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$X$ independent of itself, then $exists a in Bbb R: Bbb P[X=a] = 1$
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Let $X$ be a random variable that is independent of itself. I want to show that $exists a in Bbb R: Bbb P[X=a] = 1$.
Let $t in Bbb R$ be arbitrary. Then, $Bbb P[X le t] = Bbb P[X le t cap X le t] = (Bbb P[X le t])^2$, which implies that $Bbb P[X le t] in 0,1$. Especially there exists one $s in Bbb R: Bbb P[X le s] = 1 $.
Next, I decompose $X le s$ into $lfloorsrfloor < X le s cup bigcup_k=1^infty lfloorsrfloor -(k+1) < X le lfloorsrfloor -k$ and since they are disjoint only one of these sets has probability $1$. Continuing this idea, we get a monotone decreasing sequence of sets $(A_n)_n in Bbb N$ with $Bbb P[A_n] = 1$ for all $n in Bbb N$ and $bigcap_n in Bbb N A_n = X=a$ for some $a in Bbb R$. From the continuity of the measure we obtain $Bbb P[X=a]$ =1.
Assume, there exists another $a' in Bbb R: Bbb P[X=a'] = 1$. Then $Bbb P[X = a cup X = a'] = Bbb P[X=a] + Bbb P[X=a'] = 2$ leading to a contradiction.
probability-theory proof-verification
asked Jul 22 at 16:57
Pazu
359213
add a comment |Â
up vote
0
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Let $X$ be a random variable that is independent of itself. I want to show that $exists a in Bbb R: Bbb P[X=a] = 1$.
Let $t in Bbb R$ be arbitrary. Then, $Bbb P[X le t] = Bbb P[X le t cap X le t] = (Bbb P[X le t])^2$, which implies that $Bbb P[X le t] in 0,1$. Especially there exists one $s in Bbb R: Bbb P[X le s] = 1 $.
Next, I decompose $X le s$ into $lfloorsrfloor < X le s cup bigcup_k=1^infty lfloorsrfloor -(k+1) < X le lfloorsrfloor -k$ and since they are disjoint only one of these sets has probability $1$. Continuing this idea, we get a monotone decreasing sequence of sets $(A_n)_n in Bbb N$ with $Bbb P[A_n] = 1$ for all $n in Bbb N$ and $bigcap_n in Bbb N A_n = X=a$ for some $a in Bbb R$. From the continuity of the measure we obtain $Bbb P[X=a]$ =1.
Assume, there exists another $a' in Bbb R: Bbb P[X=a'] = 1$. Then $Bbb P[X = a cup X = a'] = Bbb P[X=a] + Bbb P[X=a'] = 2$ leading to a contradiction.
probability-theory proof-verification
asked Jul 22 at 16:57
Pazu
359213
1
You might start right away (without $s$) with $Bbb R=bigcup_kinBbb Z[k,k+1[$ and note that exactly one of the interval has probability $1$ (and then continue with interval nesting)
– Hagen von Eitzen
Jul 22 at 17:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a random variable that is independent of itself. I want to show that $exists a in Bbb R: Bbb P[X=a] = 1$.
Let $t in Bbb R$ be arbitrary. Then, $Bbb P[X le t] = Bbb P[X le t cap X le t] = (Bbb P[X le t])^2$, which implies that $Bbb P[X le t] in 0,1$. Especially there exists one $s in Bbb R: Bbb P[X le s] = 1 $.
Next, I decompose $X le s$ into $lfloorsrfloor < X le s cup bigcup_k=1^infty lfloorsrfloor -(k+1) < X le lfloorsrfloor -k$ and since they are disjoint only one of these sets has probability $1$. Continuing this idea, we get a monotone decreasing sequence of sets $(A_n)_n in Bbb N$ with $Bbb P[A_n] = 1$ for all $n in Bbb N$ and $bigcap_n in Bbb N A_n = X=a$ for some $a in Bbb R$. From the continuity of the measure we obtain $Bbb P[X=a]$ =1.
Assume, there exists another $a' in Bbb R: Bbb P[X=a'] = 1$. Then $Bbb P[X = a cup X = a'] = Bbb P[X=a] + Bbb P[X=a'] = 2$ leading to a contradiction.
probability-theory proof-verification
asked Jul 22 at 16:57
Pazu
359213
Let $X$ be a random variable that is independent of itself. I want to show that $exists a in Bbb R: Bbb P[X=a] = 1$.
