Mixing
Find $E[X_1midmax{X_1,X_2,…,X_n}]$ for $(X_k)$ i.i.d uniform on $[0,1]$
$begingroup$
Let $X_1,X_2,...,X_n$ be i.i.d uniform random variables on $[0,1]$. Define $Y=max${$X_1,X_2,...,X_n$}. Find $E[X_1|Y]$.
My answer:
$P(X_i|Y)=P(X_j|Y)$ for all $i,j=1,2,...,n$.
Since $sum_{i=n}^{n}P(X_i|Y)=1$, I get $P(X_i|Y)=1/n$.
Then $$E[X_1|Y]=int_{0}^{1}x_1times{1over n}dx_1={1over 2n}$$
Is it correct for $P(X_i|Y)=P(X_j|Y)$ and $sum_{i=n}^{n}P(X_i|Y)=1$?
probability-theory conditional-expectation
edited Jan 12 at 22:19
Did
248k23223460
asked Jan 12 at 21:39
Yibei HeYibei He
2139
$endgroup$
add a comment |
$begingroup$
Let $X_1,X_2,...,X_n$ be i.i.d uniform random variables on $[0,1]$. Define $Y=max${$X_1,X_2,...,X_n$}. Find $E[X_1|Y]$.
My answer:
$P(X_i|Y)=P(X_j|Y)$ for all $i,j=1,2,...,n$.
Since $sum_{i=n}^{n}P(X_i|Y)=1$, I get $P(X_i|Y)=1/n$.
Then $$E[X_1|Y]=int_{0}^{1}x_1times{1over n}dx_1={1over 2n}$$
Is it correct for $P(X_i|Y)=P(X_j|Y)$ and $sum_{i=n}^{n}P(X_i|Y)=1$?
probability-theory conditional-expectation
edited Jan 12 at 22:19
Did
248k23223460
asked Jan 12 at 21:39
Yibei HeYibei He
2139
$endgroup$
2
$begingroup$
None of this is correct. In fact, even passing over the misleading notations, none of this is even related to the question. What would be a definition of $E(X_1mid Y)$, according to you?
$endgroup$
– Did
Jan 12 at 22:17
$begingroup$
Possible duplicate of Conditional expectation to de maximum $E(X_1mid X_{(n)})$
$endgroup$
– StubbornAtom
Jan 14 at 13:44
$begingroup$
Also see math.stackexchange.com/questions/617415/….
$endgroup$
– StubbornAtom
Jan 14 at 13:45
add a comment |
$begingroup$
Let $X_1,X_2,...,X_n$ be i.i.d uniform random variables on $[0,1]$. Define $Y=max${$X_1,X_2,...,X_n$}. Find $E[X_1|Y]$.
My answer:
$P(X_i|Y)=P(X_j|Y)$ for all $i,j=1,2,...,n$.
Since $sum_{i=n}^{n}P(X_i|Y)=1$, I get $P(X_i|Y)=1/n$.
Then $$E[X_1|Y]=int_{0}^{1}x_1times{1over n}dx_1={1over 2n}$$
Is it correct for $P(X_i|Y)=P(X_j|Y)$ and $sum_{i=n}^{n}P(X_i|Y)=1$?
probability-theory conditional-expectation
edited Jan 12 at 22:19
Did
248k23223460
asked Jan 12 at 21:39
Yibei HeYibei He
2139
$endgroup$
Let $X_1,X_2,...,X_n$ be i.i.d uniform random variables on $[0,1]$. Define $Y=max${$X_1,X_2,...,X_n$}. Find $E[X_1|Y]$.
My answer:
$P(X_i|Y)=P(X_j|Y)$ for all $i,j=1,2,...,n$.
Since $sum_{i=n}^{n}P(X_i|Y)=1$, I get $P(X_i|Y)=1/n$.
