Where is my flaw with the calculation of this cdf?












4












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















4












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














4












4








4


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















3












$begingroup$

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 $(1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020021%2fwhere-is-my-flaw-with-the-calculation-of-this-cdf%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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 $(1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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 $(1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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 $(1)$.






share|cite|improve this answer











$endgroup$



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 $(1)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 18:02

























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




















  • $begingroup$
    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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020021%2fwhere-is-my-flaw-with-the-calculation-of-this-cdf%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more