EM algorithm and Mean residual lifeMean residual lifeDetermining the posterior distribution for an Autoregressive or order 1 modelProof of consistency of Maximum Likelihood Estimator(MLE)Ancillary statistics:Beta distribution is free of $beta$?Auxiliary random experimentNormal - $chi^2$ mixture is a $t$ distributionIf $X$ and $Y$ are independent Normal variables each with mean zero, then $fracXYsqrtX^2+Y^2$ is also a Normal variableFind the Maximum Likelihood Estimator given two pdfsMean for Lyapunov Condition on Probaiblity FunctionCheck if log-likelihood function is correctly derivedMLE, regularity conditions, finite and infinite parameter spaces
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EM algorithm and Mean residual life
Mean residual lifeDetermining the posterior distribution for an Autoregressive or order 1 modelProof of consistency of Maximum Likelihood Estimator(MLE)Ancillary statistics:Beta distribution is free of $beta$?Auxiliary random experimentNormal - $chi^2$ mixture is a $t$ distributionIf $X$ and $Y$ are independent Normal variables each with mean zero, then $fracXYsqrtX^2+Y^2$ is also a Normal variableFind the Maximum Likelihood Estimator given two pdfsMean for Lyapunov Condition on Probaiblity FunctionCheck if log-likelihood function is correctly derivedMLE, regularity conditions, finite and infinite parameter spaces
$begingroup$
I am reading Robert Hogg's (Introduction to Mathematical Statistics) EM algorithm.
In example 6.6.1 (page 370 in the 7th version), please help to explain how the following integral
$$int_a^infty(z-theta_0)frac1sqrt2pifracexp left-(z-theta_0)^2/2right 1-Phi(a-theta_0)dz$$
is equal to
$$frac11-Phi(a-theta_0)phi(a-theta_0)$$
where $phi(x)=(2pi)^-1/2expleft-x^2/2right$
or the book made some mistakes here?
I also think this post might give a little help.
Thanks
self-study expectation-maximization
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
I am reading Robert Hogg's (Introduction to Mathematical Statistics) EM algorithm.
In example 6.6.1 (page 370 in the 7th version), please help to explain how the following integral
$$int_a^infty(z-theta_0)frac1sqrt2pifracexp left-(z-theta_0)^2/2right 1-Phi(a-theta_0)dz$$
is equal to
$$frac11-Phi(a-theta_0)phi(a-theta_0)$$
where $phi(x)=(2pi)^-1/2expleft-x^2/2right$
or the book made some mistakes here?
I also think this post might give a little help.
Thanks
self-study expectation-maximization
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
I am reading Robert Hogg's (Introduction to Mathematical Statistics) EM algorithm.
In example 6.6.1 (page 370 in the 7th version), please help to explain how the following integral
$$int_a^infty(z-theta_0)frac1sqrt2pifracexp left-(z-theta_0)^2/2right 1-Phi(a-theta_0)dz$$
is equal to
$$frac11-Phi(a-theta_0)phi(a-theta_0)$$
where $phi(x)=(2pi)^-1/2expleft-x^2/2right$
or the book made some mistakes here?
I also think this post might give a little help.
Thanks
self-study expectation-maximization
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
$endgroup$
I am reading Robert Hogg's (Introduction to Mathematical Statistics) EM algorithm.
In example 6.6.1 (page 370 in the 7th version), please help to explain how the following integral
$$int_a^infty(z-theta_0)frac1sqrt2pifracexp left-(z-theta_0)^2/2right 1-Phi(a-theta_0)dz$$
is equal to
$$frac11-Phi(a-theta_0)phi(a-theta_0)$$
where $phi(x)=(2pi)^-1/2expleft-x^2/2right$
or the book made some mistakes here?
I also think this post might give a little help.
