Rigorous definition of sum of two random variables?
I've looked everywhere, but I seem unable to find a rigorous definition (in the set-theoretical sense of a random variable being a mapping from a sample space to the real number line) of the sum of two random variables (X + Y).
How do you rigorously define the sum of two random variables X + Y?
probability statistics random-variables
edited Mar 6 '18 at 20:10
nbro
2,39753171
asked Sep 30 '15 at 19:51
user2474041
92
add a comment |
I've looked everywhere, but I seem unable to find a rigorous definition (in the set-theoretical sense of a random variable being a mapping from a sample space to the real number line) of the sum of two random variables (X + Y).
How do you rigorously define the sum of two random variables X + Y?
probability statistics random-variables
edited Mar 6 '18 at 20:10
nbro
2,39753171
asked Sep 30 '15 at 19:51
user2474041
92
How do you rigorously define a random variable?
– Crostul
Sep 30 '15 at 19:56
A random variable is X is a mapping from the sample space of the experiment of interest to the real number line. So if I have a method of assigning a real number to every outcome in the sample space, I guess I would call that method a random variable.
– user2474041
Sep 30 '15 at 19:59
1
One needs to define a new sample space which is the direct product of two copies of the old sample space. As it happens, Terry Tao just wrote a great post about the foundations of probability, where this question is answered: terrytao.wordpress.com/2015/09/29/…
– Greg Martin
Sep 30 '15 at 20:27
add a comment |
I've looked everywhere, but I seem unable to find a rigorous definition (in the set-theoretical sense of a random variable being a mapping from a sample space to the real number line) of the sum of two random variables (X + Y).
How do you rigorously define the sum of two random variables X + Y?
probability statistics random-variables
edited Mar 6 '18 at 20:10
nbro
2,39753171
asked Sep 30 '15 at 19:51
user2474041
92
I've looked everywhere, but I seem unable to find a rigorous definition (in the set-theoretical sense of a random variable being a mapping from a sample space to the real number line) of the sum of two random variables (X + Y).
How do you rigorously define the sum of two random variables X + Y?
probability statistics random-variables
probability statistics random-variables
edited Mar 6 '18 at 20:10
nbro
2,39753171
asked Sep 30 '15 at 19:51
user2474041
92
edited Mar 6 '18 at 20:10
nbro
2,39753171
asked Sep 30 '15 at 19:51
user2474041
92
edited Mar 6 '18 at 20:10
nbro
2,39753171
edited Mar 6 '18 at 20:10
nbro
2,39753171
edited Mar 6 '18 at 20:10
nbro
2,39753171
2,39753171
asked Sep 30 '15 at 19:51
user2474041
92
asked Sep 30 '15 at 19:51
user2474041
92
asked Sep 30 '15 at 19:51
user2474041
92
92
How do you rigorously define a random variable?
– Crostul
Sep 30 '15 at 19:56
A random variable is X is a mapping from the sample space of the experiment of interest to the real number line. So if I have a method of assigning a real number to every outcome in the sample space, I guess I would call that method a random variable.
– user2474041
Sep 30 '15 at 19:59
1
One needs to define a new sample space which is the direct product of two copies of the old sample space. As it happens, Terry Tao just wrote a great post about the foundations of probability, where this question is answered: terrytao.wordpress.com/2015/09/29/…
– Greg Martin
Sep 30 '15 at 20:27
add a comment |
How do you rigorously define a random variable?
– Crostul
Sep 30 '15 at 19:56
A random variable is X is a mapping from the sample space of the experiment of interest to the real number line. So if I have a method of assigning a real number to every outcome in the sample space, I guess I would call that method a random variable.
– user2474041
Sep 30 '15 at 19:59
1
One needs to define a new sample space which is the direct product of two copies of the old sample space. As it happens, Terry Tao just wrote a great post about the foundations of probability, where this question is answered: terrytao.wordpress.com/2015/09/29/…
– Greg Martin
Sep 30 '15 at 20:27
How do you rigorously define a random variable?
– Crostul
Sep 30 '15 at 19:56
How do you rigorously define a random variable?
– Crostul
Sep 30 '15 at 19:56
A random variable is X is a mapping from the sample space of the experiment of interest to the real number line. So if I have a method of assigning a real number to every outcome in the sample space, I guess I would call that method a random variable.
– user2474041
Sep 30 '15 at 19:59
A random variable is X is a mapping from the sample space of the experiment of interest to the real number line. So if I have a method of assigning a real number to every outcome in the sample space, I guess I would call that method a random variable.
