$P(x>0|x+y>0);$ of two independently distributed Gaussian distributions?
$begingroup$
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
edited Dec 17 '18 at 3:27
user376343
3,7883827
asked Dec 17 '18 at 0:07
user107224user107224
442314
$endgroup$
|
show 7 more comments
$begingroup$
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
edited Dec 17 '18 at 3:27
user376343
3,7883827
asked Dec 17 '18 at 0:07
user107224user107224
442314
$endgroup$
$begingroup$
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
|
show 7 more comments
$begingroup$
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
edited Dec 17 '18 at 3:27
user376343
3,7883827
asked Dec 17 '18 at 0:07
user107224user107224
442314
$endgroup$
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
edited Dec 17 '18 at 3:27
user376343
3,7883827
asked Dec 17 '18 at 0:07
user107224user107224
442314
edited Dec 17 '18 at 3:27
user376343
3,7883827
asked Dec 17 '18 at 0:07
user107224user107224
442314
edited Dec 17 '18 at 3:27
user376343
3,7883827
edited Dec 17 '18 at 3:27
user376343
3,7883827
edited Dec 17 '18 at 3:27
user376343
3,7883827
3,7883827
asked Dec 17 '18 at 0:07
user107224user107224
442314
asked Dec 17 '18 at 0:07
user107224user107224
442314
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
|
show 7 more comments
$begingroup$
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
|
show 7 more comments
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$begingroup$
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