Clarifying the statement $f(x,y) = f_X(x) f_Y(y)$ for all $x, y$ (when $X$ and $Y$ are independent)
The textbook A First Course in Probability by Ross defines random variables $X$ and $Y$ to be independent when the following condition is satisfied: if $A$ and $B$ are subsets of $mathbb R$, then
$$
tag{0}P(X in A, Y in B) = P(X in A) P(Y in B).
$$
(I think technically we should assume that $X$ and $Y$ are Borel, but Ross glosses over that issue.)
The textbook goes on to state that if $X$ and $Y$ are jointly continuous random variables with PDFs $f_X$ and $f_Y$ and joint PDF $f$, then
$$
tag{1} f(x,y) = f_X(x) f_Y(y) quad text{for all } x,y.
$$
However, I think this statement is not correct because $f_X$ and $f_Y$ are only defined up to sets of measure $0$. (For example, I think you could redefine $f_X$ arbitrarily on a set of measure $0$, and you would still have a perfectly valid PDF for $X$.)
Question: So how can we state equation (1) in a way that is correct (so that we are not speaking nonsense to students) but which is also understandable to undergrads?
Further details: Let $F$ be the joint CDF for $X$ and $Y$,
and let $F_X$ and $F_Y$ be the CDFs for $X$ and $Y$ respectively.
It can be shown that equation (0) is equivalent to
$$
F(a,b) = F_X(a) F_Y(b)
$$
for all $a, b in mathbb R$. In other words,
$$
F(a,b) = int_{-infty}^a f_X(x) , dx
int_{-infty}^b f_Y(y) , dy
$$
for all $a,b in mathbb R$.
If $f_X$ is continuous at $a$, then differentiating both sides with respect to $a$ yields
$$
frac{partial F}{partial a} = f_X(a) int_{-infty}^b f_Y(y) , dy.
$$
If $f_Y$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b}
= f_X(a) f_Y(b).
$$
If we knew that
$$
tag{2} frac{partial^2 F}{partial a partial b} = f(a,b)
$$ then we would have established that $f(a,b) = f_X(a) f_Y(b)$.
So let's see what assumptions we need in order to conclude that $frac{partial^2 F}{partial a partial b} = f(a,b)$. We know that
$$
tag{3} F(a,b) = int_{-infty}^a int_{-infty}^b f(x,y) , dy , dx.
$$
If the function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$, then by the fundamental theorem of calculus differentiating both sides of (3) with respect to $a$ yields
$$
frac{partial F}{partial a} = int_{-infty}^b f(a,y) , dy.
$$
If the function $y mapsto f(a,y)$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b} F(a,b) = f(a,b).
$$
So, in order to establish (2), we needed the following two assumptions:
- The function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$. - The function $y mapsto f(a,y)$ is continuous at $b$.
Question: What is a nice, simple assumption I can state which will guarantee that these two conditions are met? If $f$ is continuous at $(a,b)$, then the second assumption is satisfied. But I don't see that the continuity of $f$ at $(a,b)$ would guarantee that the first condition is satisfied.
probability
asked Nov 26 at 1:28
eternalGoldenBraid
712314
add a comment |
The textbook A First Course in Probability by Ross defines random variables $X$ and $Y$ to be independent when the following condition is satisfied: if $A$ and $B$ are subsets of $mathbb R$, then
$$
tag{0}P(X in A, Y in B) = P(X in A) P(Y in B).
$$
(I think technically we should assume that $X$ and $Y$ are Borel, but Ross glosses over that issue.)
The textbook goes on to state that if $X$ and $Y$ are jointly continuous random variables with PDFs $f_X$ and $f_Y$ and joint PDF $f$, then
$$
tag{1} f(x,y) = f_X(x) f_Y(y) quad text{for all } x,y.
$$
However, I think this statement is not correct because $f_X$ and $f_Y$ are only defined up to sets of measure $0$. (For example, I think you could redefine $f_X$ arbitrarily on a set of measure $0$, and you would still have a perfectly valid PDF for $X$.)
Question: So how can we state equation (1) in a way that is correct (so that we are not speaking nonsense to students) but which is also understandable to undergrads?