– user2474041
Sep 30 '15 at 19:59
1
1
One needs to define a new sample space which is the direct product of two copies of the old sample space. As it happens, Terry Tao just wrote a great post about the foundations of probability, where this question is answered: terrytao.wordpress.com/2015/09/29/…
– Greg Martin
Sep 30 '15 at 20:27
One needs to define a new sample space which is the direct product of two copies of the old sample space. As it happens, Terry Tao just wrote a great post about the foundations of probability, where this question is answered: terrytao.wordpress.com/2015/09/29/…
– Greg Martin
Sep 30 '15 at 20:27
add a comment |
2 Answers
2
active
oldest
votes
Given random variables $X, Y : Omegatomathbb R$ defined on the same probability space $Omega$, the definition of $X+Y$ is simply the pointwise sum:
$$(X+Y)(omega) = X(omega)+Y(omega)$$
If $X$ and $Y$ are defined on different probability spaces, $X:Omega_1 to mathbb R$ and $Y:Omega_2 to mathbb R$, then $X+Y$ is undefined. However, in this case we can define a new probability space $Omega = Omega_1 times Omega_2$ and random variables $X_1, Y_1$ on $Omega$ by
begin{align*}
X_1(omega_1,omega_2) &= X(omega_1)\
Y_1(omega_1,omega_2) &= Y(omega_2)
end{align*}
in which case $X_1$ has the same distribution as $X$, $Y_1$ has the same distribution as $Y$, and $X_1$ and $Y_1$ are independent. In this case, the sum $X_1+Y_1$, which is defined pointwise as above, may be considered as a replacement for the non-existent sum $X+Y$, as long as independence between the two summands is what we were looking for.
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
Could you clarify why $X_1$ and $Y_1$ need to be independent in the second case? Imagine $X: {H, T} to {0, 1}$ records the outcome $x$ of a coin toss, then we produce another random variable $Y: {0, 1} to {0, -1}$ by deterministically defining its value to be $-x$. of How would you define $X+Y$?
– Yibo Yang
Aug 19 '18 at 0:40
add a comment |
It actually defined as an analogue of the product of the polynomial with $X, Y$ independent and P was defined in double dimension
$$(P(x + y) = t) = sum_{I = 0}^t f_x(I)g_x(t - I)$$
answered Dec 11 '18 at 2:59
user577443
92
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Given random variables $X, Y : Omegatomathbb R$ defined on the same probability space $Omega$, the definition of $X+Y$ is simply the pointwise sum:
$$(X+Y)(omega) = X(omega)+Y(omega)$$
If $X$ and $Y$ are defined on different probability spaces, $X:Omega_1 to mathbb R$ and $Y:Omega_2 to mathbb R$, then $X+Y$ is undefined. However, in this case we can define a new probability space $Omega = Omega_1 times Omega_2$ and random variables $X_1, Y_1$ on $Omega$ by
begin{align*}
X_1(omega_1,omega_2) &= X(omega_1)\
Y_1(omega_1,omega_2) &= Y(omega_2)
end{align*}
in which case $X_1$ has the same distribution as $X$, $Y_1$ has the same distribution as $Y$, and $X_1$ and $Y_1$ are independent. In this case, the sum $X_1+Y_1$, which is defined pointwise as above, may be considered as a replacement for the non-existent sum $X+Y$, as long as independence between the two summands is what we were looking for.
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
Could you clarify why $X_1$ and $Y_1$ need to be independent in the second case? Imagine $X: {H, T} to {0, 1}$ records the outcome $x$ of a coin toss, then we produce another random variable $Y: {0, 1} to {0, -1}$ by deterministically defining its value to be $-x$. of How would you define $X+Y$?
– Yibo Yang
Aug 19 '18 at 0:40
add a comment |
Given random variables $X, Y : Omegatomathbb R$ defined on the same probability space $Omega$, the definition of $X+Y$ is simply the pointwise sum:
$$(X+Y)(omega) = X(omega)+Y(omega)$$
If $X$ and $Y$ are defined on different probability spaces, $X:Omega_1 to mathbb R$ and $Y:Omega_2 to mathbb R$, then $X+Y$ is undefined. However, in this case we can define a new probability space $Omega = Omega_1 times Omega_2$ and random variables $X_1, Y_1$ on $Omega$ by
begin{align*}
X_1(omega_1,omega_2) &= X(omega_1)\
Y_1(omega_1,omega_2) &= Y(omega_2)
end{align*}
in which case $X_1$ has the same distribution as $X$, $Y_1$ has the same distribution as $Y$, and $X_1$ and $Y_1$ are independent. In this case, the sum $X_1+Y_1$, which is defined pointwise as above, may be considered as a replacement for the non-existent sum $X+Y$, as long as independence between the two summands is what we were looking for.
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
Could you clarify why $X_1$ and $Y_1$ need to be independent in the second case? Imagine $X: {H, T} to {0, 1}$ records the outcome $x$ of a coin toss, then we produce another random variable $Y: {0, 1} to {0, -1}$ by deterministically defining its value to be $-x$. of How would you define $X+Y$?
