_2))+mathsf{P}(Y_1+Y_2<3(Y_2-Y_1)) \
&=mathsf{P}(2Y_2<Y_1)+mathsf{P}(2Y_1<Y_2)=1/2.
end{align}




$$
mathsf{P}(2Y_2<Y_1)=int_0^1mathsf{P}(Y_2<y/2)dy=int_0^1(y/2)dy=1/4.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Alright, that makes sense. I'm still not too sure how to continue, however.
    $endgroup$
    – Bryden C
    Jan 2 at 2:57
















1












$begingroup$

Since $avee b=(a+b)/2+|a-b|/2$ and $awedge b=(a+b)/2-|a-b|/2$,



begin{align}
mathsf{P}(2Y_{(1)}<Y_{(2)})&=mathsf{P}(Y_1+Y_2<3|Y_1-Y_2|) \
&=mathsf{P}(Y_1+Y_2<3(Y_1-Y_2))+mathsf{P}(Y_1+Y_2<3(Y_2-Y_1)) \
&=mathsf{P}(2Y_2<Y_1)+mathsf{P}(2Y_1<Y_2)=1/2.
end{align}




$$
mathsf{P}(2Y_2<Y_1)=int_0^1mathsf{P}(Y_2<y/2)dy=int_0^1(y/2)dy=1/4.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Alright, that makes sense. I'm still not too sure how to continue, however.
    $endgroup$
    – Bryden C
    Jan 2 at 2:57














1












1








1





$begingroup$

Since $avee b=(a+b)/2+|a-b|/2$ and $awedge b=(a+b)/2-|a-b|/2$,



begin{align}
mathsf{P}(2Y_{(1)}<Y_{(2)})&=mathsf{P}(Y_1+Y_2<3|Y_1-Y_2|) \
&=mathsf{P}(Y_1+Y_2<3(Y_1-Y_2))+mathsf{P}(Y_1+Y_2<3(Y_2-Y_1)) \
&=mathsf{P}(2Y_2<Y_1)+mathsf{P}(2Y_1<Y_2)=1/2.
end{align}




$$
mathsf{P}(2Y_2<Y_1)=int_0^1mathsf{P}(Y_2<y/2)dy=int_0^1(y/2)dy=1/4.
$$






share|cite|improve this answer











$endgroup$



Since $avee b=(a+b)/2+|a-b|/2$ and $awedge b=(a+b)/2-|a-b|/2$,



begin{align}
mathsf{P}(2Y_{(1)}<Y_{(2)})&=mathsf{P}(Y_1+Y_2<3|Y_1-Y_2|) \
&=mathsf{P}(Y_1+Y_2<3(Y_1-Y_2))+mathsf{P}(Y_1+Y_2<3(Y_2-Y_1)) \
&=mathsf{P}(2Y_2<Y_1)+mathsf{P}(2Y_1<Y_2)=1/2.
end{align}




$$
mathsf{P}(2Y_2<Y_1)=int_0^1mathsf{P}(Y_2<y/2)dy=int_0^1(y/2)dy=1/4.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 3:04

























answered Jan 2 at 2:38









d.k.o.d.k.o.

8,857628




8,857628












  • $begingroup$
    Alright, that makes sense. I'm still not too sure how to continue, however.
    $endgroup$
    – Bryden C
    Jan 2 at 2:57


















  • $begingroup$
    Alright, that makes sense. I'm still not too sure how to continue, however.
    $endgroup$
    – Bryden C
    Jan 2 at 2:57
















$begingroup$
Alright, that makes sense. I'm still not too sure how to continue, however.
$endgroup$
– Bryden C
Jan 2 at 2:57




$begingroup$
Alright, that makes sense. I'm still not too sure how to continue, however.
$endgroup$
– Bryden C
Jan 2 at 2:57


















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