Where is my flaw with the calculation of this cdf?
$begingroup$
I want to calculate the ratio distribution $X/Y$ of two continuous random variables $X$ and $Y$ with each having support $(0, infty)$
I was starting like that:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty f_Y(y) F_{Xmid Y}(zy mid y)dy$$
Now:
$$F_{Xmid Y}(zy mid y)=int_{0}^{zy} frac{f_{X,Y}(x,y)}{f_Y(y)}dx =frac{frac{dF_{X,Y}(zy,y)}{dy}}{f_Y(y)}$$
and hence:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty frac{dF_{X,Y}(zy,y)}{dy}dy=left[F(zy,y)right]_0^infty=1,$$
which of course is not correct. Anyone can see my mistake? Thank you very much in advance
probability integration probability-distributions
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
$endgroup$
add a comment |
$begingroup$
I want to calculate the ratio distribution $X/Y$ of two continuous random variables $X$ and $Y$ with each having support $(0, infty)$
I was starting like that:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty f_Y(y) F_{Xmid Y}(zy mid y)dy$$
Now:
$$F_{Xmid Y}(zy mid y)=int_{0}^{zy} frac{f_{X,Y}(x,y)}{f_Y(y)}dx =frac{frac{dF_{X,Y}(zy,y)}{dy}}{f_Y(y)}$$
and hence:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty frac{dF_{X,Y}(zy,y)}{dy}dy=left[F(zy,y)right]_0^infty=1,$$
which of course is not correct. Anyone can see my mistake? Thank you very much in advance
probability integration probability-distributions
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
$endgroup$
$begingroup$
I do not understand the second equality of the second line.
$endgroup$
– drhab
Nov 30 '18 at 13:46
$begingroup$
Then maybe there is my flaw... :-) The idea is that $frac{d}{dy}F(x,y)=frac{d}{dy}int_0^y int_0^x f(x,y)dx dy= int_0^x f(x,y)dx$ but here it seems wrong....
$endgroup$
– J.Doe
Nov 30 '18 at 14:08
$begingroup$
But I do not have a joint pdf but only a joint cdf and the pdf of $X$ and $Y$... Any idea how to get past that?
$endgroup$
– J.Doe
Nov 30 '18 at 14:14
add a comment |
$begingroup$
I want to calculate the ratio distribution $X/Y$ of two continuous random variables $X$ and $Y$ with each having support $(0, infty)$
I was starting like that:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty f_Y(y) F_{Xmid Y}(zy mid y)dy$$
Now:
$$F_{Xmid Y}(zy mid y)=int_{0}^{zy} frac{f_{X,Y}(x,y)}{f_Y(y)}dx =frac{frac{dF_{X,Y}(zy,y)}{dy}}{f_Y(y)}$$
and hence:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty frac{dF_{X,Y}(zy,y)}{dy}dy=left[F(zy,y)right]_0^infty=1,$$
which of course is not correct. Anyone can see my mistake? Thank you very much in advance
probability integration probability-distributions
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
$endgroup$
I want to calculate the ratio distribution $X/Y$ of two continuous random variables $X$ and $Y$ with each having support $(0, infty)$
I was starting like that:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty f_Y(y) F_{Xmid Y}(zy mid y)dy$$
Now:
$$F_{Xmid Y}(zy mid y)=int_{0}^{zy} frac{f_{X,Y}(x,y)}{f_Y(y)}dx =frac{frac{dF_{X,Y}(zy,y)}{dy}}{f_Y(y)}$$
and hence:
$$mathbb Pleft(frac{X}{Y}leq zright)=int_{0}^infty frac{dF_{X,Y}(zy,y)}{dy}dy=left[F(zy,y)right]_0^infty=1,$$
which of course is not correct. Anyone can see my mistake? Thank you very much in advance
probability integration probability-distributions
probability integration probability-distributions
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
asked Nov 30 '18 at 12:11
J.DoeJ.Doe
19910
19910
$begingroup$
I do not understand the second equality of the second line.
$endgroup$
– drhab
Nov 30 '18 at 13:46
$begingroup$
Then maybe there is my flaw... :-) The idea is that $frac{d}{dy}F(x,y)=frac{d}{dy}int_0^y int_0^x f(x,y)dx dy= int_0^x f(x,y)dx$ but here it seems wrong....
