A Conditional Probability Problem
$begingroup$
I am interested in finding the following problem:
Let $tau_1$ and $tau_2$ are ordered statistics from a set of 2 independent uniform $(0,t)$ R.V. and let $Y_1,Y_2,Y_3$ are nonnegative iid R.V. that are also independent of $tau_1 & tau_2$. Then we want to show that
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=frac{1}{3}$.
For a fixed values of $tau_1=u_1 & tau_2=u_2$ we have,
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1, t-Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3>t-tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
u_1/t*u_2/t=u_1u_2/t^2$
Then taking expectaion with respect to joint distribution of $(tau_1,tau_2)$ which is $f(tau_1,tau_2)=2/t^2, 0<tau_1<tau_2<t $, we have
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=E(tau_1tau_2/t^2)=frac{1}{4}
$
which is not desired answer 1/3. I am not sure if my justification is correct. I appreciate any help for this problem.
probability statistics stochastic-processes conditional-probability order-statistics
asked Dec 11 '18 at 16:46
DavidDavid
976
$endgroup$
add a comment |
$begingroup$
I am interested in finding the following problem:
Let $tau_1$ and $tau_2$ are ordered statistics from a set of 2 independent uniform $(0,t)$ R.V. and let $Y_1,Y_2,Y_3$ are nonnegative iid R.V. that are also independent of $tau_1 & tau_2$. Then we want to show that
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=frac{1}{3}$.
For a fixed values of $tau_1=u_1 & tau_2=u_2$ we have,
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1, t-Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3>t-tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
u_1/t*u_2/t=u_1u_2/t^2$
Then taking expectaion with respect to joint distribution of $(tau_1,tau_2)$ which is $f(tau_1,tau_2)=2/t^2, 0<tau_1<tau_2<t $, we have
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=E(tau_1tau_2/t^2)=frac{1}{4}
$
which is not desired answer 1/3. I am not sure if my justification is correct. I appreciate any help for this problem.
probability statistics stochastic-processes conditional-probability order-statistics
asked Dec 11 '18 at 16:46
DavidDavid
976
$endgroup$
add a comment |
$begingroup$
I am interested in finding the following problem:
Let $tau_1$ and $tau_2$ are ordered statistics from a set of 2 independent uniform $(0,t)$ R.V. and let $Y_1,Y_2,Y_3$ are nonnegative iid R.V. that are also independent of $tau_1 & tau_2$. Then we want to show that
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=frac{1}{3}$.
For a fixed values of $tau_1=u_1 & tau_2=u_2$ we have,
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1, t-Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3>t-tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
u_1/t*u_2/t=u_1u_2/t^2$
Then taking expectaion with respect to joint distribution of $(tau_1,tau_2)$ which is $f(tau_1,tau_2)=2/t^2, 0<tau_1<tau_2<t $, we have
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=E(tau_1tau_2/t^2)=frac{1}{4}
$
which is not desired answer 1/3. I am not sure if my justification is correct. I appreciate any help for this problem.
probability statistics stochastic-processes conditional-probability order-statistics
asked Dec 11 '18 at 16:46
DavidDavid
976
$endgroup$
I am interested in finding the following problem:
Let $tau_1$ and $tau_2$ are ordered statistics from a set of 2 independent uniform $(0,t)$ R.V. and let $Y_1,Y_2,Y_3$ are nonnegative iid R.V. that are also independent of $tau_1 & tau_2$. Then we want to show that
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=frac{1}{3}$.
For a fixed values of $tau_1=u_1 & tau_2=u_2$ we have,
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1, t-Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_1=u_1, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3>t-tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
P(Y_1<tau_1|Y_1+Y_2+Y_3=t, tau_1=u_1)P(Y_3<tau_2|Y_1+Y_2+Y_3=t, tau_2=u_2)=\
u_1/t*u_2/t=u_1u_2/t^2$
Then taking expectaion with respect to joint distribution of $(tau_1,tau_2)$ which is $f(tau_1,tau_2)=2/t^2, 0<tau_1<tau_2<t $, we have
$P(Y_1<tau_1, Y_1+Y_2<tau_2|Y_1+Y_2+Y_3=t)=E(tau_1tau_2/t^2)=frac{1}{4}
$
which is not desired answer 1/3. I am not sure if my justification is correct. I appreciate any help for this problem.
probability statistics stochastic-processes conditional-probability order-statistics
probability statistics stochastic-processes conditional-probability order-statistics
asked Dec 11 '18 at 16:46
DavidDavid
976
asked Dec 11 '18 at 16:46
DavidDavid
976
asked Dec 11 '18 at 16:46
DavidDavid
976
asked Dec 11 '18 at 16:46
DavidDavid
976
asked Dec 11 '18 at 16:46
DavidDavid
976
976
add a comment |
add a comment |
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