$P(x>0|x+y>0);$ of two independently distributed Gaussian distributions?












1












$begingroup$



I am given two independently distributed Gaussian distributions, $X$ and $Y$, with unknown means and variances. You randomly choose an $x$ and $y$ from these two distributions respectively. Given that $x+y>0$, what is the probability that $x>0$ in terms of the unknown parameters?




Suppose $Xsim N(mu_1, sigma_1^2)$ and $Ysim N(mu_2, sigma_2^2)$.



I know that $X+Ysim N(mu_1+mu_2, sigma_1^2+sigma_2^2)$. I considered using Bayes’ theorem.



$$P(x>0|x+y>0)=frac{P(x>0)P(x+y>0|x>0)}{P(x+y>0)}.$$



Since I know the distributions of $X$ and $X+Y$, I know that the forms of $P(x>0)$ and $P(x+y>0)$ are that of integrating the Gaussian. However, I am unsure of the form of $P(x+y>0|x>0)$. I feel that this is much easier to calculate when observing the distribution of $X+Y$ on a 2D-plane, but I am not exactly sure how to write it in terms of the unknowns. Could someone please shed some insight on this? Thanks!



Edit: this was meant to be a conceptual question I believe, and should be left as a combination of integrals.






statistics probability-distributions normal-distribution conditional-probability bayes-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    One insight is that if these are supposed to be i.i.d., you have four parameters where you should have only two, since $mu_1=mu_2$ and $sigma_1=sigma_2$.
    $endgroup$
    – C Monsour
    Dec 17 '18 at 0:10










  • $begingroup$
    @CMonsour oh gosh sorry it should just be independently distributed, not i.i.d!
    $endgroup$
    – user107224
    Dec 17 '18 at 0:12






  • 1




    $begingroup$
    Even covariances are not enough in general, you are rather after formulas similar to $$P(X+Y>0,X>0)=int_0^inftyint_{-x}^infty f_Y(y)dyf_X(x)dx$$ But are you sure you need to solve the question for any values of the $mu_i$s and $sigma^2_i$s? 'Cause the answer shall be a nightmare to write down, even if rather simple conceptually.
    $endgroup$
    – Did
    Dec 17 '18 at 0:27








  • 1




    $begingroup$
    Ah, then you already have $P(X+Y>0,X>0)$, and $$P(X>0)=int_0^infty f_X(x)dx$$ and since you know $f_X$ and $f_Y$, you are done.
    $endgroup$
    – Did
    Dec 17 '18 at 0:33






  • 2




    $begingroup$
    Because ${X+Y>0,X>0}={X>0,Y>-X}$.
    $endgroup$
    – Did
    Dec 17 '18 at 0:37
















1












$begingroup$



I am given two independently distributed Gaussian distributions, $X$ and $Y$, with unknown means and variances. You randomly choose an $x$ and $y$ from these two distributions respectively. Given that $x+y>0$, what is the probability that $x>0$ in terms of the unknown parameters?




Suppose $Xsim N(mu_1, sigma_1^2)$ and $Ysim N(mu_2, sigma_2^2)$.



I know that $X+Ysim N(mu_1+mu_2, sigma_1^2+sigma_2^2)$. I considered using Bayes’ theorem.



$$P(x>0|x+y>0)=frac{P(x>0)P(x+y>0|x>0)}{P(x+y>0)}.$$



Since I know the distributions of $X$ and $X+Y$, I know that the forms of $P(x>0)$ and $P(x+y>0)$ are that of integrating the Gaussian. However, I am unsure of the form of $P(x+y>0|x>0)$. I feel that this is much easier to calculate when observing the distribution of $X+Y$ on a 2D-plane, but I am not exactly sure how to write it in terms of the unknowns. Could someone please shed some insight on this? Thanks!



Edit: this was meant to be a conceptual question I believe, and should be left as a combination of integrals.






statistics probability-distributions normal-distribution conditional-probability bayes-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    One insight is that if these are supposed to be i.i.d., you have four parameters where you should have only two, since $mu_1=mu_2$ and $sigma_1=sigma_2$.
    $endgroup$
    – C Monsour
    Dec 17 '18 at 0:10










  • $begingroup$
    @CMonsour oh gosh sorry it should just be independently distributed, not i.i.d!
    $endgroup$
    – user107224
    Dec 17 '18 at 0:12






