Identically distributed vs P(X > Y) = P(Y > X)
I've two related propositions which seem correct intuitively, but I struggle to prove them properly.
Question 1
Prove or disprove: If $X$ and $Y$ are independent and have identical marginal distributions, then $mathbb{P} (Y > X) = mathbb{P} (X > Y) = 1/2$
Due to independence, the joint PDF of $X$ and $Y$ is the product of their marginal PDF:
$$ begin{align}
mathbb{P} (Y > X) &= int_{-infty}^infty int_x^infty p(x) , p(y) , dy , dx \
mathbb{P} (X > Y) &= int_{-infty}^infty int_y^infty p(x) , p(y) , dx , dy
= int_{-infty}^infty int_x^infty p(y) , p(x) , dy , dx
end{align} $$
The last step is based on the fact that the integral won't change if we simply rename the integration parameters $x$ and $y$ consistently. So we have shown that $mathbb{P} (Y > X) = mathbb{P} (X > Y)$
Side note: Even if $X$ and $Y$ are dependent, this result still holds so long as their joint PDF is exchangeable i.e. $p(x, y) = p(y, x)$
Let $u = y - x$ so that
$$ mathbb{P} (Y > X) = int_{-infty}^infty int_0^infty p(x) , p(u + x) , du , dx $$
I thought of applying Fubini's theorem but it doesn't help to show that it's equal to 1/2, so maybe it's not 1/2?
Alternatively, consider that
$$ mathbb{P} (Y > X) + mathbb{P} (X > Y) + mathbb{P} (Y = X) = 1 $$
If we assume that $mathbb{P} (Y = X) = 0$ then we can conclude that $mathbb{P} (Y > X) = 1/2$. But is this assumption justified?
Question 2
Prove or disprove: If $X$ and $Y$ are independent and $mathbb{P} (Y > X) = mathbb{P} (X > Y)$, then they have identical marginal distributions. If this statement is true, then is it still true if $X$ and $Y$ are dependent?
@Xi'an provided a counter-example. Suppose that
$$ begin{bmatrix} X \ Y end{bmatrix}
sim mathcal{N} left(
begin{bmatrix} mu \ mu end{bmatrix},
begin{bmatrix} sigma_1^2 & c \ c & sigma_2^2 end{bmatrix}
right)$$
Then $X-Y$ and $Y-X$ have the same distribution: $mathcal{N} left(0, sigma_1^2 + sigma_2^2 - 2c right)$ and hence $mathbb{P} (Y - X > 0) = mathbb{P} (X - Y > 0)$
However the marginal distributions of $X sim mathcal{N} left(mu, sigma_1^2right)$ and $Y sim mathcal{N} left(mu, sigma_2^2right)$ may be different. This result holds regardless of whether $X$ and $Y$ are independent.
joint-distribution iid symmetry
add a comment |
I've two related propositions which seem correct intuitively, but I struggle to prove them properly.
Question 1
Prove or disprove: If $X$ and $Y$ are independent and have identical marginal distributions, then $mathbb{P} (Y > X) = mathbb{P} (X > Y) = 1/2$
Due to independence, the joint PDF of $X$ and $Y$ is the product of their marginal PDF:
$$ begin{align}
mathbb{P} (Y > X) &= int_{-infty}^infty int_x^infty p(x) , p(y) , dy , dx \
mathbb{P} (X > Y) &= int_{-infty}^infty int_y^infty p(x) , p(y) , dx , dy
= int_{-infty}^infty int_x^infty p(y) , p(x) , dy , dx
end{align} $$
The last step is based on the fact that the integral won't change if we simply rename the integration parameters $x$ and $y$ consistently. So we have shown that $mathbb{P} (Y > X) = mathbb{P} (X > Y)$
Side note: Even if $X$ and $Y$ are dependent, this result still holds so long as their joint PDF is exchangeable i.e. $p(x, y) = p(y, x)$
Let $u = y - x$ so that
$$ mathbb{P} (Y > X) = int_{-infty}^infty int_0^infty p(x) , p(u + x) , du , dx $$
I thought of applying Fubini's theorem but it doesn't help to show that it's equal to 1/2, so maybe it's not 1/2?