$endgroup$
– J.Doe
Nov 30 '18 at 14:08
$begingroup$
But I do not have a joint pdf but only a joint cdf and the pdf of $X$ and $Y$... Any idea how to get past that?
$endgroup$
– J.Doe
Nov 30 '18 at 14:14
add a comment |
$begingroup$
I do not understand the second equality of the second line.
$endgroup$
– drhab
Nov 30 '18 at 13:46
$begingroup$
Then maybe there is my flaw... :-) The idea is that $frac{d}{dy}F(x,y)=frac{d}{dy}int_0^y int_0^x f(x,y)dx dy= int_0^x f(x,y)dx$ but here it seems wrong....
$endgroup$
– J.Doe
Nov 30 '18 at 14:08
$begingroup$
But I do not have a joint pdf but only a joint cdf and the pdf of $X$ and $Y$... Any idea how to get past that?
$endgroup$
– J.Doe
Nov 30 '18 at 14:14
$begingroup$
I do not understand the second equality of the second line.
$endgroup$
– drhab
Nov 30 '18 at 13:46
$begingroup$
I do not understand the second equality of the second line.
$endgroup$
– drhab
Nov 30 '18 at 13:46
$begingroup$
Then maybe there is my flaw... :-) The idea is that $frac{d}{dy}F(x,y)=frac{d}{dy}int_0^y int_0^x f(x,y)dx dy= int_0^x f(x,y)dx$ but here it seems wrong....
$endgroup$
– J.Doe
Nov 30 '18 at 14:08
$begingroup$
Then maybe there is my flaw... :-) The idea is that $frac{d}{dy}F(x,y)=frac{d}{dy}int_0^y int_0^x f(x,y)dx dy= int_0^x f(x,y)dx$ but here it seems wrong....
$endgroup$
– J.Doe
Nov 30 '18 at 14:08
$begingroup$
But I do not have a joint pdf but only a joint cdf and the pdf of $X$ and $Y$... Any idea how to get past that?
$endgroup$
– J.Doe
Nov 30 '18 at 14:14
$begingroup$
But I do not have a joint pdf but only a joint cdf and the pdf of $X$ and $Y$... Any idea how to get past that?
$endgroup$
– J.Doe
Nov 30 '18 at 14:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Putting together your first equation and the first equality in the second, you are esentially getting
$$ p=P( X/Y le z)= int_0^infty int_0^u f_{X,Y}(x,y) , dx , dy, hspace{1cm} u=u(y)=zy tag{1}$$
which is right, of course (there was no need to use a conditional for that, though).
Your problem in what follows is that you are mixing total derivatives with partial derivatives.
When we write (in general) the formula
$$ frac{d F_{X,Y}(x,y)}{dy } = int_{-infty}^x f_{X,Y}(x',y) dx' tag{2}$$
we are implicity assuming that the derivative is done by varying $y$ and keeping $x$ constant, i.e., it's actually a partial derivative. When the variables have some arbitrary functional dependence, we should be more careful and write that partial derivative explicitly:
$$ frac{partial F_{X,Y}(u,v)}{partial v } = int_{-infty}^u f_{X,Y}(u',v) , du' tag{3}$$
In our case, plugging $(3)$ into $(1)$ we get
$$p= int_0^infty frac{partial F_{X,Y}(u,y)}{partial y } dy tag{4} $$
with $u=u(y)=zy $. Now, the integrand is not a total derivative, hence you cannot apply the fundamental theorem of calculus directly. The relation is
$$ frac{d F_{X,Y}(u,y)}{d y } = frac{partial F_{X,Y}(u,y)}{partial u } frac{du}{dy}+
frac{partial F_{X,Y}(u,y)}{partial y } = frac{partial F_{X,Y}(u,y)}{partial u } z +
frac{partial F_{X,Y}(u,y)}{partial y } tag{5}$$
You could isolate from this equation the integral in $(4)$ and replace it there... but you will get something similar (probably not simpler) than $(1)$.
