Mixing
What is the joint CDF of $f(x,y)=2(x+y), 0leq xleq yleq 1$
$begingroup$
I am trying to find the joint CDF of $f(x,y)=2(x+y) : 0leq xleq yleq 1$.
There are five different answers for the CDF depending on the restrictions of $x$ and $y$ that you use. I
found the CDF $F(x,y)=y^2x+x^2y-x^3mbox{ if }0leq xleq yleq 1$ and also I found the CDF $F(x,y)=0mbox{ if }x<0 land y<0$.
Now I need to find the CDF if $x>y$ and $0>y>1$ and then if $y>1$ and $0<x<1$. and for $y>1$ and $x>1$. I know how to do the integral and everything I am just unsure of what my bounds for each of these three integrals where we use double integrals would be.
probability integration probability-distributions
edited Jan 13 at 9:42
rtybase
11k21533
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
$endgroup$
add a comment |
$begingroup$
I am trying to find the joint CDF of $f(x,y)=2(x+y) : 0leq xleq yleq 1$.
There are five different answers for the CDF depending on the restrictions of $x$ and $y$ that you use. I
found the CDF $F(x,y)=y^2x+x^2y-x^3mbox{ if }0leq xleq yleq 1$ and also I found the CDF $F(x,y)=0mbox{ if }x<0 land y<0$.
Now I need to find the CDF if $x>y$ and $0>y>1$ and then if $y>1$ and $0<x<1$. and for $y>1$ and $x>1$. I know how to do the integral and everything I am just unsure of what my bounds for each of these three integrals where we use double integrals would be.
probability integration probability-distributions
edited Jan 13 at 9:42
rtybase
11k21533
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
$endgroup$
add a comment |
$begingroup$
I am trying to find the joint CDF of $f(x,y)=2(x+y) : 0leq xleq yleq 1$.
There are five different answers for the CDF depending on the restrictions of $x$ and $y$ that you use. I
found the CDF $F(x,y)=y^2x+x^2y-x^3mbox{ if }0leq xleq yleq 1$ and also I found the CDF $F(x,y)=0mbox{ if }x<0 land y<0$.
Now I need to find the CDF if $x>y$ and $0>y>1$ and then if $y>1$ and $0<x<1$. and for $y>1$ and $x>1$. I know how to do the integral and everything I am just unsure of what my bounds for each of these three integrals where we use double integrals would be.
probability integration probability-distributions
edited Jan 13 at 9:42
rtybase
11k21533
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
$endgroup$
I am trying to find the joint CDF of $f(x,y)=2(x+y) : 0leq xleq yleq 1$.
There are five different answers for the CDF depending on the restrictions of $x$ and $y$ that you use. I
found the CDF $F(x,y)=y^2x+x^2y-x^3mbox{ if }0leq xleq yleq 1$ and also I found the CDF $F(x,y)=0mbox{ if }x<0 land y<0$.
Now I need to find the CDF if $x>y$ and $0>y>1$ and then if $y>1$ and $0<x<1$. and for $y>1$ and $x>1$. I know how to do the integral and everything I am just unsure of what my bounds for each of these three integrals where we use double integrals would be.
probability integration probability-distributions
probability integration probability-distributions
edited Jan 13 at 9:42
rtybase
11k21533
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
edited Jan 13 at 9:42
rtybase
11k21533
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
edited Jan 13 at 9:42
rtybase
11k21533
edited Jan 13 at 9:42
rtybase
11k21533
edited Jan 13 at 9:42
rtybase
11k21533
11k21533
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
asked Nov 7 '14 at 5:41
juniormathjuniormath
155215
155215
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It looks like :
$$begin{align}F(x,y) ~&=~ begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^yint_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^1 int_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t f(s,t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^xint_0^t 2(s+t)operatorname d soperatorname d t+
int_x^yint_0^x 2(s+t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^xint_0^t2(s+t)operatorname d soperatorname d t+
int_x^1 int_0^x 2(s+t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t 2(s+t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] x y^2+ x^2 y -x^3 & : 0 leq x leq y leq 1
\[4pt] x + x^2 -x^3 & : 0leq xleq 1, y> 1
\[4pt] y^3 & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
end{align}$$
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
$endgroup$
$begingroup$
for the fourth one I happen to know the answer is y^3 so those bounds do not work for that.
