Find $P(x_2/x_3 leq a)$ where $x_1,x_2,dots,x_n $ are iid $operatorname{Uniform}(0,1)$
$begingroup$
Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.
How to find $P(x_2/x_3 leq a)$?
Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below
$$f(x_2, x_3) = 1$$
$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$
Edit, I need to find the probability by conditioning. S my approach is not right.
Update: after reviewing the comments below, I summarized them into:
$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$
these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results
I'm still thinking of the best way to get these results using conditional probability
probability-theory uniform-distribution
edited Dec 16 '18 at 9:03
user10354138
7,4422925
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
$endgroup$
|
show 1 more comment
$begingroup$
Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.
How to find $P(x_2/x_3 leq a)$?
Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below
$$f(x_2, x_3) = 1$$
$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$
Edit, I need to find the probability by conditioning. S my approach is not right.
Update: after reviewing the comments below, I summarized them into:
$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$
these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results
I'm still thinking of the best way to get these results using conditional probability
probability-theory uniform-distribution
edited Dec 16 '18 at 9:03
user10354138
7,4422925
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
$endgroup$
1
$begingroup$
Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
$endgroup$
– wolfies
Dec 10 '13 at 7:11
$begingroup$
$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
$endgroup$
– achille hui
Dec 10 '13 at 7:24
$begingroup$
Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
$endgroup$
– Macavity
Dec 10 '13 at 7:37
$begingroup$
@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
$endgroup$
– dark blue
Dec 10 '13 at 13:02
$begingroup$
@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
$endgroup$
– dark blue
Dec 10 '13 at 13:03
|
show 1 more comment
$begingroup$
Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.
How to find $P(x_2/x_3 leq a)$?
Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below
$$f(x_2, x_3) = 1$$
$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$
Edit, I need to find the probability by conditioning. S my approach is not right.
Update: after reviewing the comments below, I summarized them into:
$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$
these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results
I'm still thinking of the best way to get these results using conditional probability
probability-theory uniform-distribution
edited Dec 16 '18 at 9:03
user10354138
7,4422925
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
$endgroup$
Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.
How to find $P(x_2/x_3 leq a)$?
Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below
$$f(x_2, x_3) = 1$$
$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$
Edit, I need to find the probability by conditioning. S my approach is not right.
Update: after reviewing the comments below, I summarized them into:
$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$
these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results
I'm still thinking of the best way to get these results using conditional probability
probability-theory uniform-distribution
probability-theory uniform-distribution
edited Dec 16 '18 at 9:03
user10354138
7,4422925
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
edited Dec 16 '18 at 9:03
user10354138
7,4422925
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
edited Dec 16 '18 at 9:03
user10354138
7,4422925
edited Dec 16 '18 at 9:03
user10354138
7,4422925
edited Dec 16 '18 at 9:03
user10354138
7,4422925
7,4422925
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
asked Dec 10 '13 at 6:15
dark bluedark blue
130110
130110
1
$begingroup$
Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
$endgroup$
– wolfies
Dec 10 '13 at 7:11
$begingroup$
$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
$endgroup$
– achille hui
Dec 10 '13 at 7:24
$begingroup$
Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
$endgroup$
– Macavity
Dec 10 '13 at 7:37
$begingroup$
@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
$endgroup$
– dark blue
Dec 10 '13 at 13:02
$begingroup$
@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
$endgroup$
– dark blue
Dec 10 '13 at 13:03
|
show 1 more comment
1
$begingroup$
Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
$endgroup$
– wolfies
Dec 10 '13 at 7:11
$begingroup$
$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
$endgroup$
– achille hui
Dec 10 '13 at 7:24
$begingroup$
Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
$endgroup$
– Macavity
Dec 10 '13 at 7:37
$begingroup$
@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
$endgroup$
– dark blue
Dec 10 '13 at 13:02
$begingroup$
@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
$endgroup$
– dark blue
Dec 10 '13 at 13:03
1
1
$begingroup$
Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
$endgroup$
– wolfies
Dec 10 '13 at 7:11
$begingroup$
Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
$endgroup$
– wolfies
Dec 10 '13 at 7:11
$begingroup$
$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
$endgroup$
– achille hui
Dec 10 '13 at 7:24
$begingroup$
$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
$endgroup$
– achille hui
Dec 10 '13 at 7:24
$begingroup$
Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
$endgroup$
– Macavity
Dec 10 '13 at 7:37
$begingroup$
Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
$endgroup$
– Macavity
Dec 10 '13 at 7:37
$begingroup$
@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
$endgroup$
– dark blue
Dec 10 '13 at 13:02
$begingroup$
@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
$endgroup$
– dark blue
Dec 10 '13 at 13:02
$begingroup$
@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
$endgroup$
– dark blue
Dec 10 '13 at 13:03
$begingroup$
@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
$endgroup$
– dark blue
Dec 10 '13 at 13:03
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Hint: You should reason that (for positive a)
$$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
int_0^1 a , x_3 , dx_3 && a le 1\
int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
end{cases}
$$
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
$endgroup$
add a comment |
$begingroup$
For $xinleft(0,1right)$ we have by independency:
$Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $
So if function $f$ is defined by $fleft(xright)=begin{cases}
1 & text{if }axgeq1\
0 & text{if }axleq0\
ax & text{otherwise}end{cases}$
Then:
$Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
$endgroup$
$begingroup$
thank you, but why is the answer different from the above?
