What does this notation ($Y = mathrm dx$) in probability mean?
0
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I'm reading this research article and can't seem to understand what this notation mean,
$P(T = i) = int P(X>x)P(Y = mathrm dx) $ where $T$, $X$, $Y$ are all obviously random variables. Irrespective of their particular definition, what would a RV equal to $mathrm dx$ possibly mean ?
probability-theory self-learning
edited Dec 6 '18 at 14:40
Christoph
12k1642
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
$endgroup$
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show 2 more comments
0
$begingroup$
I'm reading this research article and can't seem to understand what this notation mean,
$P(T = i) = int P(X>x)P(Y = mathrm dx) $ where $T$, $X$, $Y$ are all obviously random variables. Irrespective of their particular definition, what would a RV equal to $mathrm dx$ possibly mean ?
probability-theory self-learning
edited Dec 6 '18 at 14:40
Christoph
12k1642
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
$endgroup$
$begingroup$
I don't know. Maybe $int f(x)P(X=dx)$ is meant to be a notation for $int f(x)dF_X(x)$. The same integral can also be written as $int f(x)P_X(dx)$ and that comes close to $int f(x)P(X=dx)$.
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– drhab
Dec 6 '18 at 14:35
$begingroup$
Does $P(xleq Y<x+dx)$ make more sense to you than $P(Y = dx)$?
$endgroup$
– Arthur
Dec 6 '18 at 14:36
$begingroup$
You are integrating the function $xmapsto P(X>x)$ with respect to the probability measure of $Y$.
$endgroup$
– Christoph
Dec 6 '18 at 14:36
2
$begingroup$
Terrible notation. Even $P(Yin dx)$ would be better, but only slightly so. Yes the authors mean $dP_Y(x)$.
$endgroup$
– Did
Dec 6 '18 at 14:38
$begingroup$
Right, I do get some idea regarding it. Also noticed something else, since we are integrating $P(X>x)$, it must mean that this is not a constant rather a RV under the probability measure of Y ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 14:54
|
show 2 more comments
0
0
0
$begingroup$
I'm reading this research article and can't seem to understand what this notation mean,
$P(T = i) = int P(X>x)P(Y = mathrm dx) $ where $T$, $X$, $Y$ are all obviously random variables. Irrespective of their particular definition, what would a RV equal to $mathrm dx$ possibly mean ?
probability-theory self-learning
edited Dec 6 '18 at 14:40
Christoph
12k1642
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
$endgroup$
I'm reading this research article and can't seem to understand what this notation mean,
$P(T = i) = int P(X>x)P(Y = mathrm dx) $ where $T$, $X$, $Y$ are all obviously random variables. Irrespective of their particular definition, what would a RV equal to $mathrm dx$ possibly mean ?
probability-theory self-learning
probability-theory self-learning
edited Dec 6 '18 at 14:40
Christoph
12k1642
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
edited Dec 6 '18 at 14:40
Christoph
12k1642
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
edited Dec 6 '18 at 14:40
Christoph
12k1642
edited Dec 6 '18 at 14:40
Christoph
12k1642
edited Dec 6 '18 at 14:40
Christoph
12k1642
12k1642
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
asked Dec 6 '18 at 14:30
vortex_sparrowvortex_sparrow
11
11
$begingroup$
I don't know. Maybe $int f(x)P(X=dx)$ is meant to be a notation for $int f(x)dF_X(x)$. The same integral can also be written as $int f(x)P_X(dx)$ and that comes close to $int f(x)P(X=dx)$.
$endgroup$
– drhab
Dec 6 '18 at 14:35
$begingroup$
Does $P(xleq Y<x+dx)$ make more sense to you than $P(Y = dx)$?
$endgroup$
– Arthur
Dec 6 '18 at 14:36
$begingroup$
You are integrating the function $xmapsto P(X>x)$ with respect to the probability measure of $Y$.
$endgroup$
– Christoph
Dec 6 '18 at 14:36
2
$begingroup$
Terrible notation. Even $P(Yin dx)$ would be better, but only slightly so. Yes the authors mean $dP_Y(x)$.
$endgroup$
– Did
Dec 6 '18 at 14:38
$begingroup$
Right, I do get some idea regarding it. Also noticed something else, since we are integrating $P(X>x)$, it must mean that this is not a constant rather a RV under the probability measure of Y ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 14:54
|
show 2 more comments
$begingroup$
I don't know. Maybe $int f(x)P(X=dx)$ is meant to be a notation for $int f(x)dF_X(x)$. The same integral can also be written as $int f(x)P_X(dx)$ and that comes close to $int f(x)P(X=dx)$.