$endgroup$
– juniormath
Nov 7 '14 at 7:43
$begingroup$
same with the third one. The answer is x+x^2-x^3 and I am getting just x+x^2 with those bounds @GrahamKemp
$endgroup$
– juniormath
Nov 7 '14 at 7:53
$begingroup$
@Graham Kemp What do you mean by min(x,t)?
$endgroup$
– AmaC
Mar 24 '17 at 13:34
$begingroup$
The minimum of $x$ and $t$.
$endgroup$
– Graham Kemp
Mar 24 '17 at 13:42
add a comment |
Your Answer
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1 Answer
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$begingroup$
It looks like :
$$begin{align}F(x,y) ~&=~ begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^yint_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^1 int_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t f(s,t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^xint_0^t 2(s+t)operatorname d soperatorname d t+
int_x^yint_0^x 2(s+t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^xint_0^t2(s+t)operatorname d soperatorname d t+
int_x^1 int_0^x 2(s+t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t 2(s+t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] x y^2+ x^2 y -x^3 & : 0 leq x leq y leq 1
\[4pt] x + x^2 -x^3 & : 0leq xleq 1, y> 1
\[4pt] y^3 & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
end{align}$$
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
$endgroup$
$begingroup$
for the fourth one I happen to know the answer is y^3 so those bounds do not work for that.
$endgroup$
– juniormath
Nov 7 '14 at 7:43
$begingroup$
same with the third one. The answer is x+x^2-x^3 and I am getting just x+x^2 with those bounds @GrahamKemp
$endgroup$
– juniormath
Nov 7 '14 at 7:53
$begingroup$
@Graham Kemp What do you mean by min(x,t)?
$endgroup$
– AmaC
Mar 24 '17 at 13:34
$begingroup$
The minimum of $x$ and $t$.
$endgroup$
– Graham Kemp
Mar 24 '17 at 13:42
add a comment |
$begingroup$
It looks like :
$$begin{align}F(x,y) ~&=~ begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^yint_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^1 int_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t f(s,t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^xint_0^t 2(s+t)operatorname d soperatorname d t+
int_x^yint_0^x 2(s+t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^xint_0^t2(s+t)operatorname d soperatorname d t+
int_x^1 int_0^x 2(s+t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t 2(s+t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] x y^2+ x^2 y -x^3 & : 0 leq x leq y leq 1
\[4pt] x + x^2 -x^3 & : 0leq xleq 1, y> 1
\[4pt] y^3 & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
end{align}$$
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
$endgroup$
$begingroup$
for the fourth one I happen to know the answer is y^3 so those bounds do not work for that.
$endgroup$
– juniormath
Nov 7 '14 at 7:43
$begingroup$
same with the third one. The answer is x+x^2-x^3 and I am getting just x+x^2 with those bounds @GrahamKemp
$endgroup$
– juniormath
Nov 7 '14 at 7:53
$begingroup$
@Graham Kemp What do you mean by min(x,t)?
$endgroup$
– AmaC
Mar 24 '17 at 13:34
$begingroup$
The minimum of $x$ and $t$.
