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Question on probability of two random variables for example $p(X < a, X < Y-b)$
Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.
Given some constants, $a,b$, how would we compute this probability?
$p(X < a, X < Y-b)$
First, I can see that this is the same as $p(X leq min{a, Y-b})$
So we would have:
$int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$
If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?
probability integration probability-theory
asked Nov 20 '18 at 10:22
Steve
1
add a comment |
Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.
Given some constants, $a,b$, how would we compute this probability?
$p(X < a, X < Y-b)$
First, I can see that this is the same as $p(X leq min{a, Y-b})$
So we would have:
$int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$
If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?
probability integration probability-theory
asked Nov 20 '18 at 10:22
Steve
1
add a comment |
Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.
Given some constants, $a,b$, how would we compute this probability?
$p(X < a, X < Y-b)$
First, I can see that this is the same as $p(X leq min{a, Y-b})$
So we would have:
$int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$
If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?
probability integration probability-theory
asked Nov 20 '18 at 10:22
Steve
1
Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.
Given some constants, $a,b$, how would we compute this probability?
$p(X < a, X < Y-b)$
First, I can see that this is the same as $p(X leq min{a, Y-b})$
So we would have:
$int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$
If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?
probability integration probability-theory
probability integration probability-theory
asked Nov 20 '18 at 10:22
Steve
1
asked Nov 20 '18 at 10:22
Steve
1
asked Nov 20 '18 at 10:22
Steve
1
asked Nov 20 '18 at 10:22
Steve
1
asked Nov 20 '18 at 10:22
Steve
1
1
add a comment |
add a comment |
1 Answer
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Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.
Then:
$$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
& =intint[x<min(a,y-b)]dF(x)dG(y)\
& =int F(min(a,y-b))dG(y)\
& =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
& =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
end{aligned}
$$
In second equality independence is used.
In third equality it is used that $F$ is a continous CDF.
Apply this to your mentioned special case.
answered Nov 20 '18 at 11:11
drhab
97.9k544129
Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
– Steve
Nov 21 '18 at 1:23
Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
– drhab
Nov 21 '18 at 7:18
add a comment |
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1 Answer
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Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.
Then:
$$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
& =intint[x<min(a,y-b)]dF(x)dG(y)\
& =int F(min(a,y-b))dG(y)\
& =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
& =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
end{aligned}
$$
In second equality independence is used.
In third equality it is used that $F$ is a continous CDF.
Apply this to your mentioned special case.
answered Nov 20 '18 at 11:11
drhab
97.9k544129
Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
– Steve
Nov 21 '18 at 1:23
Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
– drhab
Nov 21 '18 at 7:18
add a comment |
Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.
Then:
$$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
& =intint[x<min(a,y-b)]dF(x)dG(y)\
& =int F(min(a,y-b))dG(y)\
& =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
& =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
end{aligned}
$$
In second equality independence is used.
In third equality it is used that $F$ is a continous CDF.
Apply this to your mentioned special case.
answered Nov 20 '18 at 11:11
drhab
97.9k544129
Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
– Steve
Nov 21 '18 at 1:23
Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
– drhab
Nov 21 '18 at 7:18
add a comment |
Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.
Then:
$$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
& =intint[x<min(a,y-b)]dF(x)dG(y)\
& =int F(min(a,y-b))dG(y)\
& =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
& =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
end{aligned}
$$
In second equality independence is used.
In third equality it is used that $F$ is a continous CDF.
Apply this to your mentioned special case.
answered Nov 20 '18 at 11:11
drhab
97.9k544129
Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.
Then:
$$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
& =intint[x<min(a,y-b)]dF(x)dG(y)\
& =int F(min(a,y-b))dG(y)\
& =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
& =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
end{aligned}
$$
In second equality independence is used.
In third equality it is used that $F$ is a continous CDF.
Apply this to your mentioned special case.
answered Nov 20 '18 at 11:11
drhab
97.9k544129
answered Nov 20 '18 at 11:11
drhab
97.9k544129
answered Nov 20 '18 at 11:11
drhab
97.9k544129
answered Nov 20 '18 at 11:11
drhab
97.9k544129
97.9k544129
Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
– Steve
Nov 21 '18 at 1:23
Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
– drhab
Nov 21 '18 at 7:18
add a comment |
Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
– Steve
Nov 21 '18 at 1:23
Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
– drhab
Nov 21 '18 at 7:18
Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
– Steve
Nov 21 '18 at 1:23
Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
– Steve
Nov 21 '18 at 1:23
Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
– drhab
Nov 21 '18 at 7:18
Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
– drhab
Nov 21 '18 at 7:18
add a comment |
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