Alternatively, consider that
$$ mathbb{P} (Y > X) + mathbb{P} (X > Y) + mathbb{P} (Y = X) = 1 $$
If we assume that $mathbb{P} (Y = X) = 0$ then we can conclude that $mathbb{P} (Y > X) = 1/2$. But is this assumption justified?
Question 2
Prove or disprove: If $X$ and $Y$ are independent and $mathbb{P} (Y > X) = mathbb{P} (X > Y)$, then they have identical marginal distributions. If this statement is true, then is it still true if $X$ and $Y$ are dependent?
@Xi'an provided a counter-example. Suppose that
$$ begin{bmatrix} X \ Y end{bmatrix}
sim mathcal{N} left(
begin{bmatrix} mu \ mu end{bmatrix},
begin{bmatrix} sigma_1^2 & c \ c & sigma_2^2 end{bmatrix}
right)$$
Then $X-Y$ and $Y-X$ have the same distribution: $mathcal{N} left(0, sigma_1^2 + sigma_2^2 - 2c right)$ and hence $mathbb{P} (Y - X > 0) = mathbb{P} (X - Y > 0)$
However the marginal distributions of $X sim mathcal{N} left(mu, sigma_1^2right)$ and $Y sim mathcal{N} left(mu, sigma_2^2right)$ may be different. This result holds regardless of whether $X$ and $Y$ are independent.
joint-distribution iid symmetry
2
One interesting technical point is that if the probability of $X=c$ is $0$ for every value of $c,$ that's not enough to entail that probabilities are given by integrating a density function. The standard counterexample is the Cantor distribution. But more to the point$,ldotsqquad$
– Michael Hardy
yesterday
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
yesterday
1
What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0?
– The Laconic
yesterday
When using Fubini's theorem and a density $p(cdot)$ against the Lebesgue measure, $mathbb{P} (Y = X) = 0$, necessarily.
– Xi'an
yesterday
add a comment |
I've two related propositions which seem correct intuitively, but I struggle to prove them properly.
Question 1
Prove or disprove: If $X$ and $Y$ are independent and have identical marginal distributions, then $mathbb{P} (Y > X) = mathbb{P} (X > Y) = 1/2$
Due to independence, the joint PDF of $X$ and $Y$ is the product of their marginal PDF:
$$ begin{align}
mathbb{P} (Y > X) &= int_{-infty}^infty int_x^infty p(x) , p(y) , dy , dx \
mathbb{P} (X > Y) &= int_{-infty}^infty int_y^infty p(x) , p(y) , dx , dy
= int_{-infty}^infty int_x^infty p(y) , p(x) , dy , dx
end{align} $$
The last step is based on the fact that the integral won't change if we simply rename the integration parameters $x$ and $y$ consistently. So we have shown that $mathbb{P} (Y > X) = mathbb{P} (X > Y)$
Side note: Even if $X$ and $Y$ are dependent, this result still holds so long as their joint PDF is exchangeable i.e. $p(x, y) = p(y, x)$
Let $u = y - x$ so that
$$ mathbb{P} (Y > X) = int_{-infty}^infty int_0^infty p(x) , p(u + x) , du , dx $$
I thought of applying Fubini's theorem but it doesn't help to show that it's equal to 1/2, so maybe it's not 1/2?
Alternatively, consider that
$$ mathbb{P} (Y > X) + mathbb{P} (X > Y) + mathbb{P} (Y = X) = 1 $$
If we assume that $mathbb{P} (Y = X) = 0$ then we can conclude that $mathbb{P} (Y > X) = 1/2$. But is this assumption justified?
Question 2
Prove or disprove: If $X$ and $Y$ are independent and $mathbb{P} (Y > X) = mathbb{P} (X > Y)$, then they have identical marginal distributions. If this statement is true, then is it still true if $X$ and $Y$ are dependent?
