Timeline for Integration of the product of pdf & cdf of normal distribution
Current License: CC BY-SA 3.0
15 events
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| Oct 2, 2017 at 6:49 | comment | added | Hugh Perkins |
Thinking this through, I guess the answer is: the answer given above assumes we are integrating from -\infty up to \+infty? which kind of makes sense. But I think it might be worth stating this explicitly perhaps?
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| Oct 2, 2017 at 6:46 | comment | added | Hugh Perkins |
Quesiton: how are you managing to integrate -exp(-x^2)? Wouldnt this normally not be possible in closed form, except in the case of a definite integral, with bounds/limits?
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| S Jun 2, 2017 at 11:05 | history | suggested | user19797 | CC BY-SA 3.0 |
The last integral should be from ${b/\sqrt{a^2+1}}$, what makes this solution equivalent to Did's answer.
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| Jun 2, 2017 at 10:56 | review | Suggested edits | |||
| S Jun 2, 2017 at 11:05 | |||||
| Jun 5, 2013 at 19:54 | comment | added | Michael Hardy | After the last integral, instead of saying "This can be expressed with the traditional erf function.", why not just say "$=1-\Phi\left(b\sqrt{a^2+1}\right)$"? | |
| Jul 16, 2012 at 6:22 | comment | added | user9836 | Now I see that your result is correct for definite integral, but I am really expecting a closed-form result for the indefinite integral and it does not seem to be doable. Anyway thank you very much for your help, Davide. | |
| Jul 12, 2012 at 14:39 | comment | added | user9836 | It is an indefinite integral. | |
| Jul 12, 2012 at 13:33 | comment | added | Davide Giraudo | If the answer should depend on $x$, I don't understand the problem: it's like if you ask to compute $\int_0^a\sqrt{a^2-x^2}dx$ and you say that the result should depend on $x$. Where do you integrate? | |
| Jul 12, 2012 at 12:56 | comment | added | user9836 | Well, when a = 1 and b = 0, the answer should be $1/2 \Phi(x)^2$ instead of $1/2$ in your result. The answer should depend on $x$. Would you like to double-check your answer? Thank you very much. | |
| Jul 12, 2012 at 9:53 | comment | added | Davide Giraudo | The only fixed parameters are $a$ and $b$, we integrate with respect to $x$. Hence it's normal that $x$ doesn't appears in the final result. | |
| Jul 12, 2012 at 6:10 | comment | added | user9836 | Thank you but I still think that you miss something. I think the result is not correct as $\int_{b\sqrt{a^2+1}}^{+\infty}\phi(t)dt$ is a constant that does not depends on $x$ or $t$. Would you like to double check your result? Thank you very much! | |
| Jul 10, 2012 at 9:35 | comment | added | Davide Giraudo | I integrate on the whole real line with respect to $x$, and I do the substitution $\frac{a^2+1}{a^2}\left(x-\frac b{a^2+1}\right)^2=t^2$. I didn't omit the exponential term, since I wrote it using $\phi$. | |
| Jul 10, 2012 at 2:22 | comment | added | user9836 | The result of the integration step should contains some erf function of x instead of $\phi$ in the result, and the exp(-b^2/2/(a^2+1)) part should not be omitted. | |
| Jul 10, 2012 at 1:48 | comment | added | user9836 | Thank you very much for the answer but it seems that the "Integrating with respect to x" part is not correct. Can you double-check or add some more explanation on that part? Thank you! | |
| Jul 9, 2012 at 9:11 | history | answered | Davide Giraudo | CC BY-SA 3.0 |