Let $t in Bbb R$ be arbitrary. Then, $Bbb P[X le t] = Bbb P[X le t cap X le t] = (Bbb P[X le t])^2$, which implies that $Bbb P[X le t] in 0,1$. Especially there exists one $s in Bbb R: Bbb P[X le s] = 1 $.
Next, I decompose $X le s$ into $lfloorsrfloor < X le s cup bigcup_k=1^infty lfloorsrfloor -(k+1) < X le lfloorsrfloor -k$ and since they are disjoint only one of these sets has probability $1$. Continuing this idea, we get a monotone decreasing sequence of sets $(A_n)_n in Bbb N$ with $Bbb P[A_n] = 1$ for all $n in Bbb N$ and $bigcap_n in Bbb N A_n = X=a$ for some $a in Bbb R$. From the continuity of the measure we obtain $Bbb P[X=a]$ =1.
Assume, there exists another $a' in Bbb R: Bbb P[X=a'] = 1$. Then $Bbb P[X = a cup X = a'] = Bbb P[X=a] + Bbb P[X=a'] = 2$ leading to a contradiction.
probability-theory proof-verification
asked Jul 22 at 16:57
Pazu
359213
asked Jul 22 at 16:57
Pazu
359213
asked Jul 22 at 16:57
Pazu
359213
asked Jul 22 at 16:57
Pazu
359213
359213
1
You might start right away (without $s$) with $Bbb R=bigcup_kinBbb Z[k,k+1[$ and note that exactly one of the interval has probability $1$ (and then continue with interval nesting)
– Hagen von Eitzen
Jul 22 at 17:02
add a comment |Â
1
You might start right away (without $s$) with $Bbb R=bigcup_kinBbb Z[k,k+1[$ and note that exactly one of the interval has probability $1$ (and then continue with interval nesting)
– Hagen von Eitzen
Jul 22 at 17:02
1
1
You might start right away (without $s$) with $Bbb R=bigcup_kinBbb Z[k,k+1[$ and note that exactly one of the interval has probability $1$ (and then continue with interval nesting)
– Hagen von Eitzen
Jul 22 at 17:02
You might start right away (without $s$) with $Bbb R=bigcup_kinBbb Z[k,k+1[$ and note that exactly one of the interval has probability $1$ (and then continue with interval nesting)
– Hagen von Eitzen
Jul 22 at 17:02
add a comment |Â
1 Answer
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Let $F$ be the c.d.f. of $X$ (i.e., $F(x)=Pleft([Xleq x]right)$.
As pointed out by you, for each $x$, we have $F(x)=0$ or $F(x)=1$.
Let $A=xinmathbbRmid F(x)=1$. Note that $A$ is non-empty.
(For, if $A$ is empty, then $F(x)=0$ for all $x$, which contradicts
to the property $lim_xrightarrowinftyF(x)=1$.) Since $lim_xrightarrow-inftyF(x)=0$,
by the same reasoning, $A$ is bounded from below. Let $a=inf AinmathbbR$,
then there exists a sequence $(x_n)$ in $A$ such that $x_1geq x_2geqldotsgeq a$
and $x_nrightarrow a$. By the right-continuity of $F$, we have
$F(a)=lim_nrightarrowinftyF(x_n)=1$. For any $x<a,$ we have
$xnotin A$, so $F(x)=0$. Hence, $F(a-)=lim_xrightarrow a-F(x)=0$.
Finally, $Pleft([X=a]right)=F(a)-F(a-)=1$.
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
So, is my approach wrong?
– Pazu
Jul 22 at 17:45
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $F$ be the c.d.f. of $X$ (i.e., $F(x)=Pleft([Xleq x]right)$.
As pointed out by you, for each $x$, we have $F(x)=0$ or $F(x)=1$.
Let $A=xinmathbbRmid F(x)=1$. Note that $A$ is non-empty.
(For, if $A$ is empty, then $F(x)=0$ for all $x$, which contradicts
to the property $lim_xrightarrowinftyF(x)=1$.) Since $lim_xrightarrow-inftyF(x)=0$,
by the same reasoning, $A$ is bounded from below. Let $a=inf AinmathbbR$,
then there exists a sequence $(x_n)$ in $A$ such that $x_1geq x_2geqldotsgeq a$
and $x_nrightarrow a$. By the right-continuity of $F$, we have
$F(a)=lim_nrightarrowinftyF(x_n)=1$. For any $x<a,$ we have
$xnotin A$, so $F(x)=0$. Hence, $F(a-)=lim_xrightarrow a-F(x)=0$.
