$X$ independent of itself, then $exists a in Bbb R: Bbb P[X=a] = 1$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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.







share|cite|improve this question















  • 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














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.







share|cite|improve this question















  • 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












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.







share|cite|improve this question











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.









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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












  • 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










1 Answer
1






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, $Pleft([X=a]right)=F(a)-F(a-)=1$.






share|cite|improve this answer





















  • So, is my approach wrong?
    – Pazu
    Jul 22 at 17:45










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859567%2fx-independent-of-itself-then-exists-a-in-bbb-r-bbb-px-a-1%23new-answer', 'question_page');

);

Post as a guest






























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, $Pleft([X=a]right)=F(a)-F(a-)=1$.






share|cite|improve this answer





















  • So, is my approach wrong?
    – Pazu
    Jul 22 at 17:45














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, $Pleft([X=a]right)=F(a)-F(a-)=1$.






share|cite|improve this answer





















  • So, is my approach wrong?
    – Pazu
    Jul 22 at 17:45












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, $Pleft([X=a]right)=F(a)-F(a-)=1$.






share|cite|improve this answer













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, $Pleft([X=a]right)=F(a)-F(a-)=1$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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
















  • 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












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859567%2fx-independent-of-itself-then-exists-a-in-bbb-r-bbb-px-a-1%23new-answer', 'question_page');

);

Post as a guest













































































Comments

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Relationship between determinant of matrix and determinant of adjoint?

Color the edges and diagonals of a regular polygon