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 identically distributed, 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 are identically distributed.
If this is true, then is it still true if $X$ and $Y$ are dependent?
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 identically distributed, 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 are identically distributed.
If this is true, then is it still true if $X$ and $Y$ are dependent?
joint-distribution iid symmetry
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
6 hours ago
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
6 hours ago
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 identically distributed, 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 are identically distributed.
If this is true, then is it still true if $X$ and $Y$ are dependent?
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 identically distributed, 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 are identically distributed.
If this is true, then is it still true if $X$ and $Y$ are dependent?
joint-distribution iid symmetry
joint-distribution iid symmetry
edited 1 hour ago
asked 6 hours ago
farmer
566
566
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
6 hours ago
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
6 hours ago
add a comment |
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
6 hours ago
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
6 hours ago
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
6 hours ago
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
6 hours ago
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
6 hours ago
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
6 hours ago
add a comment |
2 Answers
2
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oldest
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Question 1: I assume you mean to ask whether$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$because, the way the question is written$$mathbb{P}(X<Y)=mathbb{P}(Y>X)tag{2}$$is always true since $X<Y$ and $Y>X$ are the same events. Considering (1), it is sufficient that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution for (1) to hold. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one.
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.
Thanks, I've fixed the typo in my question.
– farmer
1 hour ago
For Q1, you seem to assume that P(Y = X) = 0. Is this justified? For Q2, I can't see why that's a counter-example. To falsify it, we need to find X, Y such that (1) holds but they are not identically distributed.
– farmer
58 mins 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 $(Y,X).$
add a comment |
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2 Answers
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Question 1: I assume you mean to ask whether$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$because, the way the question is written$$mathbb{P}(X<Y)=mathbb{P}(Y>X)tag{2}$$is always true since $X<Y$ and $Y>X$ are the same events. Considering (1), it is sufficient that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution for (1) to hold. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one.
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.
Thanks, I've fixed the typo in my question.
– farmer
1 hour ago
For Q1, you seem to assume that P(Y = X) = 0. Is this justified? For Q2, I can't see why that's a counter-example. To falsify it, we need to find X, Y such that (1) holds but they are not identically distributed.
– farmer
58 mins ago
add a comment |
Question 1: I assume you mean to ask whether$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$because, the way the question is written$$mathbb{P}(X<Y)=mathbb{P}(Y>X)tag{2}$$is always true since $X<Y$ and $Y>X$ are the same events. Considering (1), it is sufficient that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution for (1) to hold. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one.
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.
Thanks, I've fixed the typo in my question.
– farmer
1 hour ago
For Q1, you seem to assume that P(Y = X) = 0. Is this justified? For Q2, I can't see why that's a counter-example. To falsify it, we need to find X, Y such that (1) holds but they are not identically distributed.
– farmer
58 mins ago
add a comment |
Question 1: I assume you mean to ask whether$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$because, the way the question is written$$mathbb{P}(X<Y)=mathbb{P}(Y>X)tag{2}$$is always true since $X<Y$ and $Y>X$ are the same events. Considering (1), it is sufficient that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution for (1) to hold. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one.
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.
Question 1: I assume you mean to ask whether$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$because, the way the question is written$$mathbb{P}(X<Y)=mathbb{P}(Y>X)tag{2}$$is always true since $X<Y$ and $Y>X$ are the same events. Considering (1), it is sufficient that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution for (1) to hold. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one.
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.
answered 5 hours ago
Xi'an
53.9k690348
53.9k690348
Thanks, I've fixed the typo in my question.
– farmer
1 hour ago
For Q1, you seem to assume that P(Y = X) = 0. Is this justified? For Q2, I can't see why that's a counter-example. To falsify it, we need to find X, Y such that (1) holds but they are not identically distributed.
– farmer
58 mins ago
add a comment |
Thanks, I've fixed the typo in my question.
– farmer
1 hour ago
For Q1, you seem to assume that P(Y = X) = 0. Is this justified? For Q2, I can't see why that's a counter-example. To falsify it, we need to find X, Y such that (1) holds but they are not identically distributed.
– farmer
58 mins ago
Thanks, I've fixed the typo in my question.
– farmer
1 hour ago
Thanks, I've fixed the typo in my question.
– farmer
1 hour ago
For Q1, you seem to assume that P(Y = X) = 0. Is this justified? For Q2, I can't see why that's a counter-example. To falsify it, we need to find X, Y such that (1) holds but they are not identically distributed.
– farmer
58 mins ago
For Q1, you seem to assume that P(Y = X) = 0. Is this justified? For Q2, I can't see why that's a counter-example. To falsify it, we need to find X, Y such that (1) holds but they are not identically distributed.
– farmer
58 mins 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 $(Y,X).$
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 $(Y,X).$
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 5 hours ago
answered 5 hours ago
Michael Hardy
3,6451430
3,6451430
add a comment |
add a comment |
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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
6 hours ago
$ldots,$is that I probably wouldn't solve this problem by considering such integrals anyway.
– Michael Hardy
6 hours ago