Further details: Let $F$ be the joint CDF for $X$ and $Y$,
and let $F_X$ and $F_Y$ be the CDFs for $X$ and $Y$ respectively.
It can be shown that equation (0) is equivalent to
$$
F(a,b) = F_X(a) F_Y(b)
$$
for all $a, b in mathbb R$. In other words,
$$
F(a,b) = int_{-infty}^a f_X(x) , dx
int_{-infty}^b f_Y(y) , dy
$$
for all $a,b in mathbb R$.
If $f_X$ is continuous at $a$, then differentiating both sides with respect to $a$ yields
$$
frac{partial F}{partial a} = f_X(a) int_{-infty}^b f_Y(y) , dy.
$$
If $f_Y$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b}
= f_X(a) f_Y(b).
$$
If we knew that
$$
tag{2} frac{partial^2 F}{partial a partial b} = f(a,b)
$$ then we would have established that $f(a,b) = f_X(a) f_Y(b)$.
So let's see what assumptions we need in order to conclude that $frac{partial^2 F}{partial a partial b} = f(a,b)$. We know that
$$
tag{3} F(a,b) = int_{-infty}^a int_{-infty}^b f(x,y) , dy , dx.
$$
If the function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$, then by the fundamental theorem of calculus differentiating both sides of (3) with respect to $a$ yields
$$
frac{partial F}{partial a} = int_{-infty}^b f(a,y) , dy.
$$
If the function $y mapsto f(a,y)$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b} F(a,b) = f(a,b).
$$
So, in order to establish (2), we needed the following two assumptions:
- The function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$. - The function $y mapsto f(a,y)$ is continuous at $b$.
Question: What is a nice, simple assumption I can state which will guarantee that these two conditions are met? If $f$ is continuous at $(a,b)$, then the second assumption is satisfied. But I don't see that the continuity of $f$ at $(a,b)$ would guarantee that the first condition is satisfied.
probability
asked Nov 26 at 1:28
eternalGoldenBraid
712314
1
You could say that (1) holds for all $(x,y)in U{times} V$, for any open $U$ and $V$ for which $f_X$, $f_Y$, and $f$ are continuous on $U$, $V$, and $U{times} V$ respectively. Most density functions seen in undergraduate courses are continuous, so you are not losing many examples this way.
– kimchi lover
Nov 26 at 2:19
@kimchilover Thank you, that is the kind of suggestion I'm looking for.
– eternalGoldenBraid
Nov 26 at 3:59
1
The correct statement is: $f(x, y) = f_X(x) f_Y(y)$ holds for almost every $x,y$ (with respect to 2-dimensional Lebesgue measure).
– Song
Nov 26 at 7:47
add a comment |
The textbook A First Course in Probability by Ross defines random variables $X$ and $Y$ to be independent when the following condition is satisfied: if $A$ and $B$ are subsets of $mathbb R$, then
$$
tag{0}P(X in A, Y in B) = P(X in A) P(Y in B).
$$
(I think technically we should assume that $X$ and $Y$ are Borel, but Ross glosses over that issue.)
The textbook goes on to state that if $X$ and $Y$ are jointly continuous random variables with PDFs $f_X$ and $f_Y$ and joint PDF $f$, then
$$
tag{1} f(x,y) = f_X(x) f_Y(y) quad text{for all } x,y.
$$
However, I think this statement is not correct because $f_X$ and $f_Y$ are only defined up to sets of measure $0$. (For example, I think you could redefine $f_X$ arbitrarily on a set of measure $0$, and you would still have a perfectly valid PDF for $X$.)
Question: So how can we state equation (1) in a way that is correct (so that we are not speaking nonsense to students) but which is also understandable to undergrads?
Further details: Let $F$ be the joint CDF for $X$ and $Y$,
and let $F_X$ and $F_Y$ be the CDFs for $X$ and $Y$ respectively.
It can be shown that equation (0) is equivalent to
$$
F(a,b) = F_X(a) F_Y(b)
$$
for all $a, b in mathbb R$. In other words,
$$
F(a,b) = int_{-infty}^a f_X(x) , dx
int_{-infty}^b f_Y(y) , dy
$$
for all $a,b in mathbb R$.