Then $$E[X_1|Y]=int_{0}^{1}x_1times{1over n}dx_1={1over 2n}$$
Is it correct for $P(X_i|Y)=P(X_j|Y)$ and $sum_{i=n}^{n}P(X_i|Y)=1$?
probability-theory conditional-expectation
probability-theory conditional-expectation
edited Jan 12 at 22:19
Did
248k23223460
asked Jan 12 at 21:39
Yibei HeYibei He
2139
edited Jan 12 at 22:19
Did
248k23223460
asked Jan 12 at 21:39
Yibei HeYibei He
2139
edited Jan 12 at 22:19
Did
248k23223460
edited Jan 12 at 22:19
Did
248k23223460
edited Jan 12 at 22:19
Did
248k23223460
248k23223460
asked Jan 12 at 21:39
Yibei HeYibei He
2139
asked Jan 12 at 21:39
Yibei HeYibei He
2139
asked Jan 12 at 21:39
Yibei HeYibei He
2139
2139
2
$begingroup$
None of this is correct. In fact, even passing over the misleading notations, none of this is even related to the question. What would be a definition of $E(X_1mid Y)$, according to you?
$endgroup$
– Did
Jan 12 at 22:17
$begingroup$
Possible duplicate of Conditional expectation to de maximum $E(X_1mid X_{(n)})$
$endgroup$
– StubbornAtom
Jan 14 at 13:44
$begingroup$
Also see math.stackexchange.com/questions/617415/….
$endgroup$
– StubbornAtom
Jan 14 at 13:45
add a comment |
2
$begingroup$
None of this is correct. In fact, even passing over the misleading notations, none of this is even related to the question. What would be a definition of $E(X_1mid Y)$, according to you?
$endgroup$
– Did
Jan 12 at 22:17
$begingroup$
Possible duplicate of Conditional expectation to de maximum $E(X_1mid X_{(n)})$
$endgroup$
– StubbornAtom
Jan 14 at 13:44
$begingroup$
Also see math.stackexchange.com/questions/617415/….
$endgroup$
– StubbornAtom
Jan 14 at 13:45
2
2
$begingroup$
None of this is correct. In fact, even passing over the misleading notations, none of this is even related to the question. What would be a definition of $E(X_1mid Y)$, according to you?
$endgroup$
– Did
Jan 12 at 22:17
$begingroup$
None of this is correct. In fact, even passing over the misleading notations, none of this is even related to the question. What would be a definition of $E(X_1mid Y)$, according to you?
$endgroup$
– Did
Jan 12 at 22:17
$begingroup$
Possible duplicate of Conditional expectation to de maximum $E(X_1mid X_{(n)})$
$endgroup$
– StubbornAtom
Jan 14 at 13:44
$begingroup$
Possible duplicate of Conditional expectation to de maximum $E(X_1mid X_{(n)})$
$endgroup$
– StubbornAtom
Jan 14 at 13:44
$begingroup$
Also see math.stackexchange.com/questions/617415/….
$endgroup$
– StubbornAtom
Jan 14 at 13:45
$begingroup$
Also see math.stackexchange.com/questions/617415/….
$endgroup$
– StubbornAtom
Jan 14 at 13:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The equality $sum_i mathbb P[X_i|Y]=1$ is false, you don't get the right result for this reason.
Observe that $mathbb P[Yleq y] = mathbb P[X_1leq yland X_2leq ylanddotsland X_nleq y]=y^n$ hence $f_Y(y)=n y^{n-1}$. Similarly $mathbb P[X_1leq x,Yleq y]=mathbb P[X_1leq min(x,y)land X_2leq ylanddotsland X_nleq y]=min(x,y) y^{n-1}$, hence for $xleq y$, $f_{X_1,Y}(x,y)=(n-1) y^{n-2}$.