  • 1




    $begingroup$
    Even covariances are not enough in general, you are rather after formulas similar to $$P(X+Y>0,X>0)=int_0^inftyint_{-x}^infty f_Y(y)dyf_X(x)dx$$ But are you sure you need to solve the question for any values of the $mu_i$s and $sigma^2_i$s? 'Cause the answer shall be a nightmare to write down, even if rather simple conceptually.
    $endgroup$
    – Did
    Dec 17 '18 at 0:27








  • 1




    $begingroup$
    Ah, then you already have $P(X+Y>0,X>0)$, and $$P(X>0)=int_0^infty f_X(x)dx$$ and since you know $f_X$ and $f_Y$, you are done.
    $endgroup$
    – Did
    Dec 17 '18 at 0:33






  • 2




    $begingroup$
    Because ${X+Y>0,X>0}={X>0,Y>-X}$.
    $endgroup$
    – Did
    Dec 17 '18 at 0:37














1












1








1





$begingroup$



I am given two independently distributed Gaussian distributions, $X$ and $Y$, with unknown means and variances. You randomly choose an $x$ and $y$ from these two distributions respectively. Given that $x+y>0$, what is the probability that $x>0$ in terms of the unknown parameters?




Suppose $Xsim N(mu_1, sigma_1^2)$ and $Ysim N(mu_2, sigma_2^2)$.



I know that $X+Ysim N(mu_1+mu_2, sigma_1^2+sigma_2^2)$. I considered using Bayes’ theorem.



$$P(x>0|x+y>0)=frac{P(x>0)P(x+y>0|x>0)}{P(x+y>0)}.$$



Since I know the distributions of $X$ and $X+Y$, I know that the forms of $P(x>0)$ and $P(x+y>0)$ are that of integrating the Gaussian. However, I am unsure of the form of $P(x+y>0|x>0)$. I feel that this is much easier to calculate when observing the distribution of $X+Y$ on a 2D-plane, but I am not exactly sure how to write it in terms of the unknowns. Could someone please shed some insight on this? Thanks!



Edit: this was meant to be a conceptual question I believe, and should be left as a combination of integrals.






statistics probability-distributions normal-distribution conditional-probability bayes-theorem







share|cite|improve this question











$endgroup$





I am given two independently distributed Gaussian distributions, $X$ and $Y$, with unknown means and variances. You randomly choose an $x$ and $y$ from these two distributions respectively. Given that $x+y>0$, what is the probability that $x>0$ in terms of the unknown parameters?




Suppose $Xsim N(mu_1, sigma_1^2)$ and $Ysim N(mu_2, sigma_2^2)$.



I know that $X+Ysim N(mu_1+mu_2, sigma_1^2+sigma_2^2)$. I considered using Bayes’ theorem.



$$P(x>0|x+y>0)=frac{P(x>0)P(x+y>0|x>0)}{P(x+y>0)}.$$



Since I know the distributions of $X$ and $X+Y$, I know that the forms of $P(x>0)$ and $P(x+y>0)$ are that of integrating the Gaussian. However, I am unsure of the form of $P(x+y>0|x>0)$. I feel that this is much easier to calculate when observing the distribution of $X+Y$ on a 2D-plane, but I am not exactly sure how to write it in terms of the unknowns. Could someone please shed some insight on this? Thanks!



Edit: this was meant to be a conceptual question I believe, and should be left as a combination of integrals.






statistics probability-distributions normal-distribution conditional-probability bayes-theorem




statistics probability-distributions normal-distribution conditional-probability bayes-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 3:27









user376343

3,7883827




3,7883827










asked Dec 17 '18 at 0:07









user107224user107224

442314




442314












  • $begingroup$
    One insight is that if these are supposed to be i.i.d., you have four parameters where you should have only two, since $mu_1=mu_2$ and $sigma_1=sigma_2$.
    $endgroup$
    – C Monsour
    Dec 17 '18 at 0:10










  • $begingroup$
    @CMonsour oh gosh sorry it should just be independently distributed, not i.i.d!
    $endgroup$
    – user107224
    Dec 17 '18 at 0:12






  • 1




    $begingroup$
    Even covariances are not enough in general, you are rather after formulas similar to $$P(X+Y>0,X>0)=int_0^inftyint_{-x}^infty f_Y(y)dyf_X(x)dx$$ But are you sure you need to solve the question for any values of the $mu_i$s and $sigma^2_i$s? 'Cause the answer shall be a nightmare to write down, even if rather simple conceptually.
    $endgroup$
    – Did
    Dec 17 '18 at 0:27








  • 1




    $begingroup$
    Ah, then you already have $P(X+Y>0,X>0)$, and $$P(X>0)=int_0^infty f_X(x)dx$$ and since you know $f_X$ and $f_Y$, you are done.
    $endgroup$
    – Did
    Dec 17 '18 at 0:33






