Any difference between $Pr(A=1, B=1)$ and $Pr(A=1 cap B=1)$?
$begingroup$
Are the two notations interchangable? Do they all mean the joint probability of $A=1$ and $B=1$? Maybe $Pr(A=1 cap B=1)$ is illegal? We can only write $Pr(Acap B)$?
probability
edited Jan 15 at 0:17
gt6989b
36k22557
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
$endgroup$
add a comment |
$begingroup$
Are the two notations interchangable? Do they all mean the joint probability of $A=1$ and $B=1$? Maybe $Pr(A=1 cap B=1)$ is illegal? We can only write $Pr(Acap B)$?
probability
edited Jan 15 at 0:17
gt6989b
36k22557
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
$endgroup$
6
$begingroup$
Yes, it means the same thing. The comma is stating another condition that must hold, so: $$P[{A=1}cap {B=1}] = P[A=1, B=1]$$ where $A$ and $B$ are random variabes and ${A=1}$ and ${B=1}$ are events.
$endgroup$
– Michael
Jan 14 at 23:01
add a comment |
$begingroup$
Are the two notations interchangable? Do they all mean the joint probability of $A=1$ and $B=1$? Maybe $Pr(A=1 cap B=1)$ is illegal? We can only write $Pr(Acap B)$?
probability
edited Jan 15 at 0:17
gt6989b
36k22557
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
$endgroup$
Are the two notations interchangable? Do they all mean the joint probability of $A=1$ and $B=1$? Maybe $Pr(A=1 cap B=1)$ is illegal? We can only write $Pr(Acap B)$?
probability
probability
edited Jan 15 at 0:17
gt6989b
36k22557
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
edited Jan 15 at 0:17
gt6989b
36k22557
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
edited Jan 15 at 0:17
gt6989b
36k22557
edited Jan 15 at 0:17
gt6989b
36k22557
edited Jan 15 at 0:17
gt6989b
36k22557
36k22557
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
asked Jan 14 at 23:00
Lerner ZhangLerner Zhang
314219
314219
6
$begingroup$
Yes, it means the same thing. The comma is stating another condition that must hold, so: $$P[{A=1}cap {B=1}] = P[A=1, B=1]$$ where $A$ and $B$ are random variabes and ${A=1}$ and ${B=1}$ are events.
$endgroup$
– Michael
Jan 14 at 23:01
add a comment |
6
$begingroup$
Yes, it means the same thing. The comma is stating another condition that must hold, so: $$P[{A=1}cap {B=1}] = P[A=1, B=1]$$ where $A$ and $B$ are random variabes and ${A=1}$ and ${B=1}$ are events.
$endgroup$
– Michael
Jan 14 at 23:01
6
6
$begingroup$
Yes, it means the same thing. The comma is stating another condition that must hold, so: $$P[{A=1}cap {B=1}] = P[A=1, B=1]$$ where $A$ and $B$ are random variabes and ${A=1}$ and ${B=1}$ are events.
$endgroup$
– Michael
Jan 14 at 23:01
$begingroup$
Yes, it means the same thing. The comma is stating another condition that must hold, so: $$P[{A=1}cap {B=1}] = P[A=1, B=1]$$ where $A$ and $B$ are random variabes and ${A=1}$ and ${B=1}$ are events.
$endgroup$
– Michael
Jan 14 at 23:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This style of notation is pretty frequent in probability.
When you see something like, where $X$ is a random variable,
$$P(X = 5)$$
you should think of "$X = 5$" in the parenthesis as the set
$$ { omega in Omega | X(omega) = 5 }$$
where $Omega$ is the sample space of $X$ (i.e., $X : Omega rightarrow mathbb{R}$). With this style of notation, commas are used to mean intersection, so
$$ P(X = 5, Y le 7)$$
is equivalent to
$$P({ omega in Omega|X(omega) = 5} cap {omega in Omega|Y(omega) le 7})$$
Notice how the former, more concise notation is easier to read and understand, while the latter, although still formally correct, is quite cluttered.
Imagine, without the convenient notation, trying to describe something like
$$ P(X = 5, Y = 12, A = 3, B le 12)$$
Ouch.
Using this notation, it is correct, as mentioned in the comments, that
$$P(A = 1 cap B = 1) = P( A = 1, B = 1)$$
although the latter is a bit more readable and more often used.
Also,
$$P(Acap B)$$
in your case probably isn't what you want, as I'm assuming your $A$ and $B$ are random variables, so their intersection is not useful in this context (it's hard to make sense of their intersection in this context, as it's a subset of $Omega times mathbb{R}$).
answered Jan 15 at 0:35
MetricMetric
1,23659
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add a comment |
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1 Answer
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$begingroup$
This style of notation is pretty frequent in probability.
When you see something like, where $X$ is a random variable,
$$P(X = 5)$$
you should think of "$X = 5$" in the parenthesis as the set
$$ { omega in Omega | X(omega) = 5 }$$
where $Omega$ is the sample space of $X$ (i.e., $X : Omega rightarrow mathbb{R}$). With this style of notation, commas are used to mean intersection, so
$$ P(X = 5, Y le 7)$$
is equivalent to
$$P({ omega in Omega|X(omega) = 5} cap {omega in Omega|Y(omega) le 7})$$
Notice how the former, more concise notation is easier to read and understand, while the latter, although still formally correct, is quite cluttered.