$endgroup$
– Graham Kemp
Mar 24 '17 at 13:42
add a comment |
$begingroup$
It looks like :
$$begin{align}F(x,y) ~&=~ begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^yint_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^1 int_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t f(s,t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^xint_0^t 2(s+t)operatorname d soperatorname d t+
int_x^yint_0^x 2(s+t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^xint_0^t2(s+t)operatorname d soperatorname d t+
int_x^1 int_0^x 2(s+t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t 2(s+t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] x y^2+ x^2 y -x^3 & : 0 leq x leq y leq 1
\[4pt] x + x^2 -x^3 & : 0leq xleq 1, y> 1
\[4pt] y^3 & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
end{align}$$
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
$endgroup$
It looks like :
$$begin{align}F(x,y) ~&=~ begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^yint_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^1 int_0^{min(x,t)} f(s,t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t f(s,t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] int_0^xint_0^t 2(s+t)operatorname d soperatorname d t+
int_x^yint_0^x 2(s+t)operatorname d soperatorname d t & : 0 leq x leq y leq 1
\[4pt] int_0^xint_0^t2(s+t)operatorname d soperatorname d t+
int_x^1 int_0^x 2(s+t)operatorname d soperatorname d t & : 0leq xleq 1, y> 1
\[4pt] int_0^y int_0^t 2(s+t)operatorname d soperatorname d t & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
\[2ex] &=~begin{cases}
0 & : x<0 lor y< 0
\[4pt] x y^2+ x^2 y -x^3 & : 0 leq x leq y leq 1
\[4pt] x + x^2 -x^3 & : 0leq xleq 1, y> 1
\[4pt] y^3 & : 0leq yleq 1, x> y
\[4pt] 1 & : x> 1 land y>1
end{cases}
end{align}$$
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
edited Mar 24 '17 at 13:57
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
answered Nov 7 '14 at 6:53
Graham KempGraham Kemp
85.7k43378
85.7k43378
$begingroup$
for the fourth one I happen to know the answer is y^3 so those bounds do not work for that.
$endgroup$
– juniormath
Nov 7 '14 at 7:43
$begingroup$
same with the third one. The answer is x+x^2-x^3 and I am getting just x+x^2 with those bounds @GrahamKemp
$endgroup$
– juniormath
Nov 7 '14 at 7:53
$begingroup$
@Graham Kemp What do you mean by min(x,t)?
$endgroup$
– AmaC
Mar 24 '17 at 13:34
$begingroup$
The minimum of $x$ and $t$.
$endgroup$
– Graham Kemp
Mar 24 '17 at 13:42
add a comment |
$begingroup$
for the fourth one I happen to know the answer is y^3 so those bounds do not work for that.
$endgroup$
– juniormath
Nov 7 '14 at 7:43
$begingroup$
same with the third one. The answer is x+x^2-x^3 and I am getting just x+x^2 with those bounds @GrahamKemp
$endgroup$
– juniormath
Nov 7 '14 at 7:53
$begingroup$
@Graham Kemp What do you mean by min(x,t)?
$endgroup$
– AmaC
Mar 24 '17 at 13:34
$begingroup$
The minimum of $x$ and $t$.
$endgroup$
– Graham Kemp
Mar 24 '17 at 13:42
$begingroup$
for the fourth one I happen to know the answer is y^3 so those bounds do not work for that.
$endgroup$
– juniormath
Nov 7 '14 at 7:43
$begingroup$
for the fourth one I happen to know the answer is y^3 so those bounds do not work for that.
$endgroup$
– juniormath
Nov 7 '14 at 7:43
$begingroup$
same with the third one. The answer is x+x^2-x^3 and I am getting just x+x^2 with those bounds @GrahamKemp
$endgroup$
– juniormath
Nov 7 '14 at 7:53
$begingroup$
same with the third one. The answer is x+x^2-x^3 and I am getting just x+x^2 with those bounds @GrahamKemp
$endgroup$
– juniormath
Nov 7 '14 at 7:53
$begingroup$
@Graham Kemp What do you mean by min(x,t)?
$endgroup$
– AmaC
Mar 24 '17 at 13:34
$begingroup$
@Graham Kemp What do you mean by min(x,t)?
$endgroup$
– AmaC
Mar 24 '17 at 13:34
$begingroup$
The minimum of $x$ and $t$.
$endgroup$
– Graham Kemp
Mar 24 '17 at 13:42
$begingroup$
The minimum of $x$ and $t$.
$endgroup$
– Graham Kemp
Mar 24 '17 at 13:42
add a comment |
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