$endgroup$
– drhab
Dec 6 '18 at 14:35
$begingroup$
Does $P(xleq Y<x+dx)$ make more sense to you than $P(Y = dx)$?
$endgroup$
– Arthur
Dec 6 '18 at 14:36
$begingroup$
You are integrating the function $xmapsto P(X>x)$ with respect to the probability measure of $Y$.
$endgroup$
– Christoph
Dec 6 '18 at 14:36
2
$begingroup$
Terrible notation. Even $P(Yin dx)$ would be better, but only slightly so. Yes the authors mean $dP_Y(x)$.
$endgroup$
– Did
Dec 6 '18 at 14:38
$begingroup$
Right, I do get some idea regarding it. Also noticed something else, since we are integrating $P(X>x)$, it must mean that this is not a constant rather a RV under the probability measure of Y ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 14:54
$begingroup$
I don't know. Maybe $int f(x)P(X=dx)$ is meant to be a notation for $int f(x)dF_X(x)$. The same integral can also be written as $int f(x)P_X(dx)$ and that comes close to $int f(x)P(X=dx)$.
$endgroup$
– drhab
Dec 6 '18 at 14:35
$begingroup$
I don't know. Maybe $int f(x)P(X=dx)$ is meant to be a notation for $int f(x)dF_X(x)$. The same integral can also be written as $int f(x)P_X(dx)$ and that comes close to $int f(x)P(X=dx)$.
$endgroup$
– drhab
Dec 6 '18 at 14:35
$begingroup$
Does $P(xleq Y<x+dx)$ make more sense to you than $P(Y = dx)$?
$endgroup$
– Arthur
Dec 6 '18 at 14:36
$begingroup$
Does $P(xleq Y<x+dx)$ make more sense to you than $P(Y = dx)$?
$endgroup$
– Arthur
Dec 6 '18 at 14:36
$begingroup$
You are integrating the function $xmapsto P(X>x)$ with respect to the probability measure of $Y$.
$endgroup$
– Christoph
Dec 6 '18 at 14:36
$begingroup$
You are integrating the function $xmapsto P(X>x)$ with respect to the probability measure of $Y$.
$endgroup$
– Christoph
Dec 6 '18 at 14:36
2
2
$begingroup$
Terrible notation. Even $P(Yin dx)$ would be better, but only slightly so. Yes the authors mean $dP_Y(x)$.
$endgroup$
– Did
Dec 6 '18 at 14:38
$begingroup$
Terrible notation. Even $P(Yin dx)$ would be better, but only slightly so. Yes the authors mean $dP_Y(x)$.
$endgroup$
– Did
Dec 6 '18 at 14:38
$begingroup$
Right, I do get some idea regarding it. Also noticed something else, since we are integrating $P(X>x)$, it must mean that this is not a constant rather a RV under the probability measure of Y ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 14:54
$begingroup$
Right, I do get some idea regarding it. Also noticed something else, since we are integrating $P(X>x)$, it must mean that this is not a constant rather a RV under the probability measure of Y ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 14:54
|
show 2 more comments
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$begingroup$
I don't know. Maybe $int f(x)P(X=dx)$ is meant to be a notation for $int f(x)dF_X(x)$. The same integral can also be written as $int f(x)P_X(dx)$ and that comes close to $int f(x)P(X=dx)$.
$endgroup$
– drhab
Dec 6 '18 at 14:35
$begingroup$
Does $P(xleq Y<x+dx)$ make more sense to you than $P(Y = dx)$?
$endgroup$
– Arthur
Dec 6 '18 at 14:36
$begingroup$
You are integrating the function $xmapsto P(X>x)$ with respect to the probability measure of $Y$.
$endgroup$
– Christoph
Dec 6 '18 at 14:36
2
$begingroup$
Terrible notation. Even $P(Yin dx)$ would be better, but only slightly so. Yes the authors mean $dP_Y(x)$.
$endgroup$
– Did
Dec 6 '18 at 14:38
$begingroup$
Right, I do get some idea regarding it. Also noticed something else, since we are integrating $P(X>x)$, it must mean that this is not a constant rather a RV under the probability measure of Y ?
$endgroup$
– vortex_sparrow
Dec 6 '18 at 14:54