@Xi'an provided a counter-example. Suppose that
$$ begin{bmatrix} X \ Y end{bmatrix}
sim mathcal{N} left(
begin{bmatrix} mu \ mu end{bmatrix},
begin{bmatrix} sigma_1^2 & c \ c & sigma_2^2 end{bmatrix}
right)$$
Then $X-Y$ and $Y-X$ have the same distribution: $mathcal{N} left(0, sigma_1^2 + sigma_2^2 - 2c right)$ and hence $mathbb{P} (Y - X > 0) = mathbb{P} (X - Y > 0)$
However the marginal distributions of $X sim mathcal{N} left(mu, sigma_1^2right)$ and $Y sim mathcal{N} left(mu, sigma_2^2right)$ may be different. This result holds regardless of whether $X$ and $Y$ are independent.
joint-distribution iid symmetry
I've two related propositions which seem correct intuitively, but I struggle to prove them properly.
Question 1
Prove or disprove: If $X$ and $Y$ are independent and have identical marginal distributions, then $mathbb{P} (Y > X) = mathbb{P} (X > Y) = 1/2$
Due to independence, the joint PDF of $X$ and $Y$ is the product of their marginal PDF:
$$ begin{align}
mathbb{P} (Y > X) &= int_{-infty}^infty int_x^infty p(x) , p(y) , dy , dx \
mathbb{P} (X > Y) &= int_{-infty}^infty int_y^infty p(x) , p(y) , dx , dy
= int_{-infty}^infty int_x^infty p(y) , p(x) , dy , dx
end{align} $$
The last step is based on the fact that the integral won't change if we simply rename the integration parameters $x$ and $y$ consistently. So we have shown that $mathbb{P} (Y > X) = mathbb{P} (X > Y)$
Side note: Even if $X$ and $Y$ are dependent, this result still holds so long as their joint PDF is exchangeable i.e. $p(x, y) = p(y, x)$
Let $u = y - x$ so that
$$ mathbb{P} (Y > X) = int_{-infty}^infty int_0^infty p(x) , p(u + x) , du , dx $$
I thought of applying Fubini's theorem but it doesn't help to show that it's equal to 1/2, so maybe it's not 1/2?
Alternatively, consider that
$$ mathbb{P} (Y > X) + mathbb{P} (X > Y) + mathbb{P} (Y = X) = 1 $$
If we assume that $mathbb{P} (Y = X) = 0$ then we can conclude that $mathbb{P} (Y > X) = 1/2$. But is this assumption justified?
Question 2
Prove or disprove: If $X$ and $Y$ are independent and $mathbb{P} (Y > X) = mathbb{P} (X > Y)$, then they have identical marginal distributions. If this statement is true, then is it still true if $X$ and $Y$ are dependent?
@Xi'an provided a counter-example. Suppose that
$$ begin{bmatrix} X \ Y end{bmatrix}
sim mathcal{N} left(
begin{bmatrix} mu \ mu end{bmatrix},
begin{bmatrix} sigma_1^2 & c \ c & sigma_2^2 end{bmatrix}
right)$$
Then $X-Y$ and $Y-X$ have the same distribution: $mathcal{N} left(0, sigma_1^2 + sigma_2^2 - 2c right)$ and hence $mathbb{P} (Y - X > 0) = mathbb{P} (X - Y > 0)$
However the marginal distributions of $X sim mathcal{N} left(mu, sigma_1^2right)$ and $Y sim mathcal{N} left(mu, sigma_2^2right)$ may be different. This result holds regardless of whether $X$ and $Y$ are independent.
joint-distribution iid symmetry
joint-distribution iid symmetry
edited yesterday
farmer
asked yesterday
farmerfarmer
637
637
2
One interesting technical point is that if the probability of $X=c$ is $0$ for every value of $c,$ that's not enough to entail that probabilities are given by integrating a density function. The standard counterexample is the Cantor distribution. But more to the point$,ldotsqquad$
– Michael Hardy
yesterday
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
yesterday
1
What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0?