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
$endgroup$
$begingroup$
Okay thank you very much, I think I understand my mistake now. So if I have given $F(x,y)=exp(-frac{1}{max(x,y)})$ and I want to calculate the ratio distribution I first say $F(zy,y)=exp(-frac{1}{ymax(z,1)})$. So if $zgeq 1$ the probability $p$ is equal to: $p =int_0^infty exp(-frac{1}{zy})(zy)^{-2}z-exp(-frac{1}{zy})(zy)^{-2}z dy=0$ and for $z<1$ we have that the derivative with respect to $u$ is zero and hence it reduces to $p= int_0^infty exp(-frac{1}{y})(y)^{-2} dy=1$, right? In other words? if $max(zy,y)=zy$ we treat the cdf function like a constant and
$endgroup$
– J.Doe
Dec 2 '18 at 1:28
$begingroup$
hence the derivative with respect to $y$ is zero and hence our probability is zero and in case $max(zy,y)=y$ we just calculate the derivative with respect to $y$, right?
$endgroup$
– J.Doe
Dec 2 '18 at 1:30
$begingroup$
mmm unless I'm confused, if $F(x,y)=exp(-frac{1}{max(x,y)})$ then the joint density is zero everywhere, except on the (non derivable) region $x=y$ , i.e., we have a degenerate joint density when $X=Y$ - and hence the problem is trivial (but should not be attacked by this approach)...
$endgroup$
– leonbloy
Dec 2 '18 at 2:33
$begingroup$
Okay, maybe I need to be more precise; If the joint cdf is given by $F(x,y)=exp(-frac{1}{max(x,y})$, then the joint cdf is only determined by the bigger of the two values. We would get such a joint cdf e.g. if we have a random variable $R$ and $X:=R$ and $Y:=R$; Then the ratio would be a constant (namely 1) and hence it really does not make sense to consider it like that; What I actually have $F(x,y)=exp(-(f(x)+g(y)+sum_{k=1}^nfrac{1}{a_k max(x,y}))$; Then we could see it like that: We have random variables $R_1,...,R_t$ and subsets
$endgroup$
– J.Doe
Dec 2 '18 at 13:57
$begingroup$
$R_X subseteq {R_1,...,R_t}$ and $R_Y subseteq {R_1,...,R_t}$ and $X=maxlimits_{i in R_X}c_{xi}R_i$ and $Y=maxlimits_{i in R_Y}c_{yi}R_i$ and then we could divide the two sets $R_X$ and $R_Y$ in random variables that are only used to define $X$, random varialbes that are only used to define $Y$ and random variables that used for both, in $X$ and $Y$ and we would end up e.g. with such a joint cdf
$endgroup$
– J.Doe
Dec 2 '18 at 14:01
add a comment |
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$begingroup$
Putting together your first equation and the first equality in the second, you are esentially getting
$$ p=P( X/Y le z)= int_0^infty int_0^u f_{X,Y}(x,y) , dx , dy, hspace{1cm} u=u(y)=zy tag{1}$$
which is right, of course (there was no need to use a conditional for that, though).
Your problem in what follows is that you are mixing total derivatives with partial derivatives.
When we write (in general) the formula
$$ frac{d F_{X,Y}(x,y)}{dy } = int_{-infty}^x f_{X,Y}(x',y) dx' tag{2}$$
we are implicity assuming that the derivative is done by varying $y$ and keeping $x$ constant, i.e., it's actually a partial derivative. When the variables have some arbitrary functional dependence, we should be more careful and write that partial derivative explicitly:
$$ frac{partial F_{X,Y}(u,v)}{partial v } = int_{-infty}^u f_{X,Y}(u',v) , du' tag{3}$$
In our case, plugging $(3)$ into $(1)$ we get
$$p= int_0^infty frac{partial F_{X,Y}(u,y)}{partial y } dy tag{4} $$
with $u=u(y)=zy $. Now, the integrand is not a total derivative, hence you cannot apply the fundamental theorem of calculus directly. The relation is
$$ frac{d F_{X,Y}(u,y)}{d y } = frac{partial F_{X,Y}(u,y)}{partial u } frac{du}{dy}+
frac{partial F_{X,Y}(u,y)}{partial y } = frac{partial F_{X,Y}(u,y)}{partial u } z +
frac{partial F_{X,Y}(u,y)}{partial y } tag{5}$$
You could isolate from this equation the integral in $(4)$ and replace it there... but you will get something similar (probably not simpler) than $(1)$.
