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Any distribution function can be expressed as a sum of absolutely continuous, singular, and discrete...
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I am trying to show that any distribution function (increasing, right continuous, $F(-infty) = 0$, $F(infty) = 1$) can be uniquely expressed as a sum of an absolutely continuous d.f., a singular continuous d.f., and a discrete d.f.
Let $F'$ be the derivative of $F$ where it exists an let the AC part be
$$g(x) = int_{-infty}^x F'(t)dt.$$
We have $g'(x) = F'(x)$ almost everywhere and thus the singular part is
$$h(x) = F(x)-g(x)$$
so that $h'(x) = 0$ almost everywhere.
Where is the discrete part, and how do we show this is a unique decomposition? How can one prove this without using extensive measure theory?
real-analysis probability-theory
asked Jan 29 at 23:01
user20354139user20354139
498211
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add a comment |
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I am trying to show that any distribution function (increasing, right continuous, $F(-infty) = 0$, $F(infty) = 1$) can be uniquely expressed as a sum of an absolutely continuous d.f., a singular continuous d.f., and a discrete d.f.
Let $F'$ be the derivative of $F$ where it exists an let the AC part be
$$g(x) = int_{-infty}^x F'(t)dt.$$
We have $g'(x) = F'(x)$ almost everywhere and thus the singular part is
$$h(x) = F(x)-g(x)$$
so that $h'(x) = 0$ almost everywhere.
Where is the discrete part, and how do we show this is a unique decomposition? How can one prove this without using extensive measure theory?
real-analysis probability-theory
asked Jan 29 at 23:01
user20354139user20354139
498211
$endgroup$
$begingroup$
See Theorem 1.3.2 of Chung's "A Course in Probability Theory'.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:19
add a comment |
$begingroup$
I am trying to show that any distribution function (increasing, right continuous, $F(-infty) = 0$, $F(infty) = 1$) can be uniquely expressed as a sum of an absolutely continuous d.f., a singular continuous d.f., and a discrete d.f.
Let $F'$ be the derivative of $F$ where it exists an let the AC part be
$$g(x) = int_{-infty}^x F'(t)dt.$$
We have $g'(x) = F'(x)$ almost everywhere and thus the singular part is
$$h(x) = F(x)-g(x)$$
so that $h'(x) = 0$ almost everywhere.
Where is the discrete part, and how do we show this is a unique decomposition? How can one prove this without using extensive measure theory?
real-analysis probability-theory
asked Jan 29 at 23:01
user20354139user20354139
498211
$endgroup$
I am trying to show that any distribution function (increasing, right continuous, $F(-infty) = 0$, $F(infty) = 1$) can be uniquely expressed as a sum of an absolutely continuous d.f., a singular continuous d.f., and a discrete d.f.
Let $F'$ be the derivative of $F$ where it exists an let the AC part be
$$g(x) = int_{-infty}^x F'(t)dt.$$
We have $g'(x) = F'(x)$ almost everywhere and thus the singular part is
$$h(x) = F(x)-g(x)$$
so that $h'(x) = 0$ almost everywhere.
Where is the discrete part, and how do we show this is a unique decomposition? How can one prove this without using extensive measure theory?
real-analysis probability-theory
real-analysis probability-theory
asked Jan 29 at 23:01
user20354139user20354139
498211
asked Jan 29 at 23:01
user20354139user20354139
498211
asked Jan 29 at 23:01
user20354139user20354139
498211
asked Jan 29 at 23:01
user20354139user20354139
498211
asked Jan 29 at 23:01
user20354139user20354139
498211
498211
$begingroup$
See Theorem 1.3.2 of Chung's "A Course in Probability Theory'.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:19
add a comment |
$begingroup$
See Theorem 1.3.2 of Chung's "A Course in Probability Theory'.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:19
$begingroup$
See Theorem 1.3.2 of Chung's "A Course in Probability Theory'.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:19
$begingroup$
See Theorem 1.3.2 of Chung's "A Course in Probability Theory'.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:19
add a comment |
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$begingroup$
See Theorem 1.3.2 of Chung's "A Course in Probability Theory'.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:19