Using that, you can obtain the conditional distribution of $X_1$ given that $Y$, for $x<y$
begin{align*}
f_{X_1|Y}(x|y) &= frac{f_{X_1,Y}(x,y)}{f_Y(y)}\
&=frac{(n-1)y^{n-2}}{n y^{n-1}}\
&=frac{n-1}{ny}
end{align*}
For $x=y$ a discontinuity happens and $mathbb P[X_1=y|Y=y]=frac{1}{n}$ by symmetry. Hence for any $xleq y$,
begin{align*}
f_{X_1|Y}(x|y) =frac{n-1}{ny}+frac{1}{n}delta(x-y)
end{align*}
finally the conditional expectation is given by
begin{align*}
mathbb E[X_1|Y=y] &= int_0^{y} x cdot f_{X_1|Y}(x|y) dx\
&=frac{n-1}{ny} cdot frac{y^2}{2}+frac{y}{n}\
&=frac{y(n+1)}{2n}
end{align*}
And hence $mathbb E[X_1|Y]=frac{Y(n+1)}{2n}$
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
$endgroup$
1
$begingroup$
But you ignored the point mass with weight $1/n$ of $f_{X_1 mid Y}(x mid y)$ at $x = y$. If $f_{X_1 mid Y} (x mid y) = (n-1)/(ny)$ then when you integrate $int_0^y f_{X_1 mid Y} (x mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$.
$endgroup$
– Alex
Jan 12 at 22:42
$begingroup$
@Alex Oh yes, something felt wrong, thank you very much.
$endgroup$
– P. Quinton
Jan 12 at 22:50
$begingroup$
@P.Quinton what does the $delta$ in case $x=y$ mean?
$endgroup$
– Yibei He
Jan 13 at 2:12
$begingroup$
@P.Quinton and why don't you consider $xgt y$? For $xge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$
$endgroup$
– Yibei He
Jan 13 at 5:39
$begingroup$
The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $int_A f(x) delta(x) dx = f(0)$. In this case it means that if you integrate $int_A f(x) delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory.
$endgroup$
– P. Quinton
Jan 13 at 7:33
|
show 3 more comments
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071418%2ffind-ex-1-mid-max-x-1-x-2-x-n-for-x-k-i-i-d-uniform-on-0-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The equality $sum_i mathbb P[X_i|Y]=1$ is false, you don't get the right result for this reason.
Observe that $mathbb P[Yleq y] = mathbb P[X_1leq yland X_2leq ylanddotsland X_nleq y]=y^n$ hence $f_Y(y)=n y^{n-1}$. Similarly $mathbb P[X_1leq x,Yleq y]=mathbb P[X_1leq min(x,y)land X_2leq ylanddotsland X_nleq y]=min(x,y) y^{n-1}$, hence for $xleq y$, $f_{X_1,Y}(x,y)=(n-1) y^{n-2}$.
Using that, you can obtain the conditional distribution of $X_1$ given that $Y$, for $x<y$
begin{align*}
f_{X_1|Y}(x|y) &= frac{f_{X_1,Y}(x,y)}{f_Y(y)}\
&=frac{(n-1)y^{n-2}}{n y^{n-1}}\
&=frac{n-1}{ny}
end{align*}
For $x=y$ a discontinuity happens and $mathbb P[X_1=y|Y=y]=frac{1}{n}$ by symmetry. Hence for any $xleq y$,
begin{align*}
f_{X_1|Y}(x|y) =frac{n-1}{ny}+frac{1}{n}delta(x-y)
end{align*}
finally the conditional expectation is given by
begin{align*}
mathbb E[X_1|Y=y] &= int_0^{y} x cdot f_{X_1|Y}(x|y) dx\
&=frac{n-1}{ny} cdot frac{y^2}{2}+frac{y}{n}\
&=frac{y(n+1)}{2n}
end{align*}
And hence $mathbb E[X_1|Y]=frac{Y(n+1)}{2n}$
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
$endgroup$
1
$begingroup$
But you ignored the point mass with weight $1/n$ of $f_{X_1 mid Y}(x mid y)$ at $x = y$. If $f_{X_1 mid Y} (x mid y) = (n-1)/(ny)$ then when you integrate $int_0^y f_{X_1 mid Y} (x mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$.
