What is the mathematical interpretation of random variable equation?
$begingroup$
Following is the equation:
$$X = Y + Z.$$
$X, Y, Z$ are random variables. This is a random variable equation. What is the meaning of this equation? Does it mean that if you take any value of $Y$ and any value of $Z$, it should equal $X$? Or does it mean that when $Y$ and $Z$ are added, then $Y+Z$ will have same distribution as $X$?
probability probability-theory probability-distributions random-variables
edited Dec 7 '18 at 2:36
timtfj
1,746418
asked Dec 7 '18 at 2:09
ShiranShiran
82
$endgroup$
|
show 6 more comments
$begingroup$
Following is the equation:
$$X = Y + Z.$$
$X, Y, Z$ are random variables. This is a random variable equation. What is the meaning of this equation? Does it mean that if you take any value of $Y$ and any value of $Z$, it should equal $X$? Or does it mean that when $Y$ and $Z$ are added, then $Y+Z$ will have same distribution as $X$?
probability probability-theory probability-distributions random-variables
edited Dec 7 '18 at 2:36
timtfj
1,746418
asked Dec 7 '18 at 2:09
ShiranShiran
82
$endgroup$
$begingroup$
It is pretty common to add random variables together to get a new random variable. For example, the binomial distribution is a sum of independent Bernoulli trials. So I think your second interpretation is correct.
$endgroup$
– gd1035
Dec 7 '18 at 2:25
$begingroup$
The equation $X=Y+Z$ usually means $X$ is a random variable defined as the sum of $Y$ and $Z$. In other contexts it may mean that a particular value of $X$ is the sum of a particular value of $Y+Z$.and you might ask what is the probability of such an equality. As you can see it all depends on context.
$endgroup$
– herb steinberg
Dec 7 '18 at 2:26
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It means that the random result $X$ equals the sum of the random results $Y$ and $Z.$
$endgroup$
– Will M.
Dec 7 '18 at 2:38
3
$begingroup$
It does not simply mean that $X$ and $Y + Z$ have the same distribution. It means that $X$ is $Y + Z$.
$endgroup$
– littleO
Dec 7 '18 at 2:48
1
$begingroup$
@herbsteinberg "As you can see it all depends on context" No it does not. Why mislead the OP?
$endgroup$
– Did
Dec 7 '18 at 9:38
|
show 6 more comments
$begingroup$
Following is the equation:
$$X = Y + Z.$$
$X, Y, Z$ are random variables. This is a random variable equation. What is the meaning of this equation? Does it mean that if you take any value of $Y$ and any value of $Z$, it should equal $X$? Or does it mean that when $Y$ and $Z$ are added, then $Y+Z$ will have same distribution as $X$?
probability probability-theory probability-distributions random-variables
edited Dec 7 '18 at 2:36
timtfj
1,746418
asked Dec 7 '18 at 2:09
ShiranShiran
82
$endgroup$
Following is the equation:
$$X = Y + Z.$$
$X, Y, Z$ are random variables. This is a random variable equation. What is the meaning of this equation? Does it mean that if you take any value of $Y$ and any value of $Z$, it should equal $X$? Or does it mean that when $Y$ and $Z$ are added, then $Y+Z$ will have same distribution as $X$?
probability probability-theory probability-distributions random-variables
probability probability-theory probability-distributions random-variables
edited Dec 7 '18 at 2:36
timtfj
1,746418
asked Dec 7 '18 at 2:09
ShiranShiran
82
edited Dec 7 '18 at 2:36
timtfj
1,746418
asked Dec 7 '18 at 2:09
ShiranShiran
82
edited Dec 7 '18 at 2:36
timtfj
1,746418
edited Dec 7 '18 at 2:36
timtfj
1,746418
edited Dec 7 '18 at 2:36
timtfj
1,746418
1,746418
asked Dec 7 '18 at 2:09
ShiranShiran
82
asked Dec 7 '18 at 2:09
ShiranShiran
82
asked Dec 7 '18 at 2:09
ShiranShiran
82
82
$begingroup$
It is pretty common to add random variables together to get a new random variable. For example, the binomial distribution is a sum of independent Bernoulli trials. So I think your second interpretation is correct.
$endgroup$
– gd1035
Dec 7 '18 at 2:25
$begingroup$
The equation $X=Y+Z$ usually means $X$ is a random variable defined as the sum of $Y$ and $Z$. In other contexts it may mean that a particular value of $X$ is the sum of a particular value of $Y+Z$.and you might ask what is the probability of such an equality. As you can see it all depends on context.