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
$endgroup$
$begingroup$
Okay thank you very much, I think I understand my mistake now. So if I have given $F(x,y)=exp(-frac{1}{max(x,y)})$ and I want to calculate the ratio distribution I first say $F(zy,y)=exp(-frac{1}{ymax(z,1)})$. So if $zgeq 1$ the probability $p$ is equal to: $p =int_0^infty exp(-frac{1}{zy})(zy)^{-2}z-exp(-frac{1}{zy})(zy)^{-2}z dy=0$ and for $z<1$ we have that the derivative with respect to $u$ is zero and hence it reduces to $p= int_0^infty exp(-frac{1}{y})(y)^{-2} dy=1$, right? In other words? if $max(zy,y)=zy$ we treat the cdf function like a constant and
$endgroup$
– J.Doe
Dec 2 '18 at 1:28
$begingroup$
hence the derivative with respect to $y$ is zero and hence our probability is zero and in case $max(zy,y)=y$ we just calculate the derivative with respect to $y$, right?
$endgroup$
– J.Doe
Dec 2 '18 at 1:30
$begingroup$
mmm unless I'm confused, if $F(x,y)=exp(-frac{1}{max(x,y)})$ then the joint density is zero everywhere, except on the (non derivable) region $x=y$ , i.e., we have a degenerate joint density when $X=Y$ - and hence the problem is trivial (but should not be attacked by this approach)...
$endgroup$
– leonbloy
Dec 2 '18 at 2:33
$begingroup$
Okay, maybe I need to be more precise; If the joint cdf is given by $F(x,y)=exp(-frac{1}{max(x,y})$, then the joint cdf is only determined by the bigger of the two values. We would get such a joint cdf e.g. if we have a random variable $R$ and $X:=R$ and $Y:=R$; Then the ratio would be a constant (namely 1) and hence it really does not make sense to consider it like that; What I actually have $F(x,y)=exp(-(f(x)+g(y)+sum_{k=1}^nfrac{1}{a_k max(x,y}))$; Then we could see it like that: We have random variables $R_1,...,R_t$ and subsets
$endgroup$
– J.Doe
Dec 2 '18 at 13:57
$begingroup$
$R_X subseteq {R_1,...,R_t}$ and $R_Y subseteq {R_1,...,R_t}$ and $X=maxlimits_{i in R_X}c_{xi}R_i$ and $Y=maxlimits_{i in R_Y}c_{yi}R_i$ and then we could divide the two sets $R_X$ and $R_Y$ in random variables that are only used to define $X$, random varialbes that are only used to define $Y$ and random variables that used for both, in $X$ and $Y$ and we would end up e.g. with such a joint cdf
$endgroup$
– J.Doe
Dec 2 '18 at 14:01
add a comment |
$begingroup$
Putting together your first equation and the first equality in the second, you are esentially getting
$$ p=P( X/Y le z)= int_0^infty int_0^u f_{X,Y}(x,y) , dx , dy, hspace{1cm} u=u(y)=zy tag{1}$$
which is right, of course (there was no need to use a conditional for that, though).
Your problem in what follows is that you are mixing total derivatives with partial derivatives.
When we write (in general) the formula
$$ frac{d F_{X,Y}(x,y)}{dy } = int_{-infty}^x f_{X,Y}(x',y) dx' tag{2}$$
we are implicity assuming that the derivative is done by varying $y$ and keeping $x$ constant, i.e., it's actually a partial derivative. When the variables have some arbitrary functional dependence, we should be more careful and write that partial derivative explicitly:
$$ frac{partial F_{X,Y}(u,v)}{partial v } = int_{-infty}^u f_{X,Y}(u',v) , du' tag{3}$$
In our case, plugging $(3)$ into $(1)$ we get
$$p= int_0^infty frac{partial F_{X,Y}(u,y)}{partial y } dy tag{4} $$
with $u=u(y)=zy $. Now, the integrand is not a total derivative, hence you cannot apply the fundamental theorem of calculus directly. The relation is
$$ frac{d F_{X,Y}(u,y)}{d y } = frac{partial F_{X,Y}(u,y)}{partial u } frac{du}{dy}+
frac{partial F_{X,Y}(u,y)}{partial y } = frac{partial F_{X,Y}(u,y)}{partial u } z +
frac{partial F_{X,Y}(u,y)}{partial y } tag{5}$$
You could isolate from this equation the integral in $(4)$ and replace it there... but you will get something similar (probably not simpler) than $(1)$.