Finally, $Pleft([X=a]right)=F(a)-F(a-)=1$.
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
So, is my approach wrong?
– Pazu
Jul 22 at 17:45
add a comment |Â
up vote
0
down vote
Let $F$ be the c.d.f. of $X$ (i.e., $F(x)=Pleft([Xleq x]right)$.
As pointed out by you, for each $x$, we have $F(x)=0$ or $F(x)=1$.
Let $A=xinmathbbRmid F(x)=1$. Note that $A$ is non-empty.
(For, if $A$ is empty, then $F(x)=0$ for all $x$, which contradicts
to the property $lim_xrightarrowinftyF(x)=1$.) Since $lim_xrightarrow-inftyF(x)=0$,
by the same reasoning, $A$ is bounded from below. Let $a=inf AinmathbbR$,
then there exists a sequence $(x_n)$ in $A$ such that $x_1geq x_2geqldotsgeq a$
and $x_nrightarrow a$. By the right-continuity of $F$, we have
$F(a)=lim_nrightarrowinftyF(x_n)=1$. For any $x<a,$ we have
$xnotin A$, so $F(x)=0$. Hence, $F(a-)=lim_xrightarrow a-F(x)=0$.
Finally, $Pleft([X=a]right)=F(a)-F(a-)=1$.
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
So, is my approach wrong?
– Pazu
Jul 22 at 17:45
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $F$ be the c.d.f. of $X$ (i.e., $F(x)=Pleft([Xleq x]right)$.
As pointed out by you, for each $x$, we have $F(x)=0$ or $F(x)=1$.
Let $A=xinmathbbRmid F(x)=1$. Note that $A$ is non-empty.
(For, if $A$ is empty, then $F(x)=0$ for all $x$, which contradicts
to the property $lim_xrightarrowinftyF(x)=1$.) Since $lim_xrightarrow-inftyF(x)=0$,
by the same reasoning, $A$ is bounded from below. Let $a=inf AinmathbbR$,
then there exists a sequence $(x_n)$ in $A$ such that $x_1geq x_2geqldotsgeq a$
and $x_nrightarrow a$. By the right-continuity of $F$, we have
$F(a)=lim_nrightarrowinftyF(x_n)=1$. For any $x<a,$ we have
$xnotin A$, so $F(x)=0$. Hence, $F(a-)=lim_xrightarrow a-F(x)=0$.
Finally, $Pleft([X=a]right)=F(a)-F(a-)=1$.
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
Let $F$ be the c.d.f. of $X$ (i.e., $F(x)=Pleft([Xleq x]right)$.
As pointed out by you, for each $x$, we have $F(x)=0$ or $F(x)=1$.
Let $A=xinmathbbRmid F(x)=1$. Note that $A$ is non-empty.
(For, if $A$ is empty, then $F(x)=0$ for all $x$, which contradicts
to the property $lim_xrightarrowinftyF(x)=1$.) Since $lim_xrightarrow-inftyF(x)=0$,
by the same reasoning, $A$ is bounded from below. Let $a=inf AinmathbbR$,
then there exists a sequence $(x_n)$ in $A$ such that $x_1geq x_2geqldotsgeq a$
and $x_nrightarrow a$. By the right-continuity of $F$, we have
$F(a)=lim_nrightarrowinftyF(x_n)=1$. For any $x<a,$ we have
$xnotin A$, so $F(x)=0$. Hence, $F(a-)=lim_xrightarrow a-F(x)=0$.
Finally, $Pleft([X=a]right)=F(a)-F(a-)=1$.
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
answered Jul 22 at 17:20
Danny Pak-Keung Chan
1,82128
1,82128
So, is my approach wrong?
– Pazu
Jul 22 at 17:45
add a comment |Â
So, is my approach wrong?
– Pazu
Jul 22 at 17:45
So, is my approach wrong?
– Pazu
Jul 22 at 17:45
So, is my approach wrong?
– Pazu
Jul 22 at 17:45
add a comment |Â
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1
You might start right away (without $s$) with $Bbb R=bigcup_kinBbb Z[k,k+1[$ and note that exactly one of the interval has probability $1$ (and then continue with interval nesting)
– Hagen von Eitzen
Jul 22 at 17:02