  • 2




    $begingroup$
    Because ${X+Y>0,X>0}={X>0,Y>-X}$.
    $endgroup$
    – Did
    Dec 17 '18 at 0:37


















  • $begingroup$
    One insight is that if these are supposed to be i.i.d., you have four parameters where you should have only two, since $mu_1=mu_2$ and $sigma_1=sigma_2$.
    $endgroup$
    – C Monsour
    Dec 17 '18 at 0:10










  • $begingroup$
    @CMonsour oh gosh sorry it should just be independently distributed, not i.i.d!
    $endgroup$
    – user107224
    Dec 17 '18 at 0:12






  • 1




    $begingroup$
    Even covariances are not enough in general, you are rather after formulas similar to $$P(X+Y>0,X>0)=int_0^inftyint_{-x}^infty f_Y(y)dyf_X(x)dx$$ But are you sure you need to solve the question for any values of the $mu_i$s and $sigma^2_i$s? 'Cause the answer shall be a nightmare to write down, even if rather simple conceptually.
    $endgroup$
    – Did
    Dec 17 '18 at 0:27








  • 1




    $begingroup$
    Ah, then you already have $P(X+Y>0,X>0)$, and $$P(X>0)=int_0^infty f_X(x)dx$$ and since you know $f_X$ and $f_Y$, you are done.
    $endgroup$
    – Did
    Dec 17 '18 at 0:33






  • 2




    $begingroup$
    Because ${X+Y>0,X>0}={X>0,Y>-X}$.
    $endgroup$
    – Did
    Dec 17 '18 at 0:37
















$begingroup$
One insight is that if these are supposed to be i.i.d., you have four parameters where you should have only two, since $mu_1=mu_2$ and $sigma_1=sigma_2$.
$endgroup$
– C Monsour
Dec 17 '18 at 0:10




$begingroup$
One insight is that if these are supposed to be i.i.d., you have four parameters where you should have only two, since $mu_1=mu_2$ and $sigma_1=sigma_2$.
$endgroup$
– C Monsour
Dec 17 '18 at 0:10












$begingroup$
@CMonsour oh gosh sorry it should just be independently distributed, not i.i.d!
$endgroup$
– user107224
Dec 17 '18 at 0:12




$begingroup$
@CMonsour oh gosh sorry it should just be independently distributed, not i.i.d!
$endgroup$
– user107224
Dec 17 '18 at 0:12




1




1




$begingroup$
Even covariances are not enough in general, you are rather after formulas similar to $$P(X+Y>0,X>0)=int_0^inftyint_{-x}^infty f_Y(y)dyf_X(x)dx$$ But are you sure you need to solve the question for any values of the $mu_i$s and $sigma^2_i$s? 'Cause the answer shall be a nightmare to write down, even if rather simple conceptually.
$endgroup$
– Did
Dec 17 '18 at 0:27






$begingroup$
Even covariances are not enough in general, you are rather after formulas similar to $$P(X+Y>0,X>0)=int_0^inftyint_{-x}^infty f_Y(y)dyf_X(x)dx$$ But are you sure you need to solve the question for any values of the $mu_i$s and $sigma^2_i$s? 'Cause the answer shall be a nightmare to write down, even if rather simple conceptually.
$endgroup$
– Did
Dec 17 '18 at 0:27






1




1




$begingroup$
Ah, then you already have $P(X+Y>0,X>0)$, and $$P(X>0)=int_0^infty f_X(x)dx$$ and since you know $f_X$ and $f_Y$, you are done.
$endgroup$
– Did
Dec 17 '18 at 0:33




$begingroup$
Ah, then you already have $P(X+Y>0,X>0)$, and $$P(X>0)=int_0^infty f_X(x)dx$$ and since you know $f_X$ and $f_Y$, you are done.
$endgroup$
– Did
Dec 17 '18 at 0:33




2




2




$begingroup$
Because ${X+Y>0,X>0}={X>0,Y>-X}$.
$endgroup$
– Did
Dec 17 '18 at 0:37




$begingroup$
Because ${X+Y>0,X>0}={X>0,Y>-X}$.
$endgroup$
– Did
Dec 17 '18 at 0:37










0






active

oldest

votes











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%2f3043372%2fpx0xy0-of-two-independently-distributed-gaussian-distributions%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















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%2f3043372%2fpx0xy0-of-two-independently-distributed-gaussian-distributions%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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more