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
$endgroup$
$begingroup$
Okay thank you very much, I think I understand my mistake now. So if I have given $F(x,y)=exp(-frac{1}{max(x,y)})$ and I want to calculate the ratio distribution I first say $F(zy,y)=exp(-frac{1}{ymax(z,1)})$. So if $zgeq 1$ the probability $p$ is equal to: $p =int_0^infty exp(-frac{1}{zy})(zy)^{-2}z-exp(-frac{1}{zy})(zy)^{-2}z dy=0$ and for $z<1$ we have that the derivative with respect to $u$ is zero and hence it reduces to $p= int_0^infty exp(-frac{1}{y})(y)^{-2} dy=1$, right? In other words? if $max(zy,y)=zy$ we treat the cdf function like a constant and
$endgroup$
– J.Doe
Dec 2 '18 at 1:28
$begingroup$
hence the derivative with respect to $y$ is zero and hence our probability is zero and in case $max(zy,y)=y$ we just calculate the derivative with respect to $y$, right?
$endgroup$
– J.Doe
Dec 2 '18 at 1:30
$begingroup$
mmm unless I'm confused, if $F(x,y)=exp(-frac{1}{max(x,y)})$ then the joint density is zero everywhere, except on the (non derivable) region $x=y$ , i.e., we have a degenerate joint density when $X=Y$ - and hence the problem is trivial (but should not be attacked by this approach)...
$endgroup$
– leonbloy
Dec 2 '18 at 2:33
$begingroup$
Okay, maybe I need to be more precise; If the joint cdf is given by $F(x,y)=exp(-frac{1}{max(x,y})$, then the joint cdf is only determined by the bigger of the two values. We would get such a joint cdf e.g. if we have a random variable $R$ and $X:=R$ and $Y:=R$; Then the ratio would be a constant (namely 1) and hence it really does not make sense to consider it like that; What I actually have $F(x,y)=exp(-(f(x)+g(y)+sum_{k=1}^nfrac{1}{a_k max(x,y}))$; Then we could see it like that: We have random variables $R_1,...,R_t$ and subsets
$endgroup$
– J.Doe
Dec 2 '18 at 13:57
$begingroup$
$R_X subseteq {R_1,...,R_t}$ and $R_Y subseteq {R_1,...,R_t}$ and $X=maxlimits_{i in R_X}c_{xi}R_i$ and $Y=maxlimits_{i in R_Y}c_{yi}R_i$ and then we could divide the two sets $R_X$ and $R_Y$ in random variables that are only used to define $X$, random varialbes that are only used to define $Y$ and random variables that used for both, in $X$ and $Y$ and we would end up e.g. with such a joint cdf
$endgroup$
– J.Doe
Dec 2 '18 at 14:01
add a comment |
$begingroup$
Putting together your first equation and the first equality in the second, you are esentially getting
$$ p=P( X/Y le z)= int_0^infty int_0^u f_{X,Y}(x,y) , dx , dy, hspace{1cm} u=u(y)=zy tag{1}$$
which is right, of course (there was no need to use a conditional for that, though).
Your problem in what follows is that you are mixing total derivatives with partial derivatives.
When we write (in general) the formula
$$ frac{d F_{X,Y}(x,y)}{dy } = int_{-infty}^x f_{X,Y}(x',y) dx' tag{2}$$
we are implicity assuming that the derivative is done by varying $y$ and keeping $x$ constant, i.e., it's actually a partial derivative. When the variables have some arbitrary functional dependence, we should be more careful and write that partial derivative explicitly:
$$ frac{partial F_{X,Y}(u,v)}{partial v } = int_{-infty}^u f_{X,Y}(u',v) , du' tag{3}$$
In our case, plugging $(3)$ into $(1)$ we get
$$p= int_0^infty frac{partial F_{X,Y}(u,y)}{partial y } dy tag{4} $$
with $u=u(y)=zy $. Now, the integrand is not a total derivative, hence you cannot apply the fundamental theorem of calculus directly. The relation is
$$ frac{d F_{X,Y}(u,y)}{d y } = frac{partial F_{X,Y}(u,y)}{partial u } frac{du}{dy}+
frac{partial F_{X,Y}(u,y)}{partial y } = frac{partial F_{X,Y}(u,y)}{partial u } z +
frac{partial F_{X,Y}(u,y)}{partial y } tag{5}$$
You could isolate from this equation the integral in $(4)$ and replace it there... but you will get something similar (probably not simpler) than $(1)$.
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
$endgroup$
Putting together your first equation and the first equality in the second, you are esentially getting
$$ p=P( X/Y le z)= int_0^infty int_0^u f_{X,Y}(x,y) , dx , dy, hspace{1cm} u=u(y)=zy tag{1}$$
which is right, of course (there was no need to use a conditional for that, though).
