How to calculate $mathbb{P}(|U_{1} - U_{2}| < frac{1}{12})$, where $U_{i} sim mathcal{U}(0, frac{1}{2})$...
$begingroup$
We have a Poisson process of intensity $lambda = 4$. We have the following event:
$A$: "Two marks appear with a separation of $frac{1}{12}$ or less".
We need to calculate the probability of exactly two "marks" appearing between $0$ and $frac{1}{2}$, and $A$ ocurring at the same time. That is, the probability of exactly two marks ocurring in $(0, frac{1}{2})$ with a separation of $frac{1}{12}$ or less.
Here's how far I've gotten:
Let $N$: Number of marks between $0$ and $frac{1}{2}$. We want:
$$mathbb{P}(N=2, A) = mathbb{P}(A|N=2)mathbb{P}(N=2) $$
$Nsim Poi(frac{1}{2}4)$, so: $$mathbb{P}(N=2) = frac{2^2}{2!}e^{-2} = 2e^{-2}$$
The arrival times for a conditioned number of marks (in this case two) is $U_{1}$, and $U_{2}$, where $U_{i} sim mathcal{U}(0, frac{1}{2})$, and $U_{i}$ are independent, so:
$$mathbb{P}(A|N=2) = mathbb{P}(max(U_{1}, U_{2}) - min(U_{1}, U_{2}) < frac{1}{12}) = mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$$
This is where I'm stuck. The answer is supposed to be $frac{11}{18}e^{-2}$, so $mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$ should be $frac{11}{36}$.
Thanks.
probability probability-theory probability-distributions poisson-distribution
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
$endgroup$
add a comment |
$begingroup$
We have a Poisson process of intensity $lambda = 4$. We have the following event:
$A$: "Two marks appear with a separation of $frac{1}{12}$ or less".
We need to calculate the probability of exactly two "marks" appearing between $0$ and $frac{1}{2}$, and $A$ ocurring at the same time. That is, the probability of exactly two marks ocurring in $(0, frac{1}{2})$ with a separation of $frac{1}{12}$ or less.
Here's how far I've gotten:
Let $N$: Number of marks between $0$ and $frac{1}{2}$. We want:
$$mathbb{P}(N=2, A) = mathbb{P}(A|N=2)mathbb{P}(N=2) $$
$Nsim Poi(frac{1}{2}4)$, so: $$mathbb{P}(N=2) = frac{2^2}{2!}e^{-2} = 2e^{-2}$$
The arrival times for a conditioned number of marks (in this case two) is $U_{1}$, and $U_{2}$, where $U_{i} sim mathcal{U}(0, frac{1}{2})$, and $U_{i}$ are independent, so:
$$mathbb{P}(A|N=2) = mathbb{P}(max(U_{1}, U_{2}) - min(U_{1}, U_{2}) < frac{1}{12}) = mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$$
This is where I'm stuck. The answer is supposed to be $frac{11}{18}e^{-2}$, so $mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$ should be $frac{11}{36}$.
Thanks.
probability probability-theory probability-distributions poisson-distribution
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
$endgroup$
add a comment |
$begingroup$
We have a Poisson process of intensity $lambda = 4$. We have the following event:
$A$: "Two marks appear with a separation of $frac{1}{12}$ or less".
We need to calculate the probability of exactly two "marks" appearing between $0$ and $frac{1}{2}$, and $A$ ocurring at the same time. That is, the probability of exactly two marks ocurring in $(0, frac{1}{2})$ with a separation of $frac{1}{12}$ or less.
Here's how far I've gotten:
Let $N$: Number of marks between $0$ and $frac{1}{2}$. We want:
$$mathbb{P}(N=2, A) = mathbb{P}(A|N=2)mathbb{P}(N=2) $$
$Nsim Poi(frac{1}{2}4)$, so: $$mathbb{P}(N=2) = frac{2^2}{2!}e^{-2} = 2e^{-2}$$
The arrival times for a conditioned number of marks (in this case two) is $U_{1}$, and $U_{2}$, where $U_{i} sim mathcal{U}(0, frac{1}{2})$, and $U_{i}$ are independent, so:
$$mathbb{P}(A|N=2) = mathbb{P}(max(U_{1}, U_{2}) - min(U_{1}, U_{2}) < frac{1}{12}) = mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$$
This is where I'm stuck. The answer is supposed to be $frac{11}{18}e^{-2}$, so $mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$ should be $frac{11}{36}$.
Thanks.
probability probability-theory probability-distributions poisson-distribution
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
$endgroup$
We have a Poisson process of intensity $lambda = 4$. We have the following event:
$A$: "Two marks appear with a separation of $frac{1}{12}$ or less".
We need to calculate the probability of exactly two "marks" appearing between $0$ and $frac{1}{2}$, and $A$ ocurring at the same time. That is, the probability of exactly two marks ocurring in $(0, frac{1}{2})$ with a separation of $frac{1}{12}$ or less.
Here's how far I've gotten:
Let $N$: Number of marks between $0$ and $frac{1}{2}$. We want:
$$mathbb{P}(N=2, A) = mathbb{P}(A|N=2)mathbb{P}(N=2) $$
$Nsim Poi(frac{1}{2}4)$, so: $$mathbb{P}(N=2) = frac{2^2}{2!}e^{-2} = 2e^{-2}$$
The arrival times for a conditioned number of marks (in this case two) is $U_{1}$, and $U_{2}$, where $U_{i} sim mathcal{U}(0, frac{1}{2})$, and $U_{i}$ are independent, so:
$$mathbb{P}(A|N=2) = mathbb{P}(max(U_{1}, U_{2}) - min(U_{1}, U_{2}) < frac{1}{12}) = mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$$
This is where I'm stuck. The answer is supposed to be $frac{11}{18}e^{-2}$, so $mathbb{P}left(|U_{1} - U_{2}| < frac{1}{12}right)$ should be $frac{11}{36}$.