– The Laconic
yesterday
When using Fubini's theorem and a density $p(cdot)$ against the Lebesgue measure, $mathbb{P} (Y = X) = 0$, necessarily.
– Xi'an
yesterday
add a comment |
2
One interesting technical point is that if the probability of $X=c$ is $0$ for every value of $c,$ that's not enough to entail that probabilities are given by integrating a density function. The standard counterexample is the Cantor distribution. But more to the point$,ldotsqquad$
– Michael Hardy
yesterday
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
yesterday
1
What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0?
– The Laconic
yesterday
When using Fubini's theorem and a density $p(cdot)$ against the Lebesgue measure, $mathbb{P} (Y = X) = 0$, necessarily.
– Xi'an
yesterday
2
2
One interesting technical point is that if the probability of $X=c$ is $0$ for every value of $c,$ that's not enough to entail that probabilities are given by integrating a density function. The standard counterexample is the Cantor distribution. But more to the point$,ldotsqquad$
– Michael Hardy
yesterday
One interesting technical point is that if the probability of $X=c$ is $0$ for every value of $c,$ that's not enough to entail that probabilities are given by integrating a density function. The standard counterexample is the Cantor distribution. But more to the point$,ldotsqquad$
– Michael Hardy
yesterday
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
yesterday
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
yesterday
1
1
What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0?
– The Laconic
yesterday
What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0?
– The Laconic
yesterday
When using Fubini's theorem and a density $p(cdot)$ against the Lebesgue measure, $mathbb{P} (Y = X) = 0$, necessarily.
– Xi'an
yesterday
When using Fubini's theorem and a density $p(cdot)$ against the Lebesgue measure, $mathbb{P} (Y = X) = 0$, necessarily.
– Xi'an
yesterday
add a comment |
2 Answers
2
active
oldest
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This answer is written under the assumption that $mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one. (In the alternative case that $mathbb{P}(Y=X)>0$ this is obviously no longer true.)
Question 2: Take a bivariate normal vector $(X,Y)$ with mean $(mu,mu)$. Then $X-Y$ and $Y-X$ are identically distributed, no matter what the correlation between $X$ and $Y$, and no matter what the variances of $X$ and $Y$ are, and therefore (1) holds. The conjecture is thus false.
add a comment |
I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[Xin A], [Yin B]$ are independent. In particular for any two numbers $x,y$ the events $[Xle x], [Yle y]$ are independent, so $F_{X,Y}(x,y) = F_X(x)cdot F_Y(y).$
And the distribution of the pair $(X,Y)$ is completely determined by the joint c.d.f.
Since it is given that $F_X=F_Y,$ we can write $F_{X,Y}(x,y) = F_X(x)cdot F_X(y).$
This is symmetric as a function of $x$ and $y,$ i.e. it remains the same if $x$ and $y$ are interchanged.
But interchanging $x$ and $y$ in $F_{X,Y}(x,y)$ is the same as interchanging $X$ and $Y,$ since
$$
F_{X,Y}(x,y) = Pr(Xle x & Yle y).
$$
Therefore (the main point):
The distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed.
– farmer
12 hours ago
@farmer : Start with $Pr(Xle x & Yle y)$ and interchange $x$ and $y,$ and you get $Pr(Xle y & Yle x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $Pr(Yle x & Xle y).$ The claim, then, is that $Pr(Xle y & Yle x)$ is the same as $Pr(Yle x & Xle y). qquad$
– Michael Hardy
4 hours ago
add a comment |
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2 Answers
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2 Answers
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This answer is written under the assumption that $mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one. (In the alternative case that $mathbb{P}(Y=X)>0$ this is obviously no longer true.)
Question 2: Take a bivariate normal vector $(X,Y)$ with mean $(mu,mu)$. Then $X-Y$ and $Y-X$ are identically distributed, no matter what the correlation between $X$ and $Y$, and no matter what the variances of $X$ and $Y$ are, and therefore (1) holds. The conjecture is thus false.
add a comment |
This answer is written under the assumption that $mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one. (In the alternative case that $mathbb{P}(Y=X)>0$ this is obviously no longer true.)