$endgroup$
– dark blue
Dec 10 '13 at 18:53
$begingroup$
It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
$endgroup$
– drhab
Dec 11 '13 at 10:14
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: You should reason that (for positive a)
$$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
int_0^1 a , x_3 , dx_3 && a le 1\
int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
end{cases}
$$
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
$endgroup$
add a comment |
$begingroup$
Hint: You should reason that (for positive a)
$$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
int_0^1 a , x_3 , dx_3 && a le 1\
int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
end{cases}
$$
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
$endgroup$
add a comment |
$begingroup$
Hint: You should reason that (for positive a)
$$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
int_0^1 a , x_3 , dx_3 && a le 1\
int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
end{cases}
$$
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
$endgroup$
Hint: You should reason that (for positive a)
$$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
int_0^1 a , x_3 , dx_3 && a le 1\
int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
end{cases}
$$
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
edited Dec 10 '13 at 14:42
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
answered Dec 10 '13 at 14:30
MacavityMacavity
35.3k52554
35.3k52554
add a comment |
add a comment |
$begingroup$
For $xinleft(0,1right)$ we have by independency:
$Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $
So if function $f$ is defined by $fleft(xright)=begin{cases}
1 & text{if }axgeq1\
0 & text{if }axleq0\
ax & text{otherwise}end{cases}$
Then:
$Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
$endgroup$
$begingroup$
thank you, but why is the answer different from the above?
$endgroup$
– dark blue
Dec 10 '13 at 18:53
$begingroup$
It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
$endgroup$
– drhab
Dec 11 '13 at 10:14
add a comment |
$begingroup$
For $xinleft(0,1right)$ we have by independency:
$Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $
So if function $f$ is defined by $fleft(xright)=begin{cases}
1 & text{if }axgeq1\
0 & text{if }axleq0\
ax & text{otherwise}end{cases}$
Then:
$Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
$endgroup$
$begingroup$
thank you, but why is the answer different from the above?
$endgroup$
– dark blue
Dec 10 '13 at 18:53
$begingroup$
It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
$endgroup$
– drhab
Dec 11 '13 at 10:14
add a comment |
$begingroup$
For $xinleft(0,1right)$ we have by independency:
$Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $
So if function $f$ is defined by $fleft(xright)=begin{cases}
1 & text{if }axgeq1\
0 & text{if }axleq0\
ax & text{otherwise}end{cases}$
Then:
$Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
$endgroup$
For $xinleft(0,1right)$ we have by independency:
$Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $
So if function $f$ is defined by $fleft(xright)=begin{cases}
1 & text{if }axgeq1\
0 & text{if }axleq0\
ax & text{otherwise}end{cases}$
Then:
$Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
answered Dec 10 '13 at 14:50
drhabdrhab
101k545136
101k545136
$begingroup$
thank you, but why is the answer different from the above?
$endgroup$
– dark blue
Dec 10 '13 at 18:53
$begingroup$
It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
$endgroup$
– drhab
Dec 11 '13 at 10:14
add a comment |
$begingroup$
thank you, but why is the answer different from the above?
$endgroup$
– dark blue
Dec 10 '13 at 18:53
$begingroup$
It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
$endgroup$
– drhab
Dec 11 '13 at 10:14
$begingroup$
thank you, but why is the answer different from the above?
$endgroup$
– dark blue
Dec 10 '13 at 18:53
$begingroup$
thank you, but why is the answer different from the above?
$endgroup$
– dark blue
Dec 10 '13 at 18:53
$begingroup$
It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
$endgroup$
– drhab
Dec 11 '13 at 10:14
$begingroup$
It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
$endgroup$
– drhab
Dec 11 '13 at 10:14
add a comment |
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1
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Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
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– wolfies
Dec 10 '13 at 7:11
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$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
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– achille hui
Dec 10 '13 at 7:24
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Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
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– Macavity
Dec 10 '13 at 7:37
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@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
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– dark blue
Dec 10 '13 at 13:02
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@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
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– dark blue
Dec 10 '13 at 13:03