Thanks.
probability probability-theory probability-distributions poisson-distribution
probability probability-theory probability-distributions poisson-distribution
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
edited Dec 12 '18 at 10:10
Davide Giraudo
127k16151264
127k16151264
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
asked Dec 11 '18 at 22:00
Juanma EloyJuanma Eloy
5611516
5611516
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Hint: saying that $U_1,U_2$ are independent uniform $[0,1/2]$ is equivalent to saying that $(U_1,U_2)$ is uniformly distributed over the square $[0,1/2]times [0,1/2]={(x,y):0le xle 1/2,0le yle 1/2}$. Draw this square, draw the region where $|x-y|le frac1{12}$, and compute the fraction of this area over the total area.
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
$endgroup$
add a comment |
$begingroup$
You can proceed e.g. as follows: Let $V_1=2U_1, V_2=2U_2$. Then $Z = |V_1-V_2|$ follows a triangular distribution with parameters $(a,b,c)=(0,1,0)$ , and you want
$$
mathbb{P}left{ |U_1-U_2| < frac{1}{12}right}
= mathbb{P}left{ Z < frac{1}{6}right} = int_0^{1/6} f_Z(z)dz = int_0^{1/6} (2-2z)dz = boxed{frac{11}{36}}
$$
as claimed.
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: saying that $U_1,U_2$ are independent uniform $[0,1/2]$ is equivalent to saying that $(U_1,U_2)$ is uniformly distributed over the square $[0,1/2]times [0,1/2]={(x,y):0le xle 1/2,0le yle 1/2}$. Draw this square, draw the region where $|x-y|le frac1{12}$, and compute the fraction of this area over the total area.
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
$endgroup$
add a comment |
$begingroup$
Hint: saying that $U_1,U_2$ are independent uniform $[0,1/2]$ is equivalent to saying that $(U_1,U_2)$ is uniformly distributed over the square $[0,1/2]times [0,1/2]={(x,y):0le xle 1/2,0le yle 1/2}$. Draw this square, draw the region where $|x-y|le frac1{12}$, and compute the fraction of this area over the total area.
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
$endgroup$
add a comment |
$begingroup$
Hint: saying that $U_1,U_2$ are independent uniform $[0,1/2]$ is equivalent to saying that $(U_1,U_2)$ is uniformly distributed over the square $[0,1/2]times [0,1/2]={(x,y):0le xle 1/2,0le yle 1/2}$. Draw this square, draw the region where $|x-y|le frac1{12}$, and compute the fraction of this area over the total area.
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
$endgroup$
Hint: saying that $U_1,U_2$ are independent uniform $[0,1/2]$ is equivalent to saying that $(U_1,U_2)$ is uniformly distributed over the square $[0,1/2]times [0,1/2]={(x,y):0le xle 1/2,0le yle 1/2}$. Draw this square, draw the region where $|x-y|le frac1{12}$, and compute the fraction of this area over the total area.
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
answered Dec 11 '18 at 22:36
Mike EarnestMike Earnest
23.4k12051
23.4k12051
add a comment |
add a comment |
$begingroup$
You can proceed e.g. as follows: Let $V_1=2U_1, V_2=2U_2$. Then $Z = |V_1-V_2|$ follows a triangular distribution with parameters $(a,b,c)=(0,1,0)$ , and you want
$$
mathbb{P}left{ |U_1-U_2| < frac{1}{12}right}
= mathbb{P}left{ Z < frac{1}{6}right} = int_0^{1/6} f_Z(z)dz = int_0^{1/6} (2-2z)dz = boxed{frac{11}{36}}
$$
as claimed.
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
$endgroup$
add a comment |
$begingroup$
You can proceed e.g. as follows: Let $V_1=2U_1, V_2=2U_2$. Then $Z = |V_1-V_2|$ follows a triangular distribution with parameters $(a,b,c)=(0,1,0)$ , and you want
$$
mathbb{P}left{ |U_1-U_2| < frac{1}{12}right}
= mathbb{P}left{ Z < frac{1}{6}right} = int_0^{1/6} f_Z(z)dz = int_0^{1/6} (2-2z)dz = boxed{frac{11}{36}}
$$
as claimed.
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
$endgroup$
add a comment |
$begingroup$
You can proceed e.g. as follows: Let $V_1=2U_1, V_2=2U_2$. Then $Z = |V_1-V_2|$ follows a triangular distribution with parameters $(a,b,c)=(0,1,0)$ , and you want
$$
mathbb{P}left{ |U_1-U_2| < frac{1}{12}right}
= mathbb{P}left{ Z < frac{1}{6}right} = int_0^{1/6} f_Z(z)dz = int_0^{1/6} (2-2z)dz = boxed{frac{11}{36}}
$$
as claimed.
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
$endgroup$
You can proceed e.g. as follows: Let $V_1=2U_1, V_2=2U_2$. Then $Z = |V_1-V_2|$ follows a triangular distribution with parameters $(a,b,c)=(0,1,0)$ , and you want
$$
mathbb{P}left{ |U_1-U_2| < frac{1}{12}right}
= mathbb{P}left{ Z < frac{1}{6}right} = int_0^{1/6} f_Z(z)dz = int_0^{1/6} (2-2z)dz = boxed{frac{11}{36}}
$$
as claimed.
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
edited Dec 12 '18 at 11:51
StubbornAtom
6,04811239
6,04811239
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
answered Dec 11 '18 at 22:39
Clement C.Clement C.
50.5k33891
50.5k33891
add a comment |
add a comment |
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