$endgroup$
– Alex
Jan 12 at 22:42
$begingroup$
@Alex Oh yes, something felt wrong, thank you very much.
$endgroup$
– P. Quinton
Jan 12 at 22:50
$begingroup$
@P.Quinton what does the $delta$ in case $x=y$ mean?
$endgroup$
– Yibei He
Jan 13 at 2:12
$begingroup$
@P.Quinton and why don't you consider $xgt y$? For $xge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$
$endgroup$
– Yibei He
Jan 13 at 5:39
$begingroup$
The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $int_A f(x) delta(x) dx = f(0)$. In this case it means that if you integrate $int_A f(x) delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory.
$endgroup$
– P. Quinton
Jan 13 at 7:33
|
show 3 more comments
$begingroup$
The equality $sum_i mathbb P[X_i|Y]=1$ is false, you don't get the right result for this reason.
Observe that $mathbb P[Yleq y] = mathbb P[X_1leq yland X_2leq ylanddotsland X_nleq y]=y^n$ hence $f_Y(y)=n y^{n-1}$. Similarly $mathbb P[X_1leq x,Yleq y]=mathbb P[X_1leq min(x,y)land X_2leq ylanddotsland X_nleq y]=min(x,y) y^{n-1}$, hence for $xleq y$, $f_{X_1,Y}(x,y)=(n-1) y^{n-2}$.
Using that, you can obtain the conditional distribution of $X_1$ given that $Y$, for $x<y$
begin{align*}
f_{X_1|Y}(x|y) &= frac{f_{X_1,Y}(x,y)}{f_Y(y)}\
&=frac{(n-1)y^{n-2}}{n y^{n-1}}\
&=frac{n-1}{ny}
end{align*}
For $x=y$ a discontinuity happens and $mathbb P[X_1=y|Y=y]=frac{1}{n}$ by symmetry. Hence for any $xleq y$,
begin{align*}
f_{X_1|Y}(x|y) =frac{n-1}{ny}+frac{1}{n}delta(x-y)
end{align*}
finally the conditional expectation is given by
begin{align*}
mathbb E[X_1|Y=y] &= int_0^{y} x cdot f_{X_1|Y}(x|y) dx\
&=frac{n-1}{ny} cdot frac{y^2}{2}+frac{y}{n}\
&=frac{y(n+1)}{2n}
end{align*}
And hence $mathbb E[X_1|Y]=frac{Y(n+1)}{2n}$
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
$endgroup$
1
$begingroup$
But you ignored the point mass with weight $1/n$ of $f_{X_1 mid Y}(x mid y)$ at $x = y$. If $f_{X_1 mid Y} (x mid y) = (n-1)/(ny)$ then when you integrate $int_0^y f_{X_1 mid Y} (x mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$.
$endgroup$
– Alex
Jan 12 at 22:42
$begingroup$
@Alex Oh yes, something felt wrong, thank you very much.
$endgroup$
– P. Quinton
Jan 12 at 22:50
$begingroup$
@P.Quinton what does the $delta$ in case $x=y$ mean?
$endgroup$
– Yibei He
Jan 13 at 2:12
$begingroup$
@P.Quinton and why don't you consider $xgt y$? For $xge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$
$endgroup$
– Yibei He
Jan 13 at 5:39
$begingroup$
The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $int_A f(x) delta(x) dx = f(0)$. In this case it means that if you integrate $int_A f(x) delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory.
$endgroup$
– P. Quinton
Jan 13 at 7:33
|
show 3 more comments
$begingroup$
The equality $sum_i mathbb P[X_i|Y]=1$ is false, you don't get the right result for this reason.
Observe that $mathbb P[Yleq y] = mathbb P[X_1leq yland X_2leq ylanddotsland X_nleq y]=y^n$ hence $f_Y(y)=n y^{n-1}$. Similarly $mathbb P[X_1leq x,Yleq y]=mathbb P[X_1leq min(x,y)land X_2leq ylanddotsland X_nleq y]=min(x,y) y^{n-1}$, hence for $xleq y$, $f_{X_1,Y}(x,y)=(n-1) y^{n-2}$.