Question 2: Take a bivariate normal vector $(X,Y)$ with mean $(mu,mu)$. Then $X-Y$ and $Y-X$ are identically distributed, no matter what the correlation between $X$ and $Y$, and no matter what the variances of $X$ and $Y$ are, and therefore (1) holds. The conjecture is thus false.
add a comment |
This answer is written under the assumption that $mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one. (In the alternative case that $mathbb{P}(Y=X)>0$ this is obviously no longer true.)
Question 2: Take a bivariate normal vector $(X,Y)$ with mean $(mu,mu)$. Then $X-Y$ and $Y-X$ are identically distributed, no matter what the correlation between $X$ and $Y$, and no matter what the variances of $X$ and $Y$ are, and therefore (1) holds. The conjecture is thus false.
This answer is written under the assumption that $mathbb{P}(Y=X)=0$
which was part of the original wording of the question.
Question 1: A sufficient condition for$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one. (In the alternative case that $mathbb{P}(Y=X)>0$ this is obviously no longer true.)
Question 2: Take a bivariate normal vector $(X,Y)$ with mean $(mu,mu)$. Then $X-Y$ and $Y-X$ are identically distributed, no matter what the correlation between $X$ and $Y$, and no matter what the variances of $X$ and $Y$ are, and therefore (1) holds. The conjecture is thus false.
edited 12 hours ago
answered yesterday
Xi'anXi'an
54.1k691350
54.1k691350
add a comment |
add a comment |
I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[Xin A], [Yin B]$ are independent. In particular for any two numbers $x,y$ the events $[Xle x], [Yle y]$ are independent, so $F_{X,Y}(x,y) = F_X(x)cdot F_Y(y).$
And the distribution of the pair $(X,Y)$ is completely determined by the joint c.d.f.
Since it is given that $F_X=F_Y,$ we can write $F_{X,Y}(x,y) = F_X(x)cdot F_X(y).$
This is symmetric as a function of $x$ and $y,$ i.e. it remains the same if $x$ and $y$ are interchanged.
But interchanging $x$ and $y$ in $F_{X,Y}(x,y)$ is the same as interchanging $X$ and $Y,$ since
$$
F_{X,Y}(x,y) = Pr(Xle x & Yle y).
$$
Therefore (the main point):
The distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed.
– farmer
12 hours ago
@farmer : Start with $Pr(Xle x & Yle y)$ and interchange $x$ and $y,$ and you get $Pr(Xle y & Yle x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $Pr(Yle x & Xle y).$ The claim, then, is that $Pr(Xle y & Yle x)$ is the same as $Pr(Yle x & Xle y). qquad$
– Michael Hardy
4 hours ago
add a comment |
I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[Xin A], [Yin B]$ are independent. In particular for any two numbers $x,y$ the events $[Xle x], [Yle y]$ are independent, so $F_{X,Y}(x,y) = F_X(x)cdot F_Y(y).$
And the distribution of the pair $(X,Y)$ is completely determined by the joint c.d.f.
Since it is given that $F_X=F_Y,$ we can write $F_{X,Y}(x,y) = F_X(x)cdot F_X(y).$
This is symmetric as a function of $x$ and $y,$ i.e. it remains the same if $x$ and $y$ are interchanged.
But interchanging $x$ and $y$ in $F_{X,Y}(x,y)$ is the same as interchanging $X$ and $Y,$ since
$$
F_{X,Y}(x,y) = Pr(Xle x & Yle y).
$$
Therefore (the main point):
The distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed.
