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Continuous distribution additivity contradiction?
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If a random variable $X$ is continuous, then by definition,
$$(1) quad P(X=x) = 0, forall x in mathbbR,.$$
Also, by additivity,
$$(2) quad P(X le c) = sum_x in mathbbR le c P(X = x),.$$
Doesn't $(1)$ imply that $(2)$ must equal $0$?
probability random-variables
asked Jul 31 at 21:51
nakamin
112
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up vote
1
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If a random variable $X$ is continuous, then by definition,
$$(1) quad P(X=x) = 0, forall x in mathbbR,.$$
Also, by additivity,
$$(2) quad P(X le c) = sum_x in mathbbR le c P(X = x),.$$
Doesn't $(1)$ imply that $(2)$ must equal $0$?
probability random-variables
asked Jul 31 at 21:51
nakamin
112
5
No, that is not the implication - you are taking an uncountable sum, and the third axiom of probability on additivity only deals with countable sums
– Henry
Jul 31 at 21:56
2
That's where integral enters the picture. It's $$P(X<c) =P(Xle c) =int_-infty^c f(x), dx$$ where $f$ is the density function of $X$ (basically defined by this formula).
– Berci
Jul 31 at 23:30
1
For example, consider $X sim mathsfExp(1).$ You have $E(X) = int_0^infty x,e^-x,dx = 1$ and $P(X <c) = int_0^c e^-x,dx = 1 - e^c,$ for $c > 0.$
– BruceET
Aug 1 at 4:26
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If a random variable $X$ is continuous, then by definition,
$$(1) quad P(X=x) = 0, forall x in mathbbR,.$$
Also, by additivity,
$$(2) quad P(X le c) = sum_x in mathbbR le c P(X = x),.$$
Doesn't $(1)$ imply that $(2)$ must equal $0$?
probability random-variables
asked Jul 31 at 21:51
nakamin
112
If a random variable $X$ is continuous, then by definition,
$$(1) quad P(X=x) = 0, forall x in mathbbR,.$$
Also, by additivity,
$$(2) quad P(X le c) = sum_x in mathbbR le c P(X = x),.$$
Doesn't $(1)$ imply that $(2)$ must equal $0$?
probability random-variables
asked Jul 31 at 21:51
nakamin
112
asked Jul 31 at 21:51
nakamin
112
asked Jul 31 at 21:51
nakamin
112
asked Jul 31 at 21:51
nakamin
112
112
5
No, that is not the implication - you are taking an uncountable sum, and the third axiom of probability on additivity only deals with countable sums
– Henry
Jul 31 at 21:56
2
That's where integral enters the picture. It's $$P(X<c) =P(Xle c) =int_-infty^c f(x), dx$$ where $f$ is the density function of $X$ (basically defined by this formula).
– Berci
Jul 31 at 23:30
1
For example, consider $X sim mathsfExp(1).$ You have $E(X) = int_0^infty x,e^-x,dx = 1$ and $P(X <c) = int_0^c e^-x,dx = 1 - e^c,$ for $c > 0.$
– BruceET
Aug 1 at 4:26
add a comment |Â
5
No, that is not the implication - you are taking an uncountable sum, and the third axiom of probability on additivity only deals with countable sums
– Henry
Jul 31 at 21:56
2
That's where integral enters the picture. It's $$P(X<c) =P(Xle c) =int_-infty^c f(x), dx$$ where $f$ is the density function of $X$ (basically defined by this formula).
– Berci
Jul 31 at 23:30
1
For example, consider $X sim mathsfExp(1).$ You have $E(X) = int_0^infty x,e^-x,dx = 1$ and $P(X <c) = int_0^c e^-x,dx = 1 - e^c,$ for $c > 0.$
– BruceET
Aug 1 at 4:26
5
5
No, that is not the implication - you are taking an uncountable sum, and the third axiom of probability on additivity only deals with countable sums
– Henry
Jul 31 at 21:56
No, that is not the implication - you are taking an uncountable sum, and the third axiom of probability on additivity only deals with countable sums
– Henry
Jul 31 at 21:56
2
2
That's where integral enters the picture. It's $$P(X<c) =P(Xle c) =int_-infty^c f(x), dx$$ where $f$ is the density function of $X$ (basically defined by this formula).
– Berci
Jul 31 at 23:30
That's where integral enters the picture. It's $$P(X<c) =P(Xle c) =int_-infty^c f(x), dx$$ where $f$ is the density function of $X$ (basically defined by this formula).
– Berci
Jul 31 at 23:30
1
1
For example, consider $X sim mathsfExp(1).$ You have $E(X) = int_0^infty x,e^-x,dx = 1$ and $P(X <c) = int_0^c e^-x,dx = 1 - e^c,$ for $c > 0.$
– BruceET
Aug 1 at 4:26
For example, consider $X sim mathsfExp(1).$ You have $E(X) = int_0^infty x,e^-x,dx = 1$ and $P(X <c) = int_0^c e^-x,dx = 1 - e^c,$ for $c > 0.$
– BruceET
Aug 1 at 4:26
add a comment |Â
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5
No, that is not the implication - you are taking an uncountable sum, and the third axiom of probability on additivity only deals with countable sums
– Henry
Jul 31 at 21:56
2
That's where integral enters the picture. It's $$P(X<c) =P(Xle c) =int_-infty^c f(x), dx$$ where $f$ is the density function of $X$ (basically defined by this formula).
– Berci
Jul 31 at 23:30
1
For example, consider $X sim mathsfExp(1).$ You have $E(X) = int_0^infty x,e^-x,dx = 1$ and $P(X <c) = int_0^c e^-x,dx = 1 - e^c,$ for $c > 0.$
– BruceET
Aug 1 at 4:26