Your problem in what follows is that you are mixing total derivatives with partial derivatives.
When we write (in general) the formula
$$ frac{d F_{X,Y}(x,y)}{dy } = int_{-infty}^x f_{X,Y}(x',y) dx' tag{2}$$
we are implicity assuming that the derivative is done by varying $y$ and keeping $x$ constant, i.e., it's actually a partial derivative. When the variables have some arbitrary functional dependence, we should be more careful and write that partial derivative explicitly:
$$ frac{partial F_{X,Y}(u,v)}{partial v } = int_{-infty}^u f_{X,Y}(u',v) , du' tag{3}$$
In our case, plugging $(3)$ into $(1)$ we get
$$p= int_0^infty frac{partial F_{X,Y}(u,y)}{partial y } dy tag{4} $$
with $u=u(y)=zy $. Now, the integrand is not a total derivative, hence you cannot apply the fundamental theorem of calculus directly. The relation is
$$ frac{d F_{X,Y}(u,y)}{d y } = frac{partial F_{X,Y}(u,y)}{partial u } frac{du}{dy}+
frac{partial F_{X,Y}(u,y)}{partial y } = frac{partial F_{X,Y}(u,y)}{partial u } z +
frac{partial F_{X,Y}(u,y)}{partial y } tag{5}$$
You could isolate from this equation the integral in $(4)$ and replace it there... but you will get something similar (probably not simpler) than $(1)$.
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
edited Nov 30 '18 at 18:02
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
answered Nov 30 '18 at 16:41
leonbloyleonbloy
40.5k645107
40.5k645107
$begingroup$
Okay thank you very much, I think I understand my mistake now. So if I have given $F(x,y)=exp(-frac{1}{max(x,y)})$ and I want to calculate the ratio distribution I first say $F(zy,y)=exp(-frac{1}{ymax(z,1)})$. So if $zgeq 1$ the probability $p$ is equal to: $p =int_0^infty exp(-frac{1}{zy})(zy)^{-2}z-exp(-frac{1}{zy})(zy)^{-2}z dy=0$ and for $z<1$ we have that the derivative with respect to $u$ is zero and hence it reduces to $p= int_0^infty exp(-frac{1}{y})(y)^{-2} dy=1$, right? In other words? if $max(zy,y)=zy$ we treat the cdf function like a constant and
$endgroup$
– J.Doe
Dec 2 '18 at 1:28
$begingroup$
hence the derivative with respect to $y$ is zero and hence our probability is zero and in case $max(zy,y)=y$ we just calculate the derivative with respect to $y$, right?
$endgroup$
– J.Doe
Dec 2 '18 at 1:30
$begingroup$
mmm unless I'm confused, if $F(x,y)=exp(-frac{1}{max(x,y)})$ then the joint density is zero everywhere, except on the (non derivable) region $x=y$ , i.e., we have a degenerate joint density when $X=Y$ - and hence the problem is trivial (but should not be attacked by this approach)...
$endgroup$
– leonbloy
Dec 2 '18 at 2:33
$begingroup$
Okay, maybe I need to be more precise; If the joint cdf is given by $F(x,y)=exp(-frac{1}{max(x,y})$, then the joint cdf is only determined by the bigger of the two values. We would get such a joint cdf e.g. if we have a random variable $R$ and $X:=R$ and $Y:=R$; Then the ratio would be a constant (namely 1) and hence it really does not make sense to consider it like that; What I actually have $F(x,y)=exp(-(f(x)+g(y)+sum_{k=1}^nfrac{1}{a_k max(x,y}))$; Then we could see it like that: We have random variables $R_1,...,R_t$ and subsets
$endgroup$
– J.Doe
Dec 2 '18 at 13:57
$begingroup$
$R_X subseteq {R_1,...,R_t}$ and $R_Y subseteq {R_1,...,R_t}$ and $X=maxlimits_{i in R_X}c_{xi}R_i$ and $Y=maxlimits_{i in R_Y}c_{yi}R_i$ and then we could divide the two sets $R_X$ and $R_Y$ in random variables that are only used to define $X$, random varialbes that are only used to define $Y$ and random variables that used for both, in $X$ and $Y$ and we would end up e.g. with such a joint cdf
$endgroup$
– J.Doe
Dec 2 '18 at 14:01
add a comment |
$begingroup$
Okay thank you very much, I think I understand my mistake now. So if I have given $F(x,y)=exp(-frac{1}{max(x,y)})$ and I want to calculate the ratio distribution I first say $F(zy,y)=exp(-frac{1}{ymax(z,1)})$. So if $zgeq 1$ the probability $p$ is equal to: $p =int_0^infty exp(-frac{1}{zy})(zy)^{-2}z-exp(-frac{1}{zy})(zy)^{-2}z dy=0$ and for $z<1$ we have that the derivative with respect to $u$ is zero and hence it reduces to $p= int_0^infty exp(-frac{1}{y})(y)^{-2} dy=1$, right? In other words? if $max(zy,y)=zy$ we treat the cdf function like a constant and
$endgroup$
– J.Doe
Dec 2 '18 at 1:28
$begingroup$
hence the derivative with respect to $y$ is zero and hence our probability is zero and in case $max(zy,y)=y$ we just calculate the derivative with respect to $y$, right?