Thanks
self-study expectation-maximization
self-study expectation-maximization
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
asked Jun 23 at 7:39
Deep NorthDeep North
3,5702 gold badges11 silver badges31 bronze badges
3,5702 gold badges11 silver badges31 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We know that $int U'exp U = exp U$ then $$int_a^infty (z-theta_0)exp-(z-theta_0)^2/2 dz = left[-exp-(z-theta_0)^2/2right]_a^infty = exp-(a-theta_0)^2/2$$
$$int_a^infty (z-theta_0)frac1sqrt2pifracexp-(z-theta_0)^2/21-Phi(a-theta_0) dz =\frac1sqrt2pifrac11-Phi(a-theta_0)exp-(a-theta_0)^2/2\ = frac11-Phi(a-theta_0)frac1sqrt2piexp-(a-theta_0)^2/2\=frac11-Phi(a-theta_0)phi(a-theta_0)$$
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Deep North
Jun 23 at 10:41
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We know that $int U'exp U = exp U$ then $$int_a^infty (z-theta_0)exp-(z-theta_0)^2/2 dz = left[-exp-(z-theta_0)^2/2right]_a^infty = exp-(a-theta_0)^2/2$$
$$int_a^infty (z-theta_0)frac1sqrt2pifracexp-(z-theta_0)^2/21-Phi(a-theta_0) dz =\frac1sqrt2pifrac11-Phi(a-theta_0)exp-(a-theta_0)^2/2\ = frac11-Phi(a-theta_0)frac1sqrt2piexp-(a-theta_0)^2/2\=frac11-Phi(a-theta_0)phi(a-theta_0)$$
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Deep North
Jun 23 at 10:41
add a comment |
$begingroup$
We know that $int U'exp U = exp U$ then $$int_a^infty (z-theta_0)exp-(z-theta_0)^2/2 dz = left[-exp-(z-theta_0)^2/2right]_a^infty = exp-(a-theta_0)^2/2$$
$$int_a^infty (z-theta_0)frac1sqrt2pifracexp-(z-theta_0)^2/21-Phi(a-theta_0) dz =\frac1sqrt2pifrac11-Phi(a-theta_0)exp-(a-theta_0)^2/2\ = frac11-Phi(a-theta_0)frac1sqrt2piexp-(a-theta_0)^2/2\=frac11-Phi(a-theta_0)phi(a-theta_0)$$
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Deep North
Jun 23 at 10:41
add a comment |
$begingroup$
We know that $int U'exp U = exp U$ then $$int_a^infty (z-theta_0)exp-(z-theta_0)^2/2 dz = left[-exp-(z-theta_0)^2/2right]_a^infty = exp-(a-theta_0)^2/2$$
$$int_a^infty (z-theta_0)frac1sqrt2pifracexp-(z-theta_0)^2/21-Phi(a-theta_0) dz =\frac1sqrt2pifrac11-Phi(a-theta_0)exp-(a-theta_0)^2/2\ = frac11-Phi(a-theta_0)frac1sqrt2piexp-(a-theta_0)^2/2\=frac11-Phi(a-theta_0)phi(a-theta_0)$$
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
$endgroup$
We know that $int U'exp U = exp U$ then $$int_a^infty (z-theta_0)exp-(z-theta_0)^2/2 dz = left[-exp-(z-theta_0)^2/2right]_a^infty = exp-(a-theta_0)^2/2$$
$$int_a^infty (z-theta_0)frac1sqrt2pifracexp-(z-theta_0)^2/21-Phi(a-theta_0) dz =\frac1sqrt2pifrac11-Phi(a-theta_0)exp-(a-theta_0)^2/2\ = frac11-Phi(a-theta_0)frac1sqrt2piexp-(a-theta_0)^2/2\=frac11-Phi(a-theta_0)phi(a-theta_0)$$
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
edited Jun 23 at 10:41
Deep North
3,5702 gold badges11 silver badges31 bronze badges
3,5702 gold badges11 silver badges31 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
answered Jun 23 at 8:44
Abdoul HakiAbdoul Haki
4309 bronze badges
4309 bronze badges
$begingroup$
Thank you very much!
$endgroup$
– Deep North
Jun 23 at 10:41
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Deep North
Jun 23 at 10:41
$begingroup$
Thank you very much!
$endgroup$
– Deep North
Jun 23 at 10:41
$begingroup$
Thank you very much!
$endgroup$
– Deep North
Jun 23 at 10:41
add a comment |
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