$endgroup$
– herb steinberg
Dec 7 '18 at 2:26
$begingroup$
It means that the random result $X$ equals the sum of the random results $Y$ and $Z.$
$endgroup$
– Will M.
Dec 7 '18 at 2:38
3
$begingroup$
It does not simply mean that $X$ and $Y + Z$ have the same distribution. It means that $X$ is $Y + Z$.
$endgroup$
– littleO
Dec 7 '18 at 2:48
1
$begingroup$
@herbsteinberg "As you can see it all depends on context" No it does not. Why mislead the OP?
$endgroup$
– Did
Dec 7 '18 at 9:38
|
show 6 more comments
$begingroup$
It is pretty common to add random variables together to get a new random variable. For example, the binomial distribution is a sum of independent Bernoulli trials. So I think your second interpretation is correct.
$endgroup$
– gd1035
Dec 7 '18 at 2:25
$begingroup$
The equation $X=Y+Z$ usually means $X$ is a random variable defined as the sum of $Y$ and $Z$. In other contexts it may mean that a particular value of $X$ is the sum of a particular value of $Y+Z$.and you might ask what is the probability of such an equality. As you can see it all depends on context.
$endgroup$
– herb steinberg
Dec 7 '18 at 2:26
$begingroup$
It means that the random result $X$ equals the sum of the random results $Y$ and $Z.$
$endgroup$
– Will M.
Dec 7 '18 at 2:38
3
$begingroup$
It does not simply mean that $X$ and $Y + Z$ have the same distribution. It means that $X$ is $Y + Z$.
$endgroup$
– littleO
Dec 7 '18 at 2:48
1
$begingroup$
@herbsteinberg "As you can see it all depends on context" No it does not. Why mislead the OP?
$endgroup$
– Did
Dec 7 '18 at 9:38
$begingroup$
It is pretty common to add random variables together to get a new random variable. For example, the binomial distribution is a sum of independent Bernoulli trials. So I think your second interpretation is correct.
$endgroup$
– gd1035
Dec 7 '18 at 2:25
$begingroup$
It is pretty common to add random variables together to get a new random variable. For example, the binomial distribution is a sum of independent Bernoulli trials. So I think your second interpretation is correct.
$endgroup$
– gd1035
Dec 7 '18 at 2:25
$begingroup$
The equation $X=Y+Z$ usually means $X$ is a random variable defined as the sum of $Y$ and $Z$. In other contexts it may mean that a particular value of $X$ is the sum of a particular value of $Y+Z$.and you might ask what is the probability of such an equality. As you can see it all depends on context.
$endgroup$
– herb steinberg
Dec 7 '18 at 2:26
$begingroup$
The equation $X=Y+Z$ usually means $X$ is a random variable defined as the sum of $Y$ and $Z$. In other contexts it may mean that a particular value of $X$ is the sum of a particular value of $Y+Z$.and you might ask what is the probability of such an equality. As you can see it all depends on context.
$endgroup$
– herb steinberg
Dec 7 '18 at 2:26
$begingroup$
It means that the random result $X$ equals the sum of the random results $Y$ and $Z.$
$endgroup$
– Will M.
Dec 7 '18 at 2:38
$begingroup$
It means that the random result $X$ equals the sum of the random results $Y$ and $Z.$
$endgroup$
– Will M.
Dec 7 '18 at 2:38
3
3
$begingroup$
It does not simply mean that $X$ and $Y + Z$ have the same distribution. It means that $X$ is $Y + Z$.
$endgroup$
– littleO
Dec 7 '18 at 2:48
$begingroup$
It does not simply mean that $X$ and $Y + Z$ have the same distribution. It means that $X$ is $Y + Z$.
$endgroup$
– littleO
Dec 7 '18 at 2:48
1
1
$begingroup$
@herbsteinberg "As you can see it all depends on context" No it does not. Why mislead the OP?
$endgroup$
– Did
Dec 7 '18 at 9:38
$begingroup$
@herbsteinberg "As you can see it all depends on context" No it does not. Why mislead the OP?