$endgroup$
– J.Doe
Dec 2 '18 at 1:30
$begingroup$
mmm unless I'm confused, if $F(x,y)=exp(-frac{1}{max(x,y)})$ then the joint density is zero everywhere, except on the (non derivable) region $x=y$ , i.e., we have a degenerate joint density when $X=Y$ - and hence the problem is trivial (but should not be attacked by this approach)...
$endgroup$
– leonbloy
Dec 2 '18 at 2:33
$begingroup$
Okay, maybe I need to be more precise; If the joint cdf is given by $F(x,y)=exp(-frac{1}{max(x,y})$, then the joint cdf is only determined by the bigger of the two values. We would get such a joint cdf e.g. if we have a random variable $R$ and $X:=R$ and $Y:=R$; Then the ratio would be a constant (namely 1) and hence it really does not make sense to consider it like that; What I actually have $F(x,y)=exp(-(f(x)+g(y)+sum_{k=1}^nfrac{1}{a_k max(x,y}))$; Then we could see it like that: We have random variables $R_1,...,R_t$ and subsets
$endgroup$
– J.Doe
Dec 2 '18 at 13:57
$begingroup$
$R_X subseteq {R_1,...,R_t}$ and $R_Y subseteq {R_1,...,R_t}$ and $X=maxlimits_{i in R_X}c_{xi}R_i$ and $Y=maxlimits_{i in R_Y}c_{yi}R_i$ and then we could divide the two sets $R_X$ and $R_Y$ in random variables that are only used to define $X$, random varialbes that are only used to define $Y$ and random variables that used for both, in $X$ and $Y$ and we would end up e.g. with such a joint cdf
$endgroup$
– J.Doe
Dec 2 '18 at 14:01
$begingroup$
Okay thank you very much, I think I understand my mistake now. So if I have given $F(x,y)=exp(-frac{1}{max(x,y)})$ and I want to calculate the ratio distribution I first say $F(zy,y)=exp(-frac{1}{ymax(z,1)})$. So if $zgeq 1$ the probability $p$ is equal to: $p =int_0^infty exp(-frac{1}{zy})(zy)^{-2}z-exp(-frac{1}{zy})(zy)^{-2}z dy=0$ and for $z<1$ we have that the derivative with respect to $u$ is zero and hence it reduces to $p= int_0^infty exp(-frac{1}{y})(y)^{-2} dy=1$, right? In other words? if $max(zy,y)=zy$ we treat the cdf function like a constant and
$endgroup$
– J.Doe
Dec 2 '18 at 1:28
$begingroup$
Okay thank you very much, I think I understand my mistake now. So if I have given $F(x,y)=exp(-frac{1}{max(x,y)})$ and I want to calculate the ratio distribution I first say $F(zy,y)=exp(-frac{1}{ymax(z,1)})$. So if $zgeq 1$ the probability $p$ is equal to: $p =int_0^infty exp(-frac{1}{zy})(zy)^{-2}z-exp(-frac{1}{zy})(zy)^{-2}z dy=0$ and for $z<1$ we have that the derivative with respect to $u$ is zero and hence it reduces to $p= int_0^infty exp(-frac{1}{y})(y)^{-2} dy=1$, right? In other words? if $max(zy,y)=zy$ we treat the cdf function like a constant and
$endgroup$
– J.Doe
Dec 2 '18 at 1:28
$begingroup$
hence the derivative with respect to $y$ is zero and hence our probability is zero and in case $max(zy,y)=y$ we just calculate the derivative with respect to $y$, right?