If $f_X$ is continuous at $a$, then differentiating both sides with respect to $a$ yields
$$
frac{partial F}{partial a} = f_X(a) int_{-infty}^b f_Y(y) , dy.
$$
If $f_Y$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b}
= f_X(a) f_Y(b).
$$
If we knew that
$$
tag{2} frac{partial^2 F}{partial a partial b} = f(a,b)
$$ then we would have established that $f(a,b) = f_X(a) f_Y(b)$.
So let's see what assumptions we need in order to conclude that $frac{partial^2 F}{partial a partial b} = f(a,b)$. We know that
$$
tag{3} F(a,b) = int_{-infty}^a int_{-infty}^b f(x,y) , dy , dx.
$$
If the function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$, then by the fundamental theorem of calculus differentiating both sides of (3) with respect to $a$ yields
$$
frac{partial F}{partial a} = int_{-infty}^b f(a,y) , dy.
$$
If the function $y mapsto f(a,y)$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b} F(a,b) = f(a,b).
$$
So, in order to establish (2), we needed the following two assumptions:
- The function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$. - The function $y mapsto f(a,y)$ is continuous at $b$.
Question: What is a nice, simple assumption I can state which will guarantee that these two conditions are met? If $f$ is continuous at $(a,b)$, then the second assumption is satisfied. But I don't see that the continuity of $f$ at $(a,b)$ would guarantee that the first condition is satisfied.
probability
asked Nov 26 at 1:28
eternalGoldenBraid
712314
The textbook A First Course in Probability by Ross defines random variables $X$ and $Y$ to be independent when the following condition is satisfied: if $A$ and $B$ are subsets of $mathbb R$, then
$$
tag{0}P(X in A, Y in B) = P(X in A) P(Y in B).
$$
(I think technically we should assume that $X$ and $Y$ are Borel, but Ross glosses over that issue.)
The textbook goes on to state that if $X$ and $Y$ are jointly continuous random variables with PDFs $f_X$ and $f_Y$ and joint PDF $f$, then
$$
tag{1} f(x,y) = f_X(x) f_Y(y) quad text{for all } x,y.
$$
However, I think this statement is not correct because $f_X$ and $f_Y$ are only defined up to sets of measure $0$. (For example, I think you could redefine $f_X$ arbitrarily on a set of measure $0$, and you would still have a perfectly valid PDF for $X$.)
Question: So how can we state equation (1) in a way that is correct (so that we are not speaking nonsense to students) but which is also understandable to undergrads?
Further details: Let $F$ be the joint CDF for $X$ and $Y$,
and let $F_X$ and $F_Y$ be the CDFs for $X$ and $Y$ respectively.
It can be shown that equation (0) is equivalent to
$$
F(a,b) = F_X(a) F_Y(b)
$$
for all $a, b in mathbb R$. In other words,
$$
F(a,b) = int_{-infty}^a f_X(x) , dx
int_{-infty}^b f_Y(y) , dy
$$
for all $a,b in mathbb R$.
If $f_X$ is continuous at $a$, then differentiating both sides with respect to $a$ yields
$$
frac{partial F}{partial a} = f_X(a) int_{-infty}^b f_Y(y) , dy.
$$
If $f_Y$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b}
= f_X(a) f_Y(b).
$$
If we knew that
$$
tag{2} frac{partial^2 F}{partial a partial b} = f(a,b)
$$ then we would have established that $f(a,b) = f_X(a) f_Y(b)$.
So let's see what assumptions we need in order to conclude that $frac{partial^2 F}{partial a partial b} = f(a,b)$. We know that
$$
tag{3} F(a,b) = int_{-infty}^a int_{-infty}^b f(x,y) , dy , dx.
$$
If the function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$, then by the fundamental theorem of calculus differentiating both sides of (3) with respect to $a$ yields
$$
frac{partial F}{partial a} = int_{-infty}^b f(a,y) , dy.
$$
If the function $y mapsto f(a,y)$ is continuous at $b$, then differentiating with respect to $b$ yields
$$
frac{partial^2 F}{partial a partial b} F(a,b) = f(a,b).
$$
So, in order to establish (2), we needed the following two assumptions:
- The function $x mapsto int_{-infty}^b f(x,y) , dy$ is continuous
at $a$. - The function $y mapsto f(a,y)$ is continuous at $b$.