$endgroup$
– Did
Dec 7 '18 at 9:38
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Random variables are defined as measurable functions from a probability space $Omega$ into some target space. Without any further context, I interpret that equality as an equality of functions, e.g. $X(omega) = Y(omega) + Z(omega)$ for every $omega in Omega$. Quite often in probability, we only care about equality on a set of measure 1, and it is common to see $X = Y+Z$ a.s. where a.s. is an abbreviation for almost surely, which means that the functions $X$ and $Y+Z$ might differ on a tiny set of measure 0, but we don't care.
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
Thanks for the explanation...
$endgroup$
– Shiran
Dec 8 '18 at 2:49
add a comment |
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1 Answer
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$begingroup$
Random variables are defined as measurable functions from a probability space $Omega$ into some target space. Without any further context, I interpret that equality as an equality of functions, e.g. $X(omega) = Y(omega) + Z(omega)$ for every $omega in Omega$. Quite often in probability, we only care about equality on a set of measure 1, and it is common to see $X = Y+Z$ a.s. where a.s. is an abbreviation for almost surely, which means that the functions $X$ and $Y+Z$ might differ on a tiny set of measure 0, but we don't care.
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
Thanks for the explanation...
$endgroup$
– Shiran
Dec 8 '18 at 2:49
add a comment |
$begingroup$
Random variables are defined as measurable functions from a probability space $Omega$ into some target space. Without any further context, I interpret that equality as an equality of functions, e.g. $X(omega) = Y(omega) + Z(omega)$ for every $omega in Omega$. Quite often in probability, we only care about equality on a set of measure 1, and it is common to see $X = Y+Z$ a.s. where a.s. is an abbreviation for almost surely, which means that the functions $X$ and $Y+Z$ might differ on a tiny set of measure 0, but we don't care.
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
Thanks for the explanation...
$endgroup$
– Shiran
Dec 8 '18 at 2:49
add a comment |
$begingroup$
Random variables are defined as measurable functions from a probability space $Omega$ into some target space. Without any further context, I interpret that equality as an equality of functions, e.g. $X(omega) = Y(omega) + Z(omega)$ for every $omega in Omega$. Quite often in probability, we only care about equality on a set of measure 1, and it is common to see $X = Y+Z$ a.s. where a.s. is an abbreviation for almost surely, which means that the functions $X$ and $Y+Z$ might differ on a tiny set of measure 0, but we don't care.
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
$endgroup$
Random variables are defined as measurable functions from a probability space $Omega$ into some target space. Without any further context, I interpret that equality as an equality of functions, e.g. $X(omega) = Y(omega) + Z(omega)$ for every $omega in Omega$. Quite often in probability, we only care about equality on a set of measure 1, and it is common to see $X = Y+Z$ a.s. where a.s. is an abbreviation for almost surely, which means that the functions $X$ and $Y+Z$ might differ on a tiny set of measure 0, but we don't care.
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
answered Dec 7 '18 at 5:47
zoidbergzoidberg
1,065113
1,065113
$begingroup$
Thanks for the explanation...
$endgroup$
– Shiran
Dec 8 '18 at 2:49
add a comment |
$begingroup$
Thanks for the explanation...
$endgroup$
– Shiran
Dec 8 '18 at 2:49
$begingroup$
Thanks for the explanation...
$endgroup$
– Shiran
Dec 8 '18 at 2:49
$begingroup$
Thanks for the explanation...
$endgroup$
– Shiran
Dec 8 '18 at 2:49
add a comment |
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$begingroup$
It is pretty common to add random variables together to get a new random variable. For example, the binomial distribution is a sum of independent Bernoulli trials. So I think your second interpretation is correct.
$endgroup$
– gd1035
Dec 7 '18 at 2:25
$begingroup$
The equation $X=Y+Z$ usually means $X$ is a random variable defined as the sum of $Y$ and $Z$. In other contexts it may mean that a particular value of $X$ is the sum of a particular value of $Y+Z$.and you might ask what is the probability of such an equality. As you can see it all depends on context.
$endgroup$
– herb steinberg
Dec 7 '18 at 2:26
$begingroup$
It means that the random result $X$ equals the sum of the random results $Y$ and $Z.$
$endgroup$
– Will M.
Dec 7 '18 at 2:38
3
$begingroup$
It does not simply mean that $X$ and $Y + Z$ have the same distribution. It means that $X$ is $Y + Z$.
$endgroup$
– littleO
Dec 7 '18 at 2:48
1
$begingroup$
@herbsteinberg "As you can see it all depends on context" No it does not. Why mislead the OP?
$endgroup$
– Did
Dec 7 '18 at 9:38