Using that, you can obtain the conditional distribution of $X_1$ given that $Y$, for $x<y$
begin{align*}
f_{X_1|Y}(x|y) &= frac{f_{X_1,Y}(x,y)}{f_Y(y)}\
&=frac{(n-1)y^{n-2}}{n y^{n-1}}\
&=frac{n-1}{ny}
end{align*}
For $x=y$ a discontinuity happens and $mathbb P[X_1=y|Y=y]=frac{1}{n}$ by symmetry. Hence for any $xleq y$,
begin{align*}
f_{X_1|Y}(x|y) =frac{n-1}{ny}+frac{1}{n}delta(x-y)
end{align*}
finally the conditional expectation is given by
begin{align*}
mathbb E[X_1|Y=y] &= int_0^{y} x cdot f_{X_1|Y}(x|y) dx\
&=frac{n-1}{ny} cdot frac{y^2}{2}+frac{y}{n}\
&=frac{y(n+1)}{2n}
end{align*}
And hence $mathbb E[X_1|Y]=frac{Y(n+1)}{2n}$
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
$endgroup$
The equality $sum_i mathbb P[X_i|Y]=1$ is false, you don't get the right result for this reason.
Observe that $mathbb P[Yleq y] = mathbb P[X_1leq yland X_2leq ylanddotsland X_nleq y]=y^n$ hence $f_Y(y)=n y^{n-1}$. Similarly $mathbb P[X_1leq x,Yleq y]=mathbb P[X_1leq min(x,y)land X_2leq ylanddotsland X_nleq y]=min(x,y) y^{n-1}$, hence for $xleq y$, $f_{X_1,Y}(x,y)=(n-1) y^{n-2}$.
Using that, you can obtain the conditional distribution of $X_1$ given that $Y$, for $x<y$
begin{align*}
f_{X_1|Y}(x|y) &= frac{f_{X_1,Y}(x,y)}{f_Y(y)}\
&=frac{(n-1)y^{n-2}}{n y^{n-1}}\
&=frac{n-1}{ny}
end{align*}
For $x=y$ a discontinuity happens and $mathbb P[X_1=y|Y=y]=frac{1}{n}$ by symmetry. Hence for any $xleq y$,
begin{align*}
f_{X_1|Y}(x|y) =frac{n-1}{ny}+frac{1}{n}delta(x-y)
end{align*}
finally the conditional expectation is given by
begin{align*}
mathbb E[X_1|Y=y] &= int_0^{y} x cdot f_{X_1|Y}(x|y) dx\
&=frac{n-1}{ny} cdot frac{y^2}{2}+frac{y}{n}\
&=frac{y(n+1)}{2n}
end{align*}
And hence $mathbb E[X_1|Y]=frac{Y(n+1)}{2n}$
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
edited Jan 12 at 22:50
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
answered Jan 12 at 22:17
P. QuintonP. Quinton
1,8011213
1,8011213
1
$begingroup$
But you ignored the point mass with weight $1/n$ of $f_{X_1 mid Y}(x mid y)$ at $x = y$. If $f_{X_1 mid Y} (x mid y) = (n-1)/(ny)$ then when you integrate $int_0^y f_{X_1 mid Y} (x mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$.
$endgroup$
– Alex
Jan 12 at 22:42
$begingroup$
@Alex Oh yes, something felt wrong, thank you very much.
$endgroup$
– P. Quinton
Jan 12 at 22:50
$begingroup$
@P.Quinton what does the $delta$ in case $x=y$ mean?
$endgroup$
– Yibei He
Jan 13 at 2:12
$begingroup$
@P.Quinton and why don't you consider $xgt y$? For $xge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$
$endgroup$
– Yibei He
Jan 13 at 5:39
$begingroup$
The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $int_A f(x) delta(x) dx = f(0)$. In this case it means that if you integrate $int_A f(x) delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory.