$endgroup$
– J.Doe
Dec 2 '18 at 1:30
$begingroup$
hence the derivative with respect to $y$ is zero and hence our probability is zero and in case $max(zy,y)=y$ we just calculate the derivative with respect to $y$, right?
$endgroup$
– J.Doe
Dec 2 '18 at 1:30
$begingroup$
mmm unless I'm confused, if $F(x,y)=exp(-frac{1}{max(x,y)})$ then the joint density is zero everywhere, except on the (non derivable) region $x=y$ , i.e., we have a degenerate joint density when $X=Y$ - and hence the problem is trivial (but should not be attacked by this approach)...
$endgroup$
– leonbloy
Dec 2 '18 at 2:33
$begingroup$
mmm unless I'm confused, if $F(x,y)=exp(-frac{1}{max(x,y)})$ then the joint density is zero everywhere, except on the (non derivable) region $x=y$ , i.e., we have a degenerate joint density when $X=Y$ - and hence the problem is trivial (but should not be attacked by this approach)...
$endgroup$
– leonbloy
Dec 2 '18 at 2:33
$begingroup$
Okay, maybe I need to be more precise; If the joint cdf is given by $F(x,y)=exp(-frac{1}{max(x,y})$, then the joint cdf is only determined by the bigger of the two values. We would get such a joint cdf e.g. if we have a random variable $R$ and $X:=R$ and $Y:=R$; Then the ratio would be a constant (namely 1) and hence it really does not make sense to consider it like that; What I actually have $F(x,y)=exp(-(f(x)+g(y)+sum_{k=1}^nfrac{1}{a_k max(x,y}))$; Then we could see it like that: We have random variables $R_1,...,R_t$ and subsets
$endgroup$
– J.Doe
Dec 2 '18 at 13:57
$begingroup$
Okay, maybe I need to be more precise; If the joint cdf is given by $F(x,y)=exp(-frac{1}{max(x,y})$, then the joint cdf is only determined by the bigger of the two values. We would get such a joint cdf e.g. if we have a random variable $R$ and $X:=R$ and $Y:=R$; Then the ratio would be a constant (namely 1) and hence it really does not make sense to consider it like that; What I actually have $F(x,y)=exp(-(f(x)+g(y)+sum_{k=1}^nfrac{1}{a_k max(x,y}))$; Then we could see it like that: We have random variables $R_1,...,R_t$ and subsets
$endgroup$
– J.Doe
Dec 2 '18 at 13:57
$begingroup$
$R_X subseteq {R_1,...,R_t}$ and $R_Y subseteq {R_1,...,R_t}$ and $X=maxlimits_{i in R_X}c_{xi}R_i$ and $Y=maxlimits_{i in R_Y}c_{yi}R_i$ and then we could divide the two sets $R_X$ and $R_Y$ in random variables that are only used to define $X$, random varialbes that are only used to define $Y$ and random variables that used for both, in $X$ and $Y$ and we would end up e.g. with such a joint cdf
$endgroup$
– J.Doe
Dec 2 '18 at 14:01
$begingroup$
$R_X subseteq {R_1,...,R_t}$ and $R_Y subseteq {R_1,...,R_t}$ and $X=maxlimits_{i in R_X}c_{xi}R_i$ and $Y=maxlimits_{i in R_Y}c_{yi}R_i$ and then we could divide the two sets $R_X$ and $R_Y$ in random variables that are only used to define $X$, random varialbes that are only used to define $Y$ and random variables that used for both, in $X$ and $Y$ and we would end up e.g. with such a joint cdf
$endgroup$
– J.Doe
Dec 2 '18 at 14:01
add a comment |
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$begingroup$
I do not understand the second equality of the second line.
$endgroup$
– drhab
Nov 30 '18 at 13:46
$begingroup$
Then maybe there is my flaw... :-) The idea is that $frac{d}{dy}F(x,y)=frac{d}{dy}int_0^y int_0^x f(x,y)dx dy= int_0^x f(x,y)dx$ but here it seems wrong....
$endgroup$
– J.Doe
Nov 30 '18 at 14:08
$begingroup$
But I do not have a joint pdf but only a joint cdf and the pdf of $X$ and $Y$... Any idea how to get past that?
$endgroup$
– J.Doe
Nov 30 '18 at 14:14