– Yibo Yang
Aug 19 '18 at 0:40
add a comment |
Given random variables $X, Y : Omegatomathbb R$ defined on the same probability space $Omega$, the definition of $X+Y$ is simply the pointwise sum:
$$(X+Y)(omega) = X(omega)+Y(omega)$$
If $X$ and $Y$ are defined on different probability spaces, $X:Omega_1 to mathbb R$ and $Y:Omega_2 to mathbb R$, then $X+Y$ is undefined. However, in this case we can define a new probability space $Omega = Omega_1 times Omega_2$ and random variables $X_1, Y_1$ on $Omega$ by
begin{align*}
X_1(omega_1,omega_2) &= X(omega_1)\
Y_1(omega_1,omega_2) &= Y(omega_2)
end{align*}
in which case $X_1$ has the same distribution as $X$, $Y_1$ has the same distribution as $Y$, and $X_1$ and $Y_1$ are independent. In this case, the sum $X_1+Y_1$, which is defined pointwise as above, may be considered as a replacement for the non-existent sum $X+Y$, as long as independence between the two summands is what we were looking for.
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
Given random variables $X, Y : Omegatomathbb R$ defined on the same probability space $Omega$, the definition of $X+Y$ is simply the pointwise sum:
$$(X+Y)(omega) = X(omega)+Y(omega)$$
If $X$ and $Y$ are defined on different probability spaces, $X:Omega_1 to mathbb R$ and $Y:Omega_2 to mathbb R$, then $X+Y$ is undefined. However, in this case we can define a new probability space $Omega = Omega_1 times Omega_2$ and random variables $X_1, Y_1$ on $Omega$ by
begin{align*}
X_1(omega_1,omega_2) &= X(omega_1)\
Y_1(omega_1,omega_2) &= Y(omega_2)
end{align*}
in which case $X_1$ has the same distribution as $X$, $Y_1$ has the same distribution as $Y$, and $X_1$ and $Y_1$ are independent. In this case, the sum $X_1+Y_1$, which is defined pointwise as above, may be considered as a replacement for the non-existent sum $X+Y$, as long as independence between the two summands is what we were looking for.
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
answered Sep 30 '15 at 23:27
Brent Kerby
4,699414
4,699414
Could you clarify why $X_1$ and $Y_1$ need to be independent in the second case? Imagine $X: {H, T} to {0, 1}$ records the outcome $x$ of a coin toss, then we produce another random variable $Y: {0, 1} to {0, -1}$ by deterministically defining its value to be $-x$. of How would you define $X+Y$?
– Yibo Yang
Aug 19 '18 at 0:40
add a comment |
Could you clarify why $X_1$ and $Y_1$ need to be independent in the second case? Imagine $X: {H, T} to {0, 1}$ records the outcome $x$ of a coin toss, then we produce another random variable $Y: {0, 1} to {0, -1}$ by deterministically defining its value to be $-x$. of How would you define $X+Y$?
– Yibo Yang
Aug 19 '18 at 0:40
Could you clarify why $X_1$ and $Y_1$ need to be independent in the second case? Imagine $X: {H, T} to {0, 1}$ records the outcome $x$ of a coin toss, then we produce another random variable $Y: {0, 1} to {0, -1}$ by deterministically defining its value to be $-x$. of How would you define $X+Y$?
– Yibo Yang
Aug 19 '18 at 0:40
Could you clarify why $X_1$ and $Y_1$ need to be independent in the second case? Imagine $X: {H, T} to {0, 1}$ records the outcome $x$ of a coin toss, then we produce another random variable $Y: {0, 1} to {0, -1}$ by deterministically defining its value to be $-x$. of How would you define $X+Y$?
– Yibo Yang
Aug 19 '18 at 0:40
add a comment |
It actually defined as an analogue of the product of the polynomial with $X, Y$ independent and P was defined in double dimension
$$(P(x + y) = t) = sum_{I = 0}^t f_x(I)g_x(t - I)$$
answered Dec 11 '18 at 2:59
user577443
92
add a comment |
It actually defined as an analogue of the product of the polynomial with $X, Y$ independent and P was defined in double dimension
$$(P(x + y) = t) = sum_{I = 0}^t f_x(I)g_x(t - I)$$
answered Dec 11 '18 at 2:59
user577443
92
add a comment |
It actually defined as an analogue of the product of the polynomial with $X, Y$ independent and P was defined in double dimension
$$(P(x + y) = t) = sum_{I = 0}^t f_x(I)g_x(t - I)$$
answered Dec 11 '18 at 2:59
user577443
92
It actually defined as an analogue of the product of the polynomial with $X, Y$ independent and P was defined in double dimension
$$(P(x + y) = t) = sum_{I = 0}^t f_x(I)g_x(t - I)$$
answered Dec 11 '18 at 2:59
user577443
92
answered Dec 11 '18 at 2:59
user577443
92
answered Dec 11 '18 at 2:59
user577443
92
answered Dec 11 '18 at 2:59
user577443
92
92
add a comment |
add a comment |
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How do you rigorously define a random variable?
– Crostul
Sep 30 '15 at 19:56
A random variable is X is a mapping from the sample space of the experiment of interest to the real number line. So if I have a method of assigning a real number to every outcome in the sample space, I guess I would call that method a random variable.
– user2474041
Sep 30 '15 at 19:59
1
One needs to define a new sample space which is the direct product of two copies of the old sample space. As it happens, Terry Tao just wrote a great post about the foundations of probability, where this question is answered: terrytao.wordpress.com/2015/09/29/…
– Greg Martin
Sep 30 '15 at 20:27