Question: What is a nice, simple assumption I can state which will guarantee that these two conditions are met? If $f$ is continuous at $(a,b)$, then the second assumption is satisfied. But I don't see that the continuity of $f$ at $(a,b)$ would guarantee that the first condition is satisfied.
probability
probability
asked Nov 26 at 1:28
eternalGoldenBraid
712314
asked Nov 26 at 1:28
eternalGoldenBraid
712314
edited Nov 26 at 6:28
asked Nov 26 at 1:28
eternalGoldenBraid
712314
asked Nov 26 at 1:28
eternalGoldenBraid
712314
asked Nov 26 at 1:28
eternalGoldenBraid
712314
712314
1
You could say that (1) holds for all $(x,y)in U{times} V$, for any open $U$ and $V$ for which $f_X$, $f_Y$, and $f$ are continuous on $U$, $V$, and $U{times} V$ respectively. Most density functions seen in undergraduate courses are continuous, so you are not losing many examples this way.
– kimchi lover
Nov 26 at 2:19
@kimchilover Thank you, that is the kind of suggestion I'm looking for.
– eternalGoldenBraid
Nov 26 at 3:59
1
The correct statement is: $f(x, y) = f_X(x) f_Y(y)$ holds for almost every $x,y$ (with respect to 2-dimensional Lebesgue measure).
– Song
Nov 26 at 7:47
add a comment |
1
You could say that (1) holds for all $(x,y)in U{times} V$, for any open $U$ and $V$ for which $f_X$, $f_Y$, and $f$ are continuous on $U$, $V$, and $U{times} V$ respectively. Most density functions seen in undergraduate courses are continuous, so you are not losing many examples this way.
– kimchi lover
Nov 26 at 2:19
@kimchilover Thank you, that is the kind of suggestion I'm looking for.
– eternalGoldenBraid
Nov 26 at 3:59
1
The correct statement is: $f(x, y) = f_X(x) f_Y(y)$ holds for almost every $x,y$ (with respect to 2-dimensional Lebesgue measure).
– Song
Nov 26 at 7:47
1
1
You could say that (1) holds for all $(x,y)in U{times} V$, for any open $U$ and $V$ for which $f_X$, $f_Y$, and $f$ are continuous on $U$, $V$, and $U{times} V$ respectively. Most density functions seen in undergraduate courses are continuous, so you are not losing many examples this way.
– kimchi lover
Nov 26 at 2:19
You could say that (1) holds for all $(x,y)in U{times} V$, for any open $U$ and $V$ for which $f_X$, $f_Y$, and $f$ are continuous on $U$, $V$, and $U{times} V$ respectively. Most density functions seen in undergraduate courses are continuous, so you are not losing many examples this way.
– kimchi lover
Nov 26 at 2:19
@kimchilover Thank you, that is the kind of suggestion I'm looking for.
– eternalGoldenBraid
Nov 26 at 3:59
@kimchilover Thank you, that is the kind of suggestion I'm looking for.
– eternalGoldenBraid
Nov 26 at 3:59
1
1
The correct statement is: $f(x, y) = f_X(x) f_Y(y)$ holds for almost every $x,y$ (with respect to 2-dimensional Lebesgue measure).
– Song
Nov 26 at 7:47
The correct statement is: $f(x, y) = f_X(x) f_Y(y)$ holds for almost every $x,y$ (with respect to 2-dimensional Lebesgue measure).
– Song
Nov 26 at 7:47
add a comment |
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1
You could say that (1) holds for all $(x,y)in U{times} V$, for any open $U$ and $V$ for which $f_X$, $f_Y$, and $f$ are continuous on $U$, $V$, and $U{times} V$ respectively. Most density functions seen in undergraduate courses are continuous, so you are not losing many examples this way.
– kimchi lover
Nov 26 at 2:19
@kimchilover Thank you, that is the kind of suggestion I'm looking for.
– eternalGoldenBraid
Nov 26 at 3:59
1
The correct statement is: $f(x, y) = f_X(x) f_Y(y)$ holds for almost every $x,y$ (with respect to 2-dimensional Lebesgue measure).
– Song
Nov 26 at 7:47