Imagine, without the convenient notation, trying to describe something like
$$ P(X = 5, Y = 12, A = 3, B le 12)$$
Ouch.
Using this notation, it is correct, as mentioned in the comments, that
$$P(A = 1 cap B = 1) = P( A = 1, B = 1)$$
although the latter is a bit more readable and more often used.
Also,
$$P(Acap B)$$
in your case probably isn't what you want, as I'm assuming your $A$ and $B$ are random variables, so their intersection is not useful in this context (it's hard to make sense of their intersection in this context, as it's a subset of $Omega times mathbb{R}$).
answered Jan 15 at 0:35
MetricMetric
1,23659
$endgroup$
add a comment |
$begingroup$
This style of notation is pretty frequent in probability.
When you see something like, where $X$ is a random variable,
$$P(X = 5)$$
you should think of "$X = 5$" in the parenthesis as the set
$$ { omega in Omega | X(omega) = 5 }$$
where $Omega$ is the sample space of $X$ (i.e., $X : Omega rightarrow mathbb{R}$). With this style of notation, commas are used to mean intersection, so
$$ P(X = 5, Y le 7)$$
is equivalent to
$$P({ omega in Omega|X(omega) = 5} cap {omega in Omega|Y(omega) le 7})$$
Notice how the former, more concise notation is easier to read and understand, while the latter, although still formally correct, is quite cluttered.
Imagine, without the convenient notation, trying to describe something like
$$ P(X = 5, Y = 12, A = 3, B le 12)$$
Ouch.
Using this notation, it is correct, as mentioned in the comments, that
$$P(A = 1 cap B = 1) = P( A = 1, B = 1)$$
although the latter is a bit more readable and more often used.
Also,
$$P(Acap B)$$
in your case probably isn't what you want, as I'm assuming your $A$ and $B$ are random variables, so their intersection is not useful in this context (it's hard to make sense of their intersection in this context, as it's a subset of $Omega times mathbb{R}$).
answered Jan 15 at 0:35
MetricMetric
1,23659
$endgroup$
add a comment |
$begingroup$
This style of notation is pretty frequent in probability.
When you see something like, where $X$ is a random variable,
$$P(X = 5)$$
you should think of "$X = 5$" in the parenthesis as the set
$$ { omega in Omega | X(omega) = 5 }$$
where $Omega$ is the sample space of $X$ (i.e., $X : Omega rightarrow mathbb{R}$). With this style of notation, commas are used to mean intersection, so
$$ P(X = 5, Y le 7)$$
is equivalent to
$$P({ omega in Omega|X(omega) = 5} cap {omega in Omega|Y(omega) le 7})$$
Notice how the former, more concise notation is easier to read and understand, while the latter, although still formally correct, is quite cluttered.
Imagine, without the convenient notation, trying to describe something like
$$ P(X = 5, Y = 12, A = 3, B le 12)$$
Ouch.
Using this notation, it is correct, as mentioned in the comments, that
$$P(A = 1 cap B = 1) = P( A = 1, B = 1)$$
although the latter is a bit more readable and more often used.
Also,
$$P(Acap B)$$
in your case probably isn't what you want, as I'm assuming your $A$ and $B$ are random variables, so their intersection is not useful in this context (it's hard to make sense of their intersection in this context, as it's a subset of $Omega times mathbb{R}$).
answered Jan 15 at 0:35
MetricMetric
1,23659
$endgroup$
This style of notation is pretty frequent in probability.
When you see something like, where $X$ is a random variable,
$$P(X = 5)$$
you should think of "$X = 5$" in the parenthesis as the set
$$ { omega in Omega | X(omega) = 5 }$$
where $Omega$ is the sample space of $X$ (i.e., $X : Omega rightarrow mathbb{R}$). With this style of notation, commas are used to mean intersection, so
$$ P(X = 5, Y le 7)$$
is equivalent to
$$P({ omega in Omega|X(omega) = 5} cap {omega in Omega|Y(omega) le 7})$$
Notice how the former, more concise notation is easier to read and understand, while the latter, although still formally correct, is quite cluttered.
Imagine, without the convenient notation, trying to describe something like
$$ P(X = 5, Y = 12, A = 3, B le 12)$$
Ouch.
Using this notation, it is correct, as mentioned in the comments, that
$$P(A = 1 cap B = 1) = P( A = 1, B = 1)$$
although the latter is a bit more readable and more often used.
Also,
$$P(Acap B)$$
in your case probably isn't what you want, as I'm assuming your $A$ and $B$ are random variables, so their intersection is not useful in this context (it's hard to make sense of their intersection in this context, as it's a subset of $Omega times mathbb{R}$).
answered Jan 15 at 0:35
MetricMetric
1,23659
edited Jan 15 at 1:38
answered Jan 15 at 0:35
MetricMetric
1,23659
answered Jan 15 at 0:35
MetricMetric
1,23659
answered Jan 15 at 0:35
MetricMetric
1,23659
1,23659
add a comment |
add a comment |
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$begingroup$
Yes, it means the same thing. The comma is stating another condition that must hold, so: $$P[{A=1}cap {B=1}] = P[A=1, B=1]$$ where $A$ and $B$ are random variabes and ${A=1}$ and ${B=1}$ are events.
$endgroup$
– Michael
Jan 14 at 23:01