– farmer
12 hours ago
@farmer : Start with $Pr(Xle x & Yle y)$ and interchange $x$ and $y,$ and you get $Pr(Xle y & Yle x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $Pr(Yle x & Xle y).$ The claim, then, is that $Pr(Xle y & Yle x)$ is the same as $Pr(Yle x & Xle y). qquad$
– Michael Hardy
4 hours ago
add a comment |
I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[Xin A], [Yin B]$ are independent. In particular for any two numbers $x,y$ the events $[Xle x], [Yle y]$ are independent, so $F_{X,Y}(x,y) = F_X(x)cdot F_Y(y).$
And the distribution of the pair $(X,Y)$ is completely determined by the joint c.d.f.
Since it is given that $F_X=F_Y,$ we can write $F_{X,Y}(x,y) = F_X(x)cdot F_X(y).$
This is symmetric as a function of $x$ and $y,$ i.e. it remains the same if $x$ and $y$ are interchanged.
But interchanging $x$ and $y$ in $F_{X,Y}(x,y)$ is the same as interchanging $X$ and $Y,$ since
$$
F_{X,Y}(x,y) = Pr(Xle x & Yle y).
$$
Therefore (the main point):
The distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[Xin A], [Yin B]$ are independent. In particular for any two numbers $x,y$ the events $[Xle x], [Yle y]$ are independent, so $F_{X,Y}(x,y) = F_X(x)cdot F_Y(y).$
And the distribution of the pair $(X,Y)$ is completely determined by the joint c.d.f.
Since it is given that $F_X=F_Y,$ we can write $F_{X,Y}(x,y) = F_X(x)cdot F_X(y).$
This is symmetric as a function of $x$ and $y,$ i.e. it remains the same if $x$ and $y$ are interchanged.
But interchanging $x$ and $y$ in $F_{X,Y}(x,y)$ is the same as interchanging $X$ and $Y,$ since
$$
F_{X,Y}(x,y) = Pr(Xle x & Yle y).
$$
Therefore (the main point):
The distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$
edited yesterday
answered yesterday
Michael HardyMichael Hardy
3,6551430
3,6551430
I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed.
– farmer
12 hours ago
@farmer : Start with $Pr(Xle x & Yle y)$ and interchange $x$ and $y,$ and you get $Pr(Xle y & Yle x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $Pr(Yle x & Xle y).$ The claim, then, is that $Pr(Xle y & Yle x)$ is the same as $Pr(Yle x & Xle y). qquad$
– Michael Hardy
4 hours ago
add a comment |
I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed.
– farmer
12 hours ago
@farmer : Start with $Pr(Xle x & Yle y)$ and interchange $x$ and $y,$ and you get $Pr(Xle y & Yle x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $Pr(Yle x & Xle y).$ The claim, then, is that $Pr(Xle y & Yle x)$ is the same as $Pr(Yle x & Xle y). qquad$
– Michael Hardy
4 hours ago
I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed.
– farmer
12 hours ago
I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed.
– farmer
12 hours ago
@farmer : Start with $Pr(Xle x & Yle y)$ and interchange $x$ and $y,$ and you get $Pr(Xle y & Yle x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $Pr(Yle x & Xle y).$ The claim, then, is that $Pr(Xle y & Yle x)$ is the same as $Pr(Yle x & Xle y). qquad$
– Michael Hardy
4 hours ago
@farmer : Start with $Pr(Xle x & Yle y)$ and interchange $x$ and $y,$ and you get $Pr(Xle y & Yle x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $Pr(Yle x & Xle y).$ The claim, then, is that $Pr(Xle y & Yle x)$ is the same as $Pr(Yle x & Xle y). qquad$
– Michael Hardy
4 hours ago
add a comment |
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2
One interesting technical point is that if the probability of $X=c$ is $0$ for every value of $c,$ that's not enough to entail that probabilities are given by integrating a density function. The standard counterexample is the Cantor distribution. But more to the point$,ldotsqquad$
– Michael Hardy
yesterday
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
yesterday
1
What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0?
– The Laconic
yesterday
When using Fubini's theorem and a density $p(cdot)$ against the Lebesgue measure, $mathbb{P} (Y = X) = 0$, necessarily.
– Xi'an
yesterday