Mean of normal CDF
Question-
Let $X$ follows $N(mu,sigma^2)$. Show that $mathbb{E}(Phi(X)) neq frac{1}{2}$ for any $mu neq 0$, where $Phi(X)$ is the cdf of $N(0,1)$ distribution.
But the theorem of Probability Integral transformation says that cdf of any random variable follows $Uniform (0,1)$ distribution. So the normal cdf should not depend on $mu$ and always follow $Uniform (0,1)$ distribution with mean $frac{1}{2}$.
Can you please point out what I am missing here?
Thanks in advance!
probability-theory statistics normal-distribution
asked Dec 9 at 14:54
user587126
155
|
show 1 more comment
Question-
Let $X$ follows $N(mu,sigma^2)$. Show that $mathbb{E}(Phi(X)) neq frac{1}{2}$ for any $mu neq 0$, where $Phi(X)$ is the cdf of $N(0,1)$ distribution.
But the theorem of Probability Integral transformation says that cdf of any random variable follows $Uniform (0,1)$ distribution. So the normal cdf should not depend on $mu$ and always follow $Uniform (0,1)$ distribution with mean $frac{1}{2}$.
Can you please point out what I am missing here?
Thanks in advance!
probability-theory statistics normal-distribution
asked Dec 9 at 14:54
user587126
155
1
"cdf of any random variable follows Uniform(0,1) distribution" No, the result is that $F_Y(Y)$ is uniform on $(0,1)$ (if $Y$ is continuous). Here one is looking at $F_Y(X)$ for $X$ and $Y$ with different distributions.
– Did
Dec 9 at 14:58
2
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
– Did
Dec 9 at 14:59
Can you show the calculation please?
– user587126
Dec 9 at 15:28
1
Start from the fact that, for every $x$, $$Phi(x)=P(mu+sigma X_0leqslant x)$$ with $X_0$ standard normal, hence, assuming that $X_0$ is independent of $X$, $$Phi(X)=P(mu+sigma X_0leqslant Xmid X)$$ Integrating, one gets $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
– Did
Dec 9 at 16:20
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
– user587126
Dec 9 at 17:49
|
show 1 more comment
Question-
Let $X$ follows $N(mu,sigma^2)$. Show that $mathbb{E}(Phi(X)) neq frac{1}{2}$ for any $mu neq 0$, where $Phi(X)$ is the cdf of $N(0,1)$ distribution.
But the theorem of Probability Integral transformation says that cdf of any random variable follows $Uniform (0,1)$ distribution. So the normal cdf should not depend on $mu$ and always follow $Uniform (0,1)$ distribution with mean $frac{1}{2}$.
Can you please point out what I am missing here?
Thanks in advance!
probability-theory statistics normal-distribution
asked Dec 9 at 14:54
user587126
155
Question-
Let $X$ follows $N(mu,sigma^2)$. Show that $mathbb{E}(Phi(X)) neq frac{1}{2}$ for any $mu neq 0$, where $Phi(X)$ is the cdf of $N(0,1)$ distribution.
But the theorem of Probability Integral transformation says that cdf of any random variable follows $Uniform (0,1)$ distribution. So the normal cdf should not depend on $mu$ and always follow $Uniform (0,1)$ distribution with mean $frac{1}{2}$.
Can you please point out what I am missing here?
Thanks in advance!
probability-theory statistics normal-distribution
probability-theory statistics normal-distribution
asked Dec 9 at 14:54
user587126
155
asked Dec 9 at 14:54
user587126
155
asked Dec 9 at 14:54
user587126
155
asked Dec 9 at 14:54
user587126
155
asked Dec 9 at 14:54
user587126
155
155
1
"cdf of any random variable follows Uniform(0,1) distribution" No, the result is that $F_Y(Y)$ is uniform on $(0,1)$ (if $Y$ is continuous). Here one is looking at $F_Y(X)$ for $X$ and $Y$ with different distributions.
