Mean of normal CDF
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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 '18 at 14:54
user587126user587126
155
$endgroup$
|
show 1 more comment
$begingroup$
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 '18 at 14:54
user587126user587126
155
$endgroup$
1
$begingroup$
"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.
$endgroup$
– Did
Dec 9 '18 at 14:58
2
$begingroup$
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
$endgroup$
– Did
Dec 9 '18 at 14:59
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Can you show the calculation please?
$endgroup$
– user587126
Dec 9 '18 at 15:28
1
$begingroup$
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 $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
$endgroup$
– Did
Dec 9 '18 at 16:20
$begingroup$
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
$endgroup$
– user587126
Dec 9 '18 at 17:49
|
show 1 more comment
$begingroup$
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 '18 at 14:54
user587126user587126
155
$endgroup$
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 '18 at 14:54
user587126user587126
155
asked Dec 9 '18 at 14:54
user587126user587126
155
asked Dec 9 '18 at 14:54
user587126user587126
155
asked Dec 9 '18 at 14:54
user587126user587126
155
asked Dec 9 '18 at 14:54
user587126user587126
155
155
1
$begingroup$
"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.
$endgroup$
– Did
Dec 9 '18 at 14:58
2
$begingroup$
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
$endgroup$
– Did
Dec 9 '18 at 14:59
$begingroup$
Can you show the calculation please?
$endgroup$
– user587126
Dec 9 '18 at 15:28
1
$begingroup$
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 $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
$endgroup$
– Did
Dec 9 '18 at 16:20
$begingroup$
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
$endgroup$
– user587126
Dec 9 '18 at 17:49
|
show 1 more comment
1
$begingroup$
"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.
$endgroup$
– Did
Dec 9 '18 at 14:58
2
$begingroup$
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
$endgroup$
– Did
Dec 9 '18 at 14:59
$begingroup$
Can you show the calculation please?
$endgroup$
– user587126
Dec 9 '18 at 15:28
1
$begingroup$
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 $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
$endgroup$
– Did
Dec 9 '18 at 16:20
$begingroup$
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
$endgroup$
– user587126
Dec 9 '18 at 17:49
1
1
$begingroup$
"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.
$endgroup$
– Did
Dec 9 '18 at 14:58
$begingroup$
"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.
$endgroup$
– Did
Dec 9 '18 at 14:58
2
2
$begingroup$
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
$endgroup$
– Did
Dec 9 '18 at 14:59
$begingroup$
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
$endgroup$
– Did
Dec 9 '18 at 14:59
$begingroup$
Can you show the calculation please?
$endgroup$
– user587126
Dec 9 '18 at 15:28
$begingroup$
Can you show the calculation please?
$endgroup$
– user587126
Dec 9 '18 at 15:28
1
1
$begingroup$
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 $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
$endgroup$
– Did
Dec 9 '18 at 16:20
$begingroup$
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 $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
$endgroup$
– Did
Dec 9 '18 at 16:20
$begingroup$
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
$endgroup$
– user587126
Dec 9 '18 at 17:49
$begingroup$
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
$endgroup$
– user587126
Dec 9 '18 at 17:49
|
show 1 more comment
0
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1
$begingroup$
"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.
$endgroup$
– Did
Dec 9 '18 at 14:58
2
$begingroup$
And it happens that $$E(Phi(X))=Phileft(-fracmu{sqrt{1+sigma^2}}right)$$ hence indeed, $E(Phi(X))=frac12$ iff $mu=0$.
$endgroup$
– Did
Dec 9 '18 at 14:59
$begingroup$
Can you show the calculation please?
$endgroup$
– user587126
Dec 9 '18 at 15:28
1
$begingroup$
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 $$E(Phi(X))=P(mu+sigma X_0leqslant X)=P(sigma X_0-Xleqslant-mu)=cdots$$
$endgroup$
– Did
Dec 9 '18 at 16:20
$begingroup$
But why $Phi(X)=P(mu+sigma X_0leq X| X)$ but not $P(mu+sigma X_0leq X)$?
$endgroup$
– user587126
Dec 9 '18 at 17:49