$endgroup$
– P. Quinton
Jan 13 at 7:33
|
show 3 more comments
1
$begingroup$
But you ignored the point mass with weight $1/n$ of $f_{X_1 mid Y}(x mid y)$ at $x = y$. If $f_{X_1 mid Y} (x mid y) = (n-1)/(ny)$ then when you integrate $int_0^y f_{X_1 mid Y} (x mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$.
$endgroup$
– Alex
Jan 12 at 22:42
$begingroup$
@Alex Oh yes, something felt wrong, thank you very much.
$endgroup$
– P. Quinton
Jan 12 at 22:50
$begingroup$
@P.Quinton what does the $delta$ in case $x=y$ mean?
$endgroup$
– Yibei He
Jan 13 at 2:12
$begingroup$
@P.Quinton and why don't you consider $xgt y$? For $xge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$
$endgroup$
– Yibei He
Jan 13 at 5:39
$begingroup$
The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $int_A f(x) delta(x) dx = f(0)$. In this case it means that if you integrate $int_A f(x) delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory.
$endgroup$
– P. Quinton
Jan 13 at 7:33
1
1
$begingroup$
But you ignored the point mass with weight $1/n$ of $f_{X_1 mid Y}(x mid y)$ at $x = y$. If $f_{X_1 mid Y} (x mid y) = (n-1)/(ny)$ then when you integrate $int_0^y f_{X_1 mid Y} (x mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$.
$endgroup$
– Alex
Jan 12 at 22:42
$begingroup$
But you ignored the point mass with weight $1/n$ of $f_{X_1 mid Y}(x mid y)$ at $x = y$. If $f_{X_1 mid Y} (x mid y) = (n-1)/(ny)$ then when you integrate $int_0^y f_{X_1 mid Y} (x mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$.
$endgroup$
– Alex
Jan 12 at 22:42
$begingroup$
@Alex Oh yes, something felt wrong, thank you very much.
$endgroup$
– P. Quinton
Jan 12 at 22:50
$begingroup$
@Alex Oh yes, something felt wrong, thank you very much.
$endgroup$
– P. Quinton
Jan 12 at 22:50
$begingroup$
@P.Quinton what does the $delta$ in case $x=y$ mean?
$endgroup$
– Yibei He
Jan 13 at 2:12
$begingroup$
@P.Quinton what does the $delta$ in case $x=y$ mean?
$endgroup$
– Yibei He
Jan 13 at 2:12
$begingroup$
@P.Quinton and why don't you consider $xgt y$? For $xge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$
$endgroup$
– Yibei He
Jan 13 at 5:39
$begingroup$
@P.Quinton and why don't you consider $xgt y$? For $xge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$
$endgroup$
– Yibei He
Jan 13 at 5:39
$begingroup$
The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $int_A f(x) delta(x) dx = f(0)$. In this case it means that if you integrate $int_A f(x) delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory.
$endgroup$
– P. Quinton
Jan 13 at 7:33
$begingroup$
The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $int_A f(x) delta(x) dx = f(0)$. In this case it means that if you integrate $int_A f(x) delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory.
$endgroup$
– P. Quinton
Jan 13 at 7:33
|
show 3 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071418%2ffind-ex-1-mid-max-x-1-x-2-x-n-for-x-k-i-i-d-uniform-on-0-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
None of this is correct. In fact, even passing over the misleading notations, none of this is even related to the question. What would be a definition of $E(X_1mid Y)$, according to you?
$endgroup$
– Did
Jan 12 at 22:17
$begingroup$
Possible duplicate of Conditional expectation to de maximum $E(X_1mid X_{(n)})$
$endgroup$
– StubbornAtom
Jan 14 at 13:44
$begingroup$
Also see math.stackexchange.com/questions/617415/….
$endgroup$
– StubbornAtom
Jan 14 at 13:45