– Did
Dec 9 at 14:58
2
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
– Did
Dec 9 at 14:59
Can you show the calculation please?
– user587126
Dec 9 at 15:28
1
Start from the fact that, for every $x$, $$Phi(x)=P(mu+sigma X_0leqslant x)$$ with $X_0$ standard normal, hence, assuming that $X_0$ is independent of $X$, $$Phi(X)=P(mu+sigma X_0leqslant Xmid X)$$ Integrating, one gets $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
– Did
Dec 9 at 16:20
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
– user587126
Dec 9 at 17:49
|
show 1 more comment
1
"cdf of any random variable follows Uniform(0,1) distribution" No, the result is that $F_Y(Y)$ is uniform on $(0,1)$ (if $Y$ is continuous). Here one is looking at $F_Y(X)$ for $X$ and $Y$ with different distributions.
– Did
Dec 9 at 14:58
2
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
– Did
Dec 9 at 14:59
Can you show the calculation please?
– user587126
Dec 9 at 15:28
1
Start from the fact that, for every $x$, $$Phi(x)=P(mu+sigma X_0leqslant x)$$ with $X_0$ standard normal, hence, assuming that $X_0$ is independent of $X$, $$Phi(X)=P(mu+sigma X_0leqslant Xmid X)$$ Integrating, one gets $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
– Did
Dec 9 at 16:20
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
– user587126
Dec 9 at 17:49
1
1
"cdf of any random variable follows Uniform(0,1) distribution" No, the result is that $F_Y(Y)$ is uniform on $(0,1)$ (if $Y$ is continuous). Here one is looking at $F_Y(X)$ for $X$ and $Y$ with different distributions.
– Did
Dec 9 at 14:58
"cdf of any random variable follows Uniform(0,1) distribution" No, the result is that $F_Y(Y)$ is uniform on $(0,1)$ (if $Y$ is continuous). Here one is looking at $F_Y(X)$ for $X$ and $Y$ with different distributions.
– Did
Dec 9 at 14:58
2
2
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
– Did
Dec 9 at 14:59
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
– Did
Dec 9 at 14:59
Can you show the calculation please?
– user587126
Dec 9 at 15:28
Can you show the calculation please?
– user587126
Dec 9 at 15:28
1
1
Start from the fact that, for every $x$, $$Phi(x)=P(mu+sigma X_0leqslant x)$$ with $X_0$ standard normal, hence, assuming that $X_0$ is independent of $X$, $$Phi(X)=P(mu+sigma X_0leqslant Xmid X)$$ Integrating, one gets $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
– Did
Dec 9 at 16:20
Start from the fact that, for every $x$, $$Phi(x)=P(mu+sigma X_0leqslant x)$$ with $X_0$ standard normal, hence, assuming that $X_0$ is independent of $X$, $$Phi(X)=P(mu+sigma X_0leqslant Xmid X)$$ Integrating, one gets $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
– Did
Dec 9 at 16:20
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
– user587126
Dec 9 at 17:49
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
– user587126
Dec 9 at 17:49
|
show 1 more comment
active
oldest
votes
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%2f3032460%2fmean-of-normal-cdf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3032460%2fmean-of-normal-cdf%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
1
"cdf of any random variable follows Uniform(0,1) distribution" No, the result is that $F_Y(Y)$ is uniform on $(0,1)$ (if $Y$ is continuous). Here one is looking at $F_Y(X)$ for $X$ and $Y$ with different distributions.
– Did
Dec 9 at 14:58
2
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
– Did
Dec 9 at 14:59
Can you show the calculation please?
– user587126
Dec 9 at 15:28
1
Start from the fact that, for every $x$, $$Phi(x)=P(mu+sigma X_0leqslant x)$$ with $X_0$ standard normal, hence, assuming that $X_0$ is independent of $X$, $$Phi(X)=P(mu+sigma X_0leqslant Xmid X)$$ Integrating, one gets $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
– Did
Dec 9 at 16:20